The alternating series test and the nesting of its partial sums, the remainder bound that comes free with it, absolute against conditional convergence, and the rearrangement theorem that gives the distinction its meaning.
Subject: Calculus II · 66 slides · symbolic lesson
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Title
Calculus II · Section 5.5
When the signs take turns, cancellation can do what shrinking alone cannot
Objectives
Every test so far needed positive terms. This lesson handles series whose terms change sign, and shows that the signs can do real work: a series whose sizes diverge can still converge once the signs alternate.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, pp. 496-504 — learning objectives 5.5.1 to 5.5.3
Sections 5.3 and 5.4 built a whole toolkit for series with positive terms. Every one of those tools quietly relied on the positive terms, and none of them can say anything about a series like one minus a half plus a third minus a quarter.
This lesson is about series whose terms take turns being positive and negative. The surprise is that alternating signs can rescue a series: the harmonic sizes diverge, but the same sizes with alternating signs converge. The reason is cancellation, and you will see it as a zig-zag of partial sums closing in on a number.
The same zig-zag also gives the easiest error estimate in the chapter. The last two parts then ask a deeper question: when the signs are doing essential work, is the sum still a sum in the ordinary sense? The answer, that reordering the terms can change the total, is one of the strangest facts in the course.
Warm-up
Discussion prompt
The integral test and the comparison tests all assumed positive terms. What property of the partial sums did their proofs rely on, and what happens to that property when the signs alternate?
Commit to an answer in writing before revealing it. The point is to notice something you have been using without saying it.
With positive terms, each partial sum is the previous one plus something positive, so the partial sums only ever go up. An increasing sequence converges exactly when it is bounded above. The integral test and both comparison tests are really machines for finding that upper bound.
Once signs alternate, the partial sums go up, then down, then up again. They are not increasing, so a bound above proves nothing on its own. You need a new mechanism, and this lesson's first part builds it from the alternating harmonic series.
Section
Part 1
Concept
Pull the sign out of each term and what is left is a size. An alternating series is a list of sizes with the signs switching at every step.
\[ \sum_{n=1}^{\infty}(-1)^{n+1}b_n = b_1 - b_2 + b_3 - b_4 + \cdots \]
\[ \sum_{n=1}^{\infty}(-1)^{n}b_n = -b_1 + b_2 - b_3 + b_4 - \cdots \]
alternating series — A series whose terms alternate between positive and negative values; written with a factor of minus one to the n plus 1 (first term positive) or minus one to the n (first term negative), times sizes that are never negative.
\[ b_n \ge 0 \quad \text{for every } n \ge 1 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 496 — definition and equations 5.13, 5.14
Separate every term into a sign and a size. The size is the term's distance from zero, which the book calls b sub n, and it is never negative. The sign is carried by a power of minus one.
There are two standard ways to write it. With minus one to the n plus 1, the first term is positive; with minus one to the n, the first term is negative. Both describe the same idea, and every theorem in this lesson applies to both.
Get into the habit of naming the sizes at the start of every problem. Both conditions of the test you are about to meet are statements about the sizes alone, and most mistakes come from checking the signed terms instead.
Worked example
Decide whether this series converges, and find its sum.
\[ \sum_{n=1}^{\infty}\left(-\frac12\right)^n = -\frac12 + \frac14 - \frac18 + \frac1{16} - \cdots \]
Read off the first term and the ratio
Why: Each term is the one before it times minus one half.
\[ a = -\frac12, \quad r = -\frac12 \]
Check the size of the ratio
Why: A geometric series converges exactly when the ratio is less than one in absolute value.
\[ |r| = \frac12 < 1 \]
Apply the geometric sum formula
Why: First term over one minus the ratio.
\[ \frac{a}{1-r} = \frac{-1/2}{1 + 1/2} = \frac{-1/2}{3/2} = -\frac13 \]
Figure (svg): Partial sums of the series minus one half plus one quarter minus one eighth and so on, for 1 to 12 terms, joined by a zig-zag line that jumps above and below a dashed line at minus one third, with the jumps shrinking by half each time.
Check against the partial sums
Why: Five and six terms already bracket the sum, one on each side.
\[ S_5 = -0.34375 < -\tfrac13 < -0.328125 = S_6 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 496 — equation 5.11
You already know how to handle this one: it is a geometric series with a negative ratio. The formula from Section 5.2 does not care about the sign of the ratio, only its size, and one half is less than one, so it converges to minus one third.
The point of doing it here is the picture. The partial sums hop above and below minus one third, and each hop is exactly half as long as the one before. That hopping pattern, with shrinking hops that straddle the sum, is the shape every convergent alternating series makes.
Look at the check: the fifth partial sum is below minus one third and the sixth is above it. Two consecutive partial sums bracket the sum. Hold on to that observation; by the end of Part 3 it becomes a theorem.
Prediction
\[ 1 - \frac12 + \frac13 - \frac14 + \frac15 - \cdots \]
Predict first
The harmonic series 1 + 1/2 + 1/3 + … diverges. This series has exactly the same sizes, with every second sign flipped. What happens to it?
Correct: It converges, to a number between 1/2 and 1
Why: The partial sums go 1, 0.5, 0.833, 0.583, 0.783, … Each step reverses direction and is shorter than the last, so they close in on a single number, ln 2, about 0.693. The sizes are the harmonic sizes, but neighbouring terms nearly cancel, and that cancellation is what converges.
Pick an option before revealing. Your answer here says a lot about how you think of series: as a list of sizes, or as a running total.
If you picked diverges, your reasoning was probably that the sizes are the harmonic sizes, and the harmonic series diverges. That is a reasonable instinct, and it is wrong, because the signs change the running total completely. Adding one half and then taking away a third is very different from adding both.
The partial sums swing back and forth, each swing shorter than the last, and settle at the natural log of 2. The next slide shows the swings, and the proof after it explains exactly why they close in.
Picture it
Figure (svg): The first 30 partial sums of the alternating harmonic series, joined by a zig-zag line: 1, then 0.5, then 0.833, then 0.583, and so on, alternately overshooting and undershooting a dashed line at ln 2, about 0.693, with the swings shrinking.
Every odd-numbered partial sum sits above the dashed line and every even-numbered one below it. The swings shrink because the sizes shrink, and the line they close in on is the sum.
This is the central picture of the lesson, so spend a moment on it. Start at the first dot, at height one. The second term pulls you down by a half, the third pushes you up by a third, the fourth pulls you down by a quarter, and so on.
Every push is smaller than the pull before it, so you never climb back as high as you were. Every pull is smaller than the push before it, so you never fall as low as you were. The odd-numbered dots form a descending staircase and the even-numbered dots an ascending one, with the dashed line squeezed between them.
The line they close in on is the natural log of 2, about 0.693. Notice that after thirty terms the swings are still about three hundredths wide: this series converges, but slowly, and Part 3 will put a number on how slowly.
Worked example
Show that the odd-numbered partial sums of the alternating harmonic series decrease and are bounded below.
Go from one odd partial sum to the next
Why: Two more terms: one subtracted, then a smaller one added.
\[ S_{2k+1} = S_{2k-1} - \frac{1}{2k} + \frac{1}{2k+1} \]
Compare the two new terms
Why: The added term has the larger denominator, so it is the smaller.
\[ \frac{1}{2k+1} < \frac{1}{2k} \;\Longrightarrow\; S_{2k+1} < S_{2k-1} \]
Group in pairs to find a floor
Why: Every bracket is a number minus a smaller one, and the last term is positive.
\[ S_{2k+1} = \left(1 - \tfrac12\right) + \cdots + \left(\tfrac{1}{2k-1} - \tfrac{1}{2k}\right) + \tfrac{1}{2k+1} \]
\[ \text{every bracket} > 0 \;\Longrightarrow\; S_{2k+1} > 0 \]
Apply the Monotone Convergence Theorem
Why: Decreasing and bounded below means convergent.
\[ S_1 > S_3 > S_5 > \cdots > 0 \;\Longrightarrow\; S_{2k+1} \to S \]
Check with the first odd partial sums
Why: They fall, and they stay positive.
\[ S_1 = 1, \quad S_3 \approx 0.8333, \quad S_5 \approx 0.7833, \quad S_7 \approx 0.7595 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 496 — proof for the alternating harmonic series
The book proves convergence of the alternating harmonic series by treating the odd-numbered and the even-numbered partial sums separately, because each of those is monotone even though the whole sequence is not. That lets you reuse the Monotone Convergence Theorem after all.
Going from one odd partial sum to the next adds two terms: minus one over 2k, then plus one over 2k plus 1. The second is smaller than the first, so the net change is negative and the odd partial sums decrease.
For a floor, group the terms in pairs, each a number minus a smaller number, so each bracket is positive, and the leftover last term is positive too. So every odd partial sum is above zero. Decreasing and bounded below means convergent. The numbers in the check, 1, 0.833, 0.783, 0.760, show exactly that behaviour.
Worked example
Now the even-numbered partial sums: show they increase and are bounded above.
Go from one even partial sum to the next
Why: This time a term is added, then a smaller one subtracted.
\[ S_{2k} = S_{2k-2} + \frac{1}{2k-1} - \frac{1}{2k} \]
Compare the two new terms
Why: One over 2k minus 1 is the larger.
\[ \frac{1}{2k-1} > \frac{1}{2k} \;\Longrightarrow\; S_{2k} > S_{2k-2} \]
Group to find a ceiling
Why: After the first term every bracket is negative, and the last term is subtracted.
\[ S_{2k} = 1 + \left(-\tfrac12 + \tfrac13\right) + \cdots + \left(-\tfrac{1}{2k-2} + \tfrac{1}{2k-1}\right) - \tfrac{1}{2k} \]
\[ \text{every bracket} < 0 \;\Longrightarrow\; S_{2k} < 1 \]
Apply the Monotone Convergence Theorem again
Why: Increasing and bounded above means convergent.
\[ S_2 < S_4 < S_6 < \cdots < 1 \;\Longrightarrow\; S_{2k} \to S' \]
Check with the first even partial sums
Why: They rise, and they stay below one.
\[ S_2 = 0.5, \quad S_4 \approx 0.5833, \quad S_6 \approx 0.6167, \quad S_8 \approx 0.6345 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 497 — proof, continued
This is the mirror image of the previous slide. Going from one even partial sum to the next adds one over 2k minus 1 and then subtracts the smaller one over 2k, so the net change is positive and the even partial sums increase.
For a ceiling, keep the first term 1 by itself and group the rest in pairs of the form minus something plus something smaller. Each such bracket is negative, and the last term is subtracted, so the total is below one.
Increasing and bounded above means convergent, again by the Monotone Convergence Theorem. So there are now two limits: one for the odd staircase and one for the even staircase. The only thing left is to show they are the same number.
Concept
Figure (svg): Two sequences of dots for k up to 40: the odd-numbered partial sums of the alternating harmonic series, starting at 1 and falling, and the even-numbered ones, starting at 0.5 and rising. Both approach a dashed line at ln 2 from opposite sides.
Two limits so far, one for each staircase. They are the same number, because consecutive partial sums differ by a single term, and the terms shrink to zero.
\[ S_{2k+1} = S_{2k} + \frac{1}{2k+1} \]
\[ \lim_{k\to\infty}S_{2k+1} = \lim_{k\to\infty}S_{2k} + \lim_{k\to\infty}\frac{1}{2k+1} \;\Longrightarrow\; S = S' + 0 \]
Odd and even partial sums share one limit, so the whole sequence of partial sums converges. The value turns out to be the natural log of 2, which Chapter 6 derives.
\[ 1 - \frac12 + \frac13 - \frac14 + \cdots = \ln 2 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 497 — Figure 5.17
The picture shows both staircases at once: the odd partial sums coming down, the even ones going up. Each converges on its own; the question is whether they converge to the same place.
They must, because an odd partial sum is just the even partial sum before it plus one more term, one over 2k plus 1. Take limits of both sides. The extra term goes to zero, so the two limits are equal. When both the odd-numbered and the even-numbered terms of a sequence approach the same number, the whole sequence approaches it.
The value itself, ln 2, is a separate fact. It comes from the power series for the logarithm in Chapter 6. For now the important thing is the argument, because nothing in it used anything special about one over n except that the sizes decrease and tend to zero.
Intuition
Group the alternating harmonic series two terms at a time and each pair collapses into one small positive number.
\[ \frac{1}{2k-1} - \frac{1}{2k} = \frac{2k - (2k-1)}{2k(2k-1)} = \frac{1}{2k(2k-1)} \]
\[ 2k(2k-1) = 4k^2 - 2k \ge k^2 \;\Longrightarrow\; \frac{1}{2k(2k-1)} \le \frac{1}{k^2} \]
Figure (svg): Bars for the paired terms of the alternating harmonic series, one minus a half, a third minus a quarter, and so on, with heights 0.5, 0.083, 0.033, drawn under the curve y equals one over x squared evaluated at each whole number k.
The pairs are no bigger than the terms of a convergent p-series. The sizes on their own add up to infinity; what converges is the near-cancellation between each pair of neighbours.
Here is another way to see the convergence, which makes the role of cancellation very concrete. Pair each odd term with the even term after it. One over 2k minus 1, minus one over 2k, combines over a common denominator into one over 2k times 2k minus 1.
That denominator is roughly four k squared, so each pair is about a quarter of one over k squared. The second line makes the comparison exact: each pair is no bigger than one over k squared. The picture shows the bars for the pairs sitting under the dots of one over k squared.
So the alternating harmonic series, grouped in pairs, is a series of positive terms dominated by a convergent p-series. The sizes one over n add up to infinity, but the differences between neighbours shrink like one over n squared, and those differences are what the partial sums actually accumulate.
Section
Part 2
Concept
Alternating series test (Theorem 5.13) — An alternating series converges if its sizes never increase and tend to zero. It is enough for the sizes to stop increasing from some index N onward.
\[ 0 \le b_{n+1} \le b_n \text{ for all } n \ge 1 \quad\text{and}\quad \lim_{n\to\infty}b_n = 0 \]
\[ \Longrightarrow\quad \sum_{n=1}^{\infty}(-1)^{n+1}b_n \text{ and } \sum_{n=1}^{\infty}(-1)^{n}b_n \text{ converge} \]
Figure (svg): A horizontal number line from 0.45 to 1.05. Starting at S1 equals 1, arrows step left by one half to S2, right by one third to S3, left by one quarter to S4, and so on down the page. The arrows alternate in direction and shrink, trapping a dashed vertical line at ln 2.
The proof is the one you just saw for the alternating harmonic series, word for word, with the sizes in place of one over n.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 498 — Theorem 5.13 and Figure 5.18
Everything from Part 1 packages into one theorem. If the sizes never increase and tend to zero, the alternating series converges. The proof is word for word the one you just did: the odd partial sums decrease, the even ones increase, and they differ by a single size that shrinks to zero.
The number line shows the mechanism directly. Each arrow is one term; it points the opposite way to the arrow before it and is no longer, so each new partial sum lands between the two before it. The intervals between consecutive partial sums are nested, one inside the next, and the sum sits inside all of them.
Two practical remarks. The sizes only need to decrease from some point on, not from the very first term. And the test has only one conclusion, converges; it never proves divergence.
Notation
Annotate
On: \( b_{n+1} \le b_n \;\text{and}\; b_n \to 0 \;\Longrightarrow\; \sum_{n=1}^{\infty}(-1)^{n+1}b_n \text{ converges} \)
Step through the notes one at a time. The first is the one that matters most in practice: b sub n is the size, with the sign already stripped off. Write it down explicitly before checking anything.
The second and third notes are the two conditions, and each has its own job. Decreasing sizes make the partial sums nest; sizes tending to zero make the nest close down to a single point. You will see a series that fails each one, and both fail to converge.
The last note is about what the theorem cannot do. There is no divergence conclusion anywhere in it. If a condition fails, the correct statement is that the alternating series test does not apply, and you go looking for another test.
Intuition
| condition | what it guarantees | what can happen without it |
|---|---|---|
| sizes do not increase | each partial sum lands between the two before it | the swings can widen again; Exercise 298 drifts to infinity |
| sizes tend to 0 | the nested intervals shrink to one point | the swings stay wide; Example 5.19(b) never settles |
Neither condition alone is enough, and the examples that break each one are coming up. Notice also what the test never asks: it never needs the sum, a comparison series or an integral. That is why it is so quick to use.
Read the table as two short stories. If the sizes stop decreasing, a later step can be longer than an earlier one, so the zig-zag can widen again and the nesting breaks. Exercise 298, coming up soon, uses exactly that to drift off to infinity.
If the sizes decrease but do not tend to zero, the zig-zag keeps a fixed minimum width, and the partial sums bounce between two levels forever. Example 5.19(b) is that failure, drawn.
Notice how little the test asks for. No integral, no comparison series, no formula for the sum: just two facts about the sizes. That is why it is usually the quickest test to run on an alternating series, once you have checked that absolute convergence does not settle things first.
Worked example
Determine whether the series converges or diverges.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} \]
Name the sizes
Why: Strip off the sign.
\[ b_n = \frac{1}{n^2} \]
Check that the sizes decrease
Why: A larger denominator gives a smaller fraction.
\[ (n+1)^2 > n^2 \;\Longrightarrow\; \frac{1}{(n+1)^2} < \frac{1}{n^2} \]
Check that the sizes tend to zero
Why: The denominator grows without bound.
\[ \lim_{n\to\infty}\frac{1}{n^2} = 0 \]
Conclude
Why: Both hypotheses of Theorem 5.13 hold.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} \text{ converges} \]
Figure (svg): Dots for the signed terms of the series with terms minus one to the n plus 1 over n squared, for n from 1 to 12: 1, minus a quarter, one ninth, and so on, alternating above and below the axis, each lying on one of two dashed envelope curves y equals plus or minus one over x squared.
Check with two partial sums
Why: Ten and eleven terms bracket the known value, pi squared over twelve.
\[ S_{10} \approx 0.81796 < \frac{\pi^2}{12} \approx 0.82247 < 0.82623 \approx S_{11} \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 498 — Example 5.19a
This is the model solution, and every line is short. Name the sizes. Check they decrease, by comparing denominators. Check they tend to zero. Then, and only then, state the conclusion.
The figure shows the signed terms as dots flipping above and below the axis, each one sitting on one of the two dashed envelope curves. The distance from the axis is the size, and you can watch it shrink toward zero steadily.
The check uses a fact you will prove later: this series sums to pi squared over twelve. Ten terms fall just below that value and eleven terms just above it, the bracketing you saw with the geometric series. It is a first taste of the error estimate in Part 3.
Worked example
Determine whether the series converges or diverges.
\[ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n+1} \]
Name the sizes
Why: Strip off the sign.
\[ b_n = \frac{n}{n+1} \]
Find their limit
Why: Divide the top and the bottom by n.
\[ \frac{n}{n+1} = \frac{1}{1 + 1/n} \to 1 \]
Read the alternating series test
Why: The second condition fails, so this test is silent.
\[ b_n \to 1 \ne 0 \;\Longrightarrow\; \text{Theorem 5.13 does not apply} \]
Switch to the divergence test
Why: The terms swing between values near 1 and near minus 1, so they have no limit, and certainly not zero.
\[ (-1)^{n+1}\frac{n}{n+1} \not\to 0 \;\Longrightarrow\; \text{diverges} \]
Figure (svg): The first 30 partial sums of the series with terms minus one to the n plus 1 times n over n plus 1, joined by a zig-zag. The odd-numbered sums climb toward about 0.69 and the even ones sink toward about minus 0.31; the vertical swings approach length 1 and never shrink.
Check with the partial sums
Why: The odd and even partial sums settle at two different levels a whole unit apart, so there is no single limit.
\[ S_{2k+1} \to 0.693\ldots, \qquad S_{2k} \to -0.307\ldots \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, pp. 498-499 — Example 5.19b
Here the sizes approach 1, not 0, so the second condition fails and the alternating series test is silent. That is not the same as saying the series diverges; it only means this test has nothing to say.
The divergence test from Section 5.3 does have something to say. The terms themselves swing between values close to 1 and values close to minus 1, so they have no limit, and a series whose terms do not tend to zero diverges.
The figure makes the failure visible. The odd partial sums creep up toward about 0.693 and the even ones down toward about minus 0.307. The gap between the two levels is the size of the next term, which approaches 1, so the zig-zag never closes. Compare this with the previous figure, where the swings shrank to nothing.
Trap
The sizes of this series do not decrease, and a quick write-up says:
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(2 + (-1)^n)}{n^2} \]
\[ b_3 = \tfrac19 < b_4 = \tfrac{3}{16} \]
\[ \Longrightarrow\; \text{diverges?} \]
Wrong. A failed hypothesis is silence, not a verdict.
The alternating series test has one verdict, converges. When a condition fails, go to another test. Here every size is at most three over n squared, so the series converges, and absolutely.
\[ |a_n| \le \frac{3}{n^2} \;\Longrightarrow\; \text{converges} \]
This mistake mirrors the one you met with the divergence test, but the other way round. There, a zero limit meant the test was silent, not that the series converged. Here, a failed condition means the test is silent, not that the series diverges.
The series on the slide is a genuine example. Its sizes jump around, one ninth followed by three sixteenths, so they are not decreasing, and the alternating test does not apply. Yet every size is at most three over n squared, so the absolute values converge by comparison with a p-series, and the series converges absolutely. Always ask what a test is entitled to conclude before you write its verdict.
Worked example
Determine whether the series converges or diverges.
\[ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{2^n} \]
Name the sizes
Why: Strip off the sign.
\[ b_n = \frac{n}{2^n} \]
Compare consecutive sizes by their ratio
Why: A ratio of at most one means the next size is no bigger.
\[ \frac{b_{n+1}}{b_n} = \frac{n+1}{2^{n+1}}\cdot\frac{2^n}{n} = \frac{n+1}{2n} \]
Show the ratio is at most one
Why: For n at least 1, n plus 1 is at most 2n; the first two sizes tie.
\[ n + 1 \le 2n \;\Longrightarrow\; b_{n+1} \le b_n \]
Find the limit of the sizes
Why: L'Hôpital's rule on x over 2 to the x: the exponential wins.
\[ \lim_{x\to\infty}\frac{x}{2^x} = \lim_{x\to\infty}\frac{1}{2^x\ln 2} = 0 \]
Figure (svg): Stems for the sizes n over 2 to the n, for n from 1 to 10: 0.5, 0.5, 0.375, 0.25, 0.156, and then shrinking toward zero. The first two stems are the same height.
Conclude
Why: Both conditions hold.
\[ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{2^n} \text{ converges} \]
Check the value numerically
Why: Ten terms give 0.21875. The sum, from a Chapter 6 power series, is two ninths, and the gap is below the next size, eleven over 2048.
\[ \left|\tfrac29 - 0.21875\right| \approx 0.00347 < \tfrac{11}{2048} \approx 0.00537 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 499 — Checkpoint 5.18
When sizes involve powers or factorials, comparing consecutive sizes by their ratio is usually cleaner than subtracting them. The powers of two cancel down to n plus 1 over 2n, and a ratio of at most one means the next size is no bigger.
Look at the first two sizes in the figure: both are one half. The ratio is exactly one when n equals 1, so the sizes tie. That is allowed, because the condition says each size is at most the one before it, not strictly less.
The limit needs L'Hôpital's rule, since both n and 2 to the n grow without bound. The exponential wins. For the check, a formula from Chapter 6 gives the exact sum, two ninths, and ten terms land within the next size of it, just as the error estimate in Part 3 will promise.
Counterexample
\[ 1 - \frac12 + \frac12 - \frac14 + \frac13 - \frac16 + \frac14 - \frac18 + \cdots \]
Figure (svg): The first 40 partial sums of 1 minus one half plus one half minus one quarter plus one third minus one sixth and so on, joined by a zig-zag that wobbles but drifts steadily upward past 1.7, following a dashed curve that grows like one half of ln k.
Discussion prompt
Exercise 298. These terms alternate in sign and tend to zero. Show that the series diverges anyway, and name the hypothesis of the test that fails.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 506 — Exercise 298
Before revealing, try to see the pattern in the terms. They come in pairs: one over k, then one over 2k with a minus sign. Every pair adds one over 2k, which is positive.
So after 2k terms the total is half of the harmonic partial sum up to k, and the harmonic series diverges. The picture shows the partial sums wobbling up and down but drifting steadily upward along the dashed curve, which grows like half of a logarithm.
The terms alternate and tend to zero, so this example shows the decreasing condition is not decoration. When sizes can jump back up, from one quarter to one third, the zig-zag can widen, the nesting breaks, and the partial sums escape.
Sorting
Sort into buckets
For each series, do the hypotheses of Theorem 5.13 hold (so it converges by that test), or not?
For each item, name the sizes and run through three questions: do the signs alternate, do the sizes decrease, do they tend to zero. You are sorting by whether the test applies, not by whether the series converges.
One over root n, one over ln of n plus 1, and one over n factorial all decrease to zero, so those three converge by the test. The other three each fail a different requirement: the sizes n over n plus 1 tend to one, the sizes in the fourth item are not decreasing, and cos n over n squared does not alternate at all.
Two of the failures still converge, by absolute convergence, which is the subject of Part 4. Only n over n plus 1 actually diverges. That is the lesson of the trap slide again: the test failing tells you nothing about the answer.
Step zero
\[ \sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n} \]
Discussion prompt
The sizes ln n over n do tend to zero. Before applying the test, check the other condition. Do the sizes decrease, and from which n?
Write your answer before revealing. It is tempting to assume the sizes decrease because they obviously head to zero, but heading to zero eventually does not mean decreasing from the start.
Compute a few sizes. At n equal to 1 the size is zero, because ln 1 is zero. Then it rises to about 0.347, then about 0.366, and only then begins to fall. So the sizes are not decreasing from the first term.
This matters because the remark after Theorem 5.13 says the test still works if the sizes decrease from some point on. The next slide finds that point with a derivative.
Worked example
Finish the test for the series on the previous slide.
Treat the sizes as a function of x
Why: Decreasing is easiest to decide with a derivative.
\[ f(x) = \frac{\ln x}{x} \]
Differentiate with the quotient rule
Why: The derivative of ln x is one over x.
\[ f'(x) = \frac{(1/x)\cdot x - \ln x}{x^2} = \frac{1 - \ln x}{x^2} \]
Find where the derivative is negative
Why: The numerator is negative once ln x exceeds one.
\[ f'(x) < 0 \iff \ln x > 1 \iff x > e \approx 2.718 \]
Find the limit of the sizes
Why: L'Hôpital's rule: differentiate the top and the bottom.
\[ \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = 0 \]
Apply the test from N equal to 3
Why: Theorem 5.13 only needs the sizes to decrease from some N onward.
\[ b_{n+1} \le b_n \text{ for } n \ge 3 \;\Longrightarrow\; \text{converges} \]
Check numerically
Why: The sum is about 0.15987. A thousand terms land within the next size of it.
\[ \left|S - S_{1000}\right| \approx |0.15987 - 0.16332| = 0.00345 < b_{1001} \approx 0.00690 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 498 — the remark after Theorem 5.13
The derivative settles the question. The quotient rule gives one minus ln x, over x squared. The denominator is always positive, so the sign is the sign of one minus ln x, which is negative exactly when x is bigger than e. From n equal to 3 onward, the sizes decrease.
The limit comes from L'Hôpital's rule: the logarithm grows much more slowly than x. So from n equal to 3 on, both conditions hold, and the series converges. The first two terms change the value of the sum, but not whether it exists.
The check compares the thousandth partial sum with the actual sum, about 0.15987. The gap is 0.00345, comfortably below the next size, about 0.0069. Keep that comparison in mind: it is the remainder estimate of the next part, in action.
Section
Part 3
Concept
Every partial sum lands between the two before it, so the true sum is trapped between any two consecutive partial sums. The distance from one of them to the sum is at most the distance between the two.
\[ S \text{ lies between } S_N \text{ and } S_{N+1} \]
\[ |R_N| = |S - S_N| \le |S_{N+1} - S_N| = b_{N+1} \]
Remainder estimate (Theorem 5.14) — If an alternating series satisfies the hypotheses of the alternating series test, the error made by stopping after N terms is at most the size of the first term left out.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 499 — Theorem 5.14
The nesting picture gives you more than convergence. Since each partial sum lands between the two before it, the true sum lies between any two consecutive partial sums, the Nth and the next one.
If a number lies between two others, its distance from either one is at most the distance between them. The distance between the Nth partial sum and the next one is exactly the size of the term that was added, b sub N plus 1. So the error in stopping after N terms is at most the size of the first term you did not add.
Compare this with Section 5.3, where bounding the error for a positive series needed two improper integrals. Here the bound is one term, read straight off the series. The book's own line uses a lowercase n on the right-hand side; it means the same capital N plus 1.
Concept
The first term you leave out also tells you which side of the sum you are on. Stop after a subtracted term and you are below; stop after an added term and you are above.
\[ S_{10} = 1 - \tfrac12 + \tfrac13 - \cdots - \tfrac1{10} \approx 0.6456 \]
\[ \text{next term: } +\tfrac1{11} \;\Longrightarrow\; S_{10} < S < S_{11} \approx 0.7365 \]
So the sum sits in an interval whose ends are two partial sums you already have. For the alternating harmonic series the true value lands inside, as it must.
\[ 0.6456 < \ln 2 \approx 0.6931 < 0.7365 \]
The theorem bounds the size of the error, but the zig-zag also tells you its direction. If the first term you leave out is positive, adding it would move you up toward the sum, so you are currently below the sum. If it is negative, you are above.
For the alternating harmonic series after ten terms, the next term is plus one eleventh, so the partial sum 0.6456 is an underestimate, and the sum lies between it and the eleventh partial sum, 0.7365. The natural log of 2, about 0.6931, sits inside that interval as promised.
This two-sided trap is something positive series never give you: there, every partial sum is an underestimate, and you need an integral to say by how much.
Notation
Annotate
On: \( |R_N| = |S - S_N| \le b_{N+1} \)
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 499 — Theorem 5.14
The remainder is the true sum minus the partial sum. For an alternating series its sign flips as N changes, which is why the theorem talks about its absolute value.
The note on b sub N plus 1 flags the most common slip on exams: using the last term you included instead of the first term you left out. Stopping after ten terms, the bound is the eleventh size, not the tenth.
The last note is the one that is easy to forget. The bound is a consequence of the nesting, so it is only available when the alternating series test's hypotheses hold. A later slide shows what goes wrong if you ignore that.
Picture it
Figure (svg): For N from 1 to 12, a dot at the partial sum S_N of the alternating reciprocal squares, with a vertical bar reaching one next-term size above and below it. Every bar crosses the dashed line at pi squared over 12, about 0.822, and the bars shrink quickly.
For each N, the bar runs one next-term size above and below the partial sum. Every bar reaches the dashed line, the true sum: that is Theorem 5.14, drawn twelve times.
Each dot is a partial sum of the alternating reciprocal squares, and each vertical bar stretches one next-term size above and below it. The dashed line is the true sum, pi squared over twelve.
Every bar crosses the dashed line. That is Theorem 5.14 drawn twelve times: the true sum is always within one next-term size of the partial sum. Notice how fast the bars shrink, because the sizes here are one over n squared.
Look more closely and you will see the dashed line always sits in the half of the bar pointing back toward the previous dot. That is the sign information from two slides ago: the true sum is between consecutive partial sums, not just near each one.
Worked example
Bound the error in approximating the sum by the tenth partial sum.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} \approx S_{10} \]
Confirm the hypotheses
Why: Example 5.19(a) showed the sizes decrease to zero.
\[ b_n = \frac{1}{n^2}, \quad b_{n+1} < b_n, \quad b_n \to 0 \]
Identify the first size left out
Why: Stopping after ten terms omits the eleventh.
\[ b_{11} = \frac{1}{11^2} = \frac{1}{121} \]
State the bound
Why: Theorem 5.14.
\[ |R_{10}| \le \frac{1}{121} \approx 0.008265 \]
Compute the partial sum
Why: Ten terms, kept to five places.
\[ S_{10} = 1 - \tfrac14 + \tfrac19 - \cdots - \tfrac1{100} \approx 0.81796 \]
Check against the exact value
Why: This sum is known to be pi squared over twelve; the actual error is about half the bound.
\[ \frac{\pi^2}{12} - S_{10} \approx 0.82247 - 0.81796 = 0.00451 < 0.008265 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, pp. 499-500 — Example 5.20
This is the book's example and it takes one line once the hypotheses are known: the first size left out is one over eleven squared, one over 121, about 0.008265. That number bounds the error, whatever the partial sum happens to be.
The rest of the slide is there to show the bound is real. The partial sum of ten terms is about 0.81796. The exact sum is known, pi squared over twelve, about 0.82247, so the actual error is about 0.0045, a little over half the bound.
Notice which way the error goes: the eleventh term is positive, so the tenth partial sum is below the sum. The check confirms it. Getting both the size and the direction of the error from one term is what makes alternating series so pleasant numerically.
Worked example
Bound the error in approximating the sum by the twentieth partial sum.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \approx S_{20} \]
Confirm the hypotheses
Why: The sizes one over n decrease to zero.
\[ b_n = \frac1n, \quad b_{n+1} < b_n, \quad b_n \to 0 \]
Identify the first size left out
Why: The twenty-first.
\[ b_{21} = \frac{1}{21} \]
State the bound
Why: Theorem 5.14.
\[ |R_{20}| \le \frac{1}{21} \approx 0.0476 \]
Compute the partial sum
Why: Twenty terms, by machine.
\[ S_{20} \approx 0.66877 \]
Use the sign as well
Why: The twenty-first term is added, so the sum is above the partial sum.
\[ 0.66877 < S < 0.66877 + 0.04762 = 0.71639 \]
Check against ln 2
Why: The true error is about half the bound.
\[ \ln 2 - S_{20} \approx 0.69315 - 0.66877 = 0.02438 < 0.0476 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 500 — Checkpoint 5.19
Same method, slower series. The first size left out is one over 21, about 0.0476, so twenty terms guarantee only about one correct decimal place.
Using the sign as well, the twenty-first term is positive, so the sum lies above the twentieth partial sum and below it plus 0.0476. The natural log of 2 falls inside that interval, about 0.024 above the partial sum.
Once again the actual error is roughly half the bound. That is not a coincidence, and the next two slides show it is a general pattern for series like this one.
Estimation
Predict first
For the alternating harmonic series after 40 terms, the bound is 1/41 ≈ 0.0244. Guess the actual error before looking at the next slide.
Correct: About half of it
Why: The actual error after 40 terms is about 0.0123, almost exactly half of 1/41. The sum sits roughly in the middle of the last swing, so the next-term bound overstates the error by a factor of about two, not by orders of magnitude.
Commit to a guess before revealing. You have now seen two examples where the actual error was around half the bound; this asks whether that was luck.
It was not. After forty terms the bound is about 0.0244 and the actual error is about 0.0123, very close to exactly half. When the sizes change slowly from one term to the next, the true sum sits close to the middle of the last swing, so the error is about half of one swing.
Intuition
Figure (svg): On a logarithmic vertical axis, for N from 1 to 40: a curve for the bound one over N plus 1, and dots for the actual error of the alternating harmonic series after N terms. The dots run parallel to the curve, about half its height.
On a logarithmic scale the dots run parallel to the curve, one steady step below it: the actual error is about half the next size, every time.
\[ |R_N| \approx \tfrac12 b_{N+1} \]
That suggests a cheap improvement: average two consecutive partial sums, the midpoint of the last swing.
\[ \frac{S_{10} + S_{11}}{2} \approx 0.6911 \quad\text{vs}\quad \ln 2 \approx 0.6931 \]
The vertical axis is logarithmic, so equal vertical gaps mean equal ratios. The dots for the actual error run parallel to the curve for the bound, a constant gap below it, and that gap is a factor of about two.
Two lessons come out of this. First, the bound is honest: it never overstates the error by a huge factor for a series like this, so choosing N from it does not waste many terms. Second, since the sum sits near the middle of the last swing, averaging two consecutive partial sums gives a much better estimate.
Averaging the tenth and eleventh partial sums gives 0.6911, only about 0.002 away from ln 2, where each partial sum on its own was off by about 0.05. This averaging idea is the seed of several acceleration methods used in real numerical work.
Worked example
Choose N so the Nth partial sum is within one thousandth of the sum.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}: \quad |R_N| < 0.001 \]
Force the bound below the tolerance
Why: If the first omitted size is small enough, the error is smaller still.
\[ b_{N+1} = \frac{1}{(N+1)^2} < 0.001 \]
Solve for N plus 1
Why: Invert both sides, then take square roots.
\[ (N+1)^2 > 1000 \;\Longrightarrow\; N + 1 > \sqrt{1000} \approx 31.62 \]
Take the smallest whole number that works
Why: N plus 1 must be at least 32.
\[ N + 1 = 32 \;\Longrightarrow\; N = 31 \]
Compare with the positive series
Why: Section 5.3's integral bound for the reciprocal squares, with no signs, needed a thousand terms.
\[ R_N < \frac1N \le 0.001 \iff N \ge 1000 \]
Check both sides of the boundary, and the real error
Why: At N equal to 31 the bound holds and at 30 it fails; the actual error at 31 is about half a thousandth.
\[ \frac{1}{32^2} \approx 0.000977 < 0.001 < 0.001041 \approx \frac{1}{31^2} \]
\[ \left|\tfrac{\pi^2}{12} - S_{31}\right| \approx 0.000504 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 506 — in the style of Exercise 285
Now run the estimate backwards. You want the error below a thousandth, and the error is at most the first omitted size, so it is enough to make that size less than a thousandth. The inequality becomes N plus 1 squared bigger than one thousand.
The square root of 1000 is about 31.62, so N plus 1 must be at least 32, and N is 31. The check tests the integers on both sides: at 31 terms the bound holds, at 30 it just fails. The actual error at 31 terms is about half a thousandth.
The comparison step is the striking part. For the same sizes without signs, Section 5.3's integral estimate needed a thousand terms for the same accuracy. The signs cut the work by a factor of about thirty, because the tail of an alternating series mostly cancels itself.
Worked example
How many terms of this series pin down pi over four to within 0.0001?
\[ \frac{\pi}{4} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} = 1 - \frac13 + \frac15 - \frac17 + \cdots \]
Name the sizes and check the test
Why: Odd reciprocals decrease to zero.
\[ b_n = \frac{1}{2n+1}, \quad b_{n+1} < b_n, \quad b_n \to 0 \]
Write the bound for stopping at n equal to N
Why: The sum starts at n equal to 0, so the first omitted term has index N plus 1.
\[ |R_N| \le b_{N+1} = \frac{1}{2N+3} \]
Force it below the tolerance
Why: Invert and solve.
\[ \frac{1}{2N+3} < 0.0001 \iff 2N + 3 > 10{,}000 \iff N > 4998.5 \]
Take the smallest N
Why: Terms 0 through 4999: five thousand of them.
\[ N = 4999 \]
Compute the partial sum
Why: By machine, then multiply by four.
\[ S_{4999} \approx 0.785348, \quad 4S_{4999} \approx 3.141393 \]
Figure (svg): The first 41 partial sums of four times the series 1 minus a third plus a fifth and so on, joined by a zig-zag around a dashed line at pi. The swings start at more than a full unit and are still about a tenth wide after 40 terms.
Check against pi
Why: The actual error in pi over four is half the bound, as usual.
\[ \frac{\pi}{4} - S_{4999} \approx 0.0000500 < 0.0001 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 507 — Exercise 306
This famous series gives pi over four, and it is an honest test of the remainder estimate because the sum starts at n equal to 0. Stopping at index N means the first omitted term has index N plus 1, and its size is one over 2N plus 3.
Making that below one ten-thousandth needs 2N plus 3 above ten thousand, so N is 4999: five thousand terms in all. Multiplying the partial sum by four gives 3.14139, which matches pi only to about three decimal places, because the error in pi is four times the error in pi over four.
The picture shows why so many terms are needed. The sizes shrink like one over 2n, so the zig-zag tightens very slowly; after forty terms it still swings by a tenth. The series is beautiful but useless for computing pi, which is why real computations use far faster series.
Error analysis
Annotate
On: \( \sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n+1}: \quad |R_{10}| \le b_{11} = \frac{11}{12} \)
Read the line before revealing the notes, and look for the step that has no right to be there. The arithmetic is fine: the eleventh size really is eleven twelfths.
The problem is the series. Its sizes tend to one, so it fails the alternating series test, and in fact it diverges. A divergent series has no sum, so the remainder, the sum minus the partial sum, does not exist, and bounding it is meaningless.
The remainder estimate is not a separate fact you can apply to any alternating series. It comes out of the nesting, and the nesting comes from the test's hypotheses. Check those first, every time.
Comparison
Comparison matrix
| integral test (§5.3) | alternating series (§5.5) | |
|---|---|---|
| the bound on |R_N| | ∫ from N to ∞ of f | b(N+1) |
| sizes 1/n²: terms for error 0.001 | 1000 | 31 |
| which side of S is S_N? | always below | below after a minus, above after a plus |
| sizes 1/n: terms for error 0.001 | no bound: the integral diverges | 1000 |
Fill the blanks before checking. The first row asks for the alternating bound itself, the next omitted size. The second row compares the two methods on the same sizes, one over n squared, for an error of a thousandth: a thousand terms with the integral bound, thirty-one with the alternating bound.
The last row is the sharpest contrast. With sizes one over n, the positive series diverges, so there is no error to bound at all. The alternating version converges, and a thousand terms make the next size one over 1001, below a thousandth.
The row about sides is worth remembering: positive partial sums always undershoot, while alternating partial sums undershoot after a minus and overshoot after a plus.
Tweak it
Parameter explorer
The graph shows the partial sums of the alternating series of 1/nᵖ, as a step function of N. Slide p from 2 down to 0. When do the swings close up, and when do they stop closing?
\[ S_N = \sum_{n=1}^{N}\frac{(-1)^{n+1}}{n^{{p}}} \]
Start with p at 2 and watch the step graph settle almost immediately. Move p down to 1 and you get the alternating harmonic series: the swings shrink, but slowly. Keep going to one half and the swings shrink more slowly still, yet they do shrink.
Now push p all the way to zero. Every size becomes 1, the partial sums flip between 1 and 0 forever, and the swings never close. That is the only value on the slider where the series diverges.
Contrast this with the p-series slider in Section 5.3, where everything at or below p equal to one diverged. With alternating signs, any positive p converges, because the only requirements are sizes that decrease and tend to zero.
Real world
\[ \cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots \]
Figure (svg): Bars on a logarithmic scale for the sizes of the terms of the cosine series at one: 1 over 0 factorial, 2 factorial, 4 factorial, 6 factorial, 8 factorial and 10 factorial, falling from 1 to about 0.0000003. A dashed line marks the tolerance 0.00001; the bar for 1 over 8 factorial is above it and the one for 1 over 10 factorial is below.
Discussion prompt
Exercise 303. Chapter 6 derives this series. At theta equal to 1 it is alternating with sizes one over (2k)!. How many terms guarantee cos 1 to within 0.00001?
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 507 — Exercise 303
Chapter 6 will show where this series for cosine comes from. At theta equal to 1 it is an alternating series whose sizes are one over the even factorials, and factorials grow so fast that the sizes collapse.
The picture shows the sizes on a logarithmic scale against the tolerance line. One over eight factorial is still above one hundred-thousandth, so the term with eight factorial must be included. One over ten factorial is below it, so once you have included the eight-factorial term you may stop: five terms in all.
Five terms give 0.5403026, and the true value of cos 1 is 0.5403023, an error of about three ten-millionths, just under the bound. This is close to how software evaluates functions: pick a series, and use a remainder bound to decide how many terms guarantee the accuracy it promises.
Section
Part 4
Concept
Any series with some negative terms has a partner: the series of absolute values. Comparing the two sorts convergent series into two kinds.
absolute convergence — The series of absolute values converges.
conditional convergence — The series converges, but the series of absolute values diverges.
Figure (svg): Three nested rounded regions. The outer one is all series; inside it, the convergent series; inside that, the absolutely convergent series. The alternating reciprocal squares and cos n over n squared sit in the innermost region; the alternating harmonic series and the alternating series of one over 3n plus 1 sit in the middle ring; the harmonic series and the alternating n over n plus 1 sit outside the convergent region.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 500 — definition
Every series with negative terms has a partner made of the absolute values. Asking whether the partner converges splits convergent series into two kinds.
Absolutely convergent means the partner converges: the sizes alone add up to something finite, and the signs are along for the ride. Conditionally convergent means the series converges but the partner does not: the signs are doing essential work, and without the cancellation there would be no sum.
The picture places the lesson's examples. The innermost region is absolute convergence, the ring around it is conditional convergence, and everything outside the middle region diverges. Theorem 5.15, coming shortly, is the statement that the inner region really does sit inside the middle one.
Concept
Strip the signs from the alternating harmonic series and you get the harmonic series, which diverges; the signed version converges. Conditional.
\[ \sum\left|\frac{(-1)^{n+1}}{n}\right| = \sum\frac1n = \infty \quad\Longrightarrow\quad \text{conditional} \]
Strip the signs from the alternating reciprocal squares and you get a p-series with p equal to 2, which converges. Absolute.
\[ \sum\left|\frac{(-1)^{n+1}}{n^2}\right| = \sum\frac{1}{n^2} = \frac{\pi^2}{6} \quad\Longrightarrow\quad \text{absolute} \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 500 — the two examples before the definition
Keep one example of each kind in your head, and use them to test every claim you meet. The alternating harmonic series is the model of conditional convergence: strip the signs and you are left with the harmonic series, which diverges.
The alternating reciprocal squares are the model of absolute convergence: strip the signs and you are left with a convergent p-series, whose sum is pi squared over six.
Both series converge, so convergence alone does not tell them apart. What tells them apart is whether the convergence survives removing the signs, and Part 5 will show that this difference has real consequences.
Worked example
Prove it: if the series of absolute values converges, so does the series itself.
Split into the two possible signs
Why: Each term is either its own absolute value or minus it.
\[ a_n = |a_n| \quad\text{or}\quad a_n = -|a_n| \]
Add the absolute value in each case
Why: The two cases give twice the size, or nothing.
\[ |a_n| + a_n = 2|a_n| \quad\text{or}\quad |a_n| + a_n = 0 \]
Sandwich the new terms
Why: Either way they are never negative and never above twice the size.
\[ 0 \le |a_n| + a_n \le 2|a_n| \]
Figure (svg): For n from 1 to 16 and the terms a sub n equal to sin n over n to the one and a half: hollow circles at height twice the absolute value of a sub n, and filled bars for the absolute value of a sub n plus a sub n. Each bar either reaches its circle, when a sub n is positive, or is flat at zero, when a sub n is negative.
Apply the comparison test
Why: Twice a convergent series converges, and the new series sits under it.
\[ \sum 2|a_n| < \infty \;\Longrightarrow\; \sum\left(|a_n| + a_n\right) < \infty \]
Subtract the absolute values back off
Why: The difference of two convergent series converges.
\[ \sum a_n = \sum\left(|a_n| + a_n\right) - \sum |a_n| \quad\text{converges} \]
Check the sandwich on the first terms of the figure
Why: With terms sin n over n to the three halves: the first is positive, the fourth negative.
\[ a_1 \approx 0.8415: \; |a_1| + a_1 \approx 1.683 = 2|a_1|; \quad a_4 \approx -0.0946: \; |a_4| + a_4 = 0 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, pp. 500-501 — Theorem 5.15 and proof
The proof is short and clever. Add each term to its own absolute value. If the term is positive you get twice its size; if it is negative you get zero. Either way the result is between zero and twice the size.
The figure shows this for terms sin n over n to the three halves: each bar either reaches the hollow circle at twice the size, when the term is positive, or is flat at zero, when the term is negative. The new series has nonnegative terms under a convergent series, so it converges by the comparison test.
Finally, the original series is the new series minus the series of absolute values, and the difference of two convergent series converges. The check simply confirms the sandwich on two terms of the figure, one of each sign.
Notation
Annotate
On: \( \sum_{n=1}^{\infty}|a_n| \text{ converges} \;\Longrightarrow\; \sum_{n=1}^{\infty}a_n \text{ converges} \)
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 500 — Theorem 5.15
The first note is the practical payoff. The series of absolute values has nonnegative terms, so all of the positive-term machinery from Sections 5.3 and 5.4 applies to it. Theorem 5.15 lets you use those tools on series with signs.
The second note is what makes the theorem so general. The alternating series test needs a strict plus, minus, plus pattern. This theorem needs no pattern at all, which is how it handles cos n over n squared, where the signs follow no rule.
The third and fourth notes are the limits of the theorem. It runs one way only, and it transfers convergence, not the value: the sum of the absolute values is usually a different, larger number.
Trap
Having shown a series converges:
\[ \sum\frac{(-1)^{n+1}}{n} \text{ converges} \]
\[ \Longrightarrow\; \sum\frac1n \text{ converges?} \]
Wrong. The implication only runs one way.
Absolute convergence implies convergence, never the reverse. The alternating harmonic series converges, yet its absolute values are the harmonic series, which diverges. That gap is exactly conditional convergence.
\[ \text{absolute} \Longrightarrow \text{convergent} \]
\[ \text{convergent} \not\Longrightarrow \text{absolute} \]
Under pressure it is easy to remember the theorem as convergence and absolute convergence being the same thing. They are not, and the alternating harmonic series is the permanent counterexample: it converges, and its absolute values diverge.
When you have shown a series converges by the alternating series test, you have learned nothing about its absolute values. You must test those separately, and if they diverge, the verdict is conditional convergence.
Worked example
Absolutely convergent, conditionally convergent, or divergent?
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{3n+1} \]
Test absolute convergence first
Why: Drop the signs.
\[ \sum_{n=1}^{\infty}\left|\frac{(-1)^{n+1}}{3n+1}\right| = \sum_{n=1}^{\infty}\frac{1}{3n+1} \]
Limit-compare with the harmonic series
Why: Divide the terms.
\[ \lim_{n\to\infty}\frac{1/(3n+1)}{1/n} = \lim_{n\to\infty}\frac{n}{3n+1} = \frac13 \]
Read the comparison
Why: A finite positive limit means a shared fate, and the harmonic series diverges.
\[ \sum\frac{1}{3n+1} = \infty \;\Longrightarrow\; \text{not absolutely convergent} \]
Now apply the alternating series test
Why: The sizes decrease and tend to zero.
\[ \frac{1}{3(n+1)+1} < \frac{1}{3n+1}, \quad \frac{1}{3n+1} \to 0 \]
Classify
Why: Convergent, but not absolutely.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{3n+1} \text{ converges conditionally} \]
Figure (svg): Two sequences of partial sums for k up to 60. The sums of one over 3n plus 1 climb steadily past 1.4 without levelling off; the sums of the alternating version zig-zag in a narrow band and settle near 0.164.
Check with partial sums
Why: Sixty absolute values already total 1.413 and keep climbing; the signed sum after 60 terms is within the next size, one over 184, of its limit 0.16435.
\[ |S - S_{60}| \approx |0.16435 - 0.16161| = 0.00274 < \tfrac{1}{184} \approx 0.00543 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 501 — Example 5.21a
The order of work matters. Test the absolute values first, because a yes ends the problem. Here the absolute values behave like one third of one over n, and the limit comparison with the harmonic series makes that precise: the ratio tends to one third, a finite positive number, so both diverge.
That rules out absolute convergence but says nothing yet about the series itself. Now the alternating series test applies: the sizes one over 3n plus 1 decrease and tend to zero, so the series converges. Convergent but not absolutely: conditional.
The figure shows both halves of that verdict. The sums of the absolute values climb past 1.4 after sixty terms and keep going, growing like a third of a logarithm. The signed sums settle into a narrow band near 0.164, and the check confirms the sixtieth partial sum is within the next size of the limit.
Worked example
Absolutely convergent, conditionally convergent, or divergent?
\[ \sum_{n=1}^{\infty}\frac{\cos n}{n^2} \]
Notice that it is not alternating
Why: The signs follow cos n, which has no regular rhythm.
\[ \cos 1 \approx 0.54, \; \cos 2 \approx -0.42, \; \cos 3 \approx -0.99, \; \cos 4 \approx -0.65 \]
Bound the absolute values
Why: Cosine is never bigger than one in size.
\[ \left|\frac{\cos n}{n^2}\right| \le \frac{1}{n^2} \]
Compare with a p-series
Why: p equals 2, which is bigger than one.
\[ \sum\frac{1}{n^2} < \infty \;\Longrightarrow\; \sum\left|\frac{\cos n}{n^2}\right| < \infty \]
Apply Theorem 5.15
Why: Absolute convergence gives convergence.
\[ \sum_{n=1}^{\infty}\frac{\cos n}{n^2} \text{ converges absolutely} \]
Figure (svg): Stems for the terms cos n over n squared, for n from 1 to 20, with signs in no regular pattern (positive, negative, negative, negative, positive, positive, …), all lying between the dashed envelope curves plus and minus one over x squared.
Check the comparison numerically
Why: The absolute values total about 0.928, below the comparison total; the signed series settles near 0.324.
\[ \sum\left|\frac{\cos n}{n^2}\right| \approx 0.928 < \frac{\pi^2}{6} \approx 1.645 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, pp. 501-502 — Example 5.21b
This series is not alternating, and it is worth seeing that for yourself. The signs of cos n go plus, minus, minus, minus, plus, plus for n from 1 to 6. The alternating series test cannot be used at all.
Absolute convergence rescues it. Cosine is never larger than one in size, so each absolute value is at most one over n squared, and comparison with that convergent p-series shows the absolute values converge. Theorem 5.15 then gives convergence.
The figure shows the terms wandering irregularly between the two envelope curves. Their signs have no pattern, but their sizes are squeezed by one over n squared, and that squeeze is all the comparison test needs.
Worked example
Absolutely convergent, conditionally convergent, or divergent?
\[ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{2n^3+1} \]
Drop the signs
Why: Test absolute convergence first.
\[ |a_n| = \frac{n}{2n^3+1} \]
Make the denominator smaller
Why: Dropping the plus one makes the fraction bigger.
\[ \frac{n}{2n^3+1} < \frac{n}{2n^3} = \frac{1}{2n^2} \]
Compare with a p-series
Why: Half of a convergent p-series converges.
\[ \sum\frac{1}{2n^2} = \frac12\sum\frac{1}{n^2} < \infty \]
Classify
Why: The absolute values converge by comparison.
\[ \sum|a_n| < \infty \;\Longrightarrow\; \text{converges absolutely} \]
Check the comparison numerically
Why: The absolute values total about 0.647, below the comparison total, pi squared over twelve.
\[ \sum\frac{n}{2n^3+1} \approx 0.647 < \frac{\pi^2}{12} \approx 0.822 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 502 — Checkpoint 5.20
Test the absolute values first. For large n the plus one in the denominator hardly matters, so the terms behave like n over 2n cubed, which is one over 2n squared. That suggests comparison with a p-series.
To make the comparison rigorous, notice that removing the plus one makes the denominator smaller and the fraction bigger. So each absolute value is less than one over 2n squared, which is half of a convergent p-series. The absolute values converge, and the series converges absolutely.
Once absolute convergence is established you are done: there is no need to check the alternating series test at all. The numerical check confirms that the absolute values total about 0.647, below pi squared over twelve.
Concept
With negative terms around, test the absolute values before anything else. Positive-term tests are the richest toolkit you have, and a yes settles everything at once.
| result for the absolute values | what it tells you | next move |
|---|---|---|
| converges | absolutely convergent (Theorem 5.15) | done |
| diverges | not absolutely convergent; the series itself is undecided | alternating series test, or divergence test |
\[ \sum|a_n| = \infty \;\not\Longrightarrow\; \sum a_n = \infty \]
A divergent series of absolute values proves nothing about the signed series by itself: Example 5.21(a) had exactly that and still converged.
Absolute convergence is the stronger property, and testing for it first is usually the efficient route. If the absolute values converge, you have convergence, you know the sum does not depend on the order, and you are finished.
If the absolute values diverge, be careful about what you conclude. The signed series might still converge, conditionally, as the alternating harmonic series does. Divergence of the absolute values only means you have more work to do: the divergence test, or the alternating series test.
The table is the whole decision in two rows. Keep it next to the flow diagram at the end of the lesson.
Discrimination
Sort into buckets
Classify each series.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 505 — Exercises 250, 254, 263, 276
For each series, first drop the signs and decide whether the absolute values converge. If they do, the answer is absolute. If they do not, decide whether the alternating test rescues it or the terms fail to tend to zero.
Two items hide their structure. The cos n pi series is alternating in disguise, because cos n pi is minus one to the n, so it is the alternating harmonic series with the sign flipped: conditional. And ln of 1 plus one over n behaves like one over n for large n, so its absolute values diverge while the alternating version converges: conditional again.
The only divergent one is n over n plus 3, whose terms tend to one. Notice that the one over n factorial series converges absolutely very comfortably, because factorials outgrow powers of two.
Section
Part 5
Concept
Reordering a finite sum never changes it. For the alternating harmonic series, reordering can send the partial sums to infinity. Spend positive terms until the total passes 10:
\[ 1 + \frac13 + \frac15 + \cdots + \frac{1}{2k-1} > 10 \]
This is possible because the odd reciprocals alone diverge; it first happens after 68,100,151 of them. Then subtract one half and spend more positives until the total passes 100:
\[ \cdots - \frac12 + \frac{1}{2k+1} + \cdots + \frac{1}{2j+1} > 100 \]
\[ \text{then } -\tfrac14, \text{ then past } 1000, \;\ldots \;\Longrightarrow\; \text{partial sums unbounded} \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 502 — rearranging to diverge
For a finite sum, changing the order never changes the total. It is natural to expect the same of infinite sums, and this slide shows that for the alternating harmonic series the expectation fails completely.
The trick is that the positive terms alone, one plus a third plus a fifth and so on, add up to infinity. So you can take enough of them to pass 10, then pay out a single negative term, then take more positives to pass 100, pay out one more negative, and so on. Every term is eventually used exactly once, but the partial sums are unbounded.
The first stage alone needs about sixty-eight million positive terms, which gives a feel for how slowly those positive terms diverge. Slow or not, they diverge, and that is all the construction needs.
Worked example
Rearrange the alternating harmonic series so that its sum is three halves of ln 2.
Start from the known sum
Why: Proved in Part 1, with the value from Chapter 6.
\[ 1 - \tfrac12 + \tfrac13 - \tfrac14 + \tfrac15 - \tfrac16 + \cdots = \ln 2 \]
Halve every term
Why: A convergent series may be multiplied by a constant.
\[ \tfrac12 - \tfrac14 + \tfrac16 - \tfrac18 + \cdots = \tfrac12\ln 2 \]
Space it out with zeros
Why: A zero before each term only repeats partial sums, so the sum is unchanged.
\[ 0 + \tfrac12 + 0 - \tfrac14 + 0 + \tfrac16 + 0 - \tfrac18 + \cdots = \tfrac12\ln 2 \]
Add the two series term by term
Why: The sum of two convergent series converges to the sum of their sums.
\[ (1+0) + \left(-\tfrac12+\tfrac12\right) + \left(\tfrac13+0\right) + \left(-\tfrac14-\tfrac14\right) \]
\[ + \left(\tfrac15+0\right) + \left(-\tfrac16+\tfrac16\right) + \left(\tfrac17+0\right) + \left(-\tfrac18-\tfrac18\right) + \cdots \]
Simplify each bracket
Why: Even terms with a plus cancel; even terms with a minus double.
\[ 1 + 0 + \tfrac13 - \tfrac12 + \tfrac15 + 0 + \tfrac17 - \tfrac14 + \cdots = \tfrac32\ln 2 \]
Drop the zeros
Why: Every odd reciprocal appears once, and every even reciprocal once with a minus sign: a rearrangement.
\[ 1 + \frac13 - \frac12 + \frac15 + \frac17 - \frac14 + \cdots = \frac{3\ln 2}{2} \]
Figure (svg): Two zig-zags of partial sums for 60 terms. The alternating harmonic series in its usual order settles on a dashed line at ln 2, about 0.693. The same terms taken two positives then one negative settle on a higher dashed line at three halves ln 2, about 1.040.
Check numerically
Why: Nine thousand terms of the rearranged series agree with three halves of ln 2 to about four decimals.
\[ S_{9000}^{\text{new}} \approx 1.03964, \qquad \tfrac32\ln 2 \approx 1.03972 \]
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 503 — Example 5.22
This construction is a small masterpiece, and every step uses only the algebra of convergent series. Halve the known series, then slide a zero in front of every term; neither step changes a sum.
Now add the original series and the spaced-out half series term by term. Every odd reciprocal appears once, from the original. Every even reciprocal appears twice: in half of the positions the two copies cancel, and in the other half they double into a single negative term. What survives is two positive odd terms followed by one negative even term, over and over.
Each reciprocal appears exactly once, so the result is a rearrangement of the alternating harmonic series, and its sum is ln 2 plus half of ln 2. The figure and the check agree: the same terms, reordered, settle at 1.0397 instead of 0.6931.
Picture it
Figure (svg): Sixty partial sums of a rearrangement of the alternating harmonic series built by one rule: add the next unused positive term while the total is below 1.5, otherwise subtract the next unused negative term. The zig-zag climbs to 1.5, then oscillates around it with shrinking swings.
One rule does it: below the target, add the next unused positive term; above it, subtract the next unused negative term. Each overshoot is at most one term, and the terms shrink to zero, so the running total homes in on the target. Any target works.
This picture shows the general idea behind the rearrangement theorem, using a target of 1.5. The rule is greedy: whenever the running total is below the target, add the next positive term you have not used; whenever it is above, subtract the next negative term you have not used.
Because the positive terms alone diverge, you can always climb back above the target, and because the negative terms alone diverge, you can always get back below it. So the rule never gets stuck, and every term is eventually used.
Each overshoot is at most the size of one term, and the sizes tend to zero, so the overshoots shrink and the total homes in on 1.5. Replace 1.5 by any number you like and the same rule reaches it.
Concept
| kind of convergence | what rearranging the terms can do |
|---|---|
| absolute | nothing: every rearrangement has the same sum |
| conditional | anything: converge to any chosen number, or diverge |
Why the difference? Split a conditionally convergent series into its positive and its negative terms. For the alternating harmonic series both halves diverge:
\[ \sum_{k=1}^{\infty}\frac{1}{2k-1} = \infty \quad\text{and}\quad \sum_{k=1}^{\infty}\frac{1}{2k} = \infty \]
An unlimited supply of both, in steps that shrink to zero, lets a steering rule aim anywhere. An absolutely convergent series has only a finite amount of each to spend, so there is nothing to steer with.
OpenStax Calculus Volume 2, §5.5 Alternating Series §5.5, p. 502 — Riemann Rearrangement Theorem
The table states Riemann's theorem, which the book mentions without proof. For absolutely convergent series, order does not matter at all. For conditionally convergent series, order can produce any sum, or divergence.
The reason is in the second half of the slide. A conditionally convergent series must have positive terms that add to infinity and negative terms that add to minus infinity; otherwise the absolute values would converge. That unlimited supply in both directions is exactly what the steering rule needs.
An absolutely convergent series has a finite total of positive parts and a finite total of negative parts. However you reorder, you eventually use all of each, so the sum cannot change. This is why later work, such as rearranging power series in Chapter 6, insists on absolute convergence.
Matching
Match the pairs
Why: Absolute convergence makes the order irrelevant, so the reciprocal squares keep their sum. The alternating harmonic series is conditionally convergent: ln 2 in its usual order, (3/2) ln 2 in Example 5.22's order, and divergent in the order that chases 10, 100, 1000. A series whose terms do not tend to 0 has none of those terms tending to 0 in any order, so it diverges however you arrange it.
Match each series before checking. Decide first whether it converges absolutely, conditionally or not at all, because that alone determines what reordering can do.
The reciprocal squares converge absolutely, so every order gives pi squared over twelve. The alternating harmonic series in its usual order gives ln 2, but a different order diverges. Example 5.22's order gives three halves of ln 2. And a series whose terms do not tend to zero diverges in every order, since reordering cannot make its terms shrink.
Anomaly
\[ \ln 2 = 1 - \tfrac12 + \tfrac13 - \tfrac14 + \tfrac15 - \cdots \]
\[ \tfrac32\ln 2 = 1 + \tfrac13 - \tfrac12 + \tfrac15 + \tfrac17 - \tfrac14 + \cdots \]
Both right-hand sides use exactly the same terms, each once. So the left-hand sides are equal, and dividing by ln 2 gives one equals three halves.
Predict first
Which step of this argument is wrong?
Correct: 'Same terms, so same sum': that fails for conditionally convergent series
Why: Both values are correct (Part 1 and Example 5.22). The broken step is the belief that an infinite sum does not depend on order. A series is the limit of its partial sums, and reordering changes the partial sums. For a conditionally convergent series it changes their limit too.
Pick an answer before revealing. Both equations on the slide are true: you proved the first in Part 1 and the second in Example 5.22. So the error must be in the reasoning that connects them.
The broken step is the silent assumption that the same terms, in a different order, must give the same sum. That is true for finite sums, and for absolutely convergent series, but the alternating harmonic series converges only conditionally.
The resolution is to remember what an infinite sum is: the limit of a sequence of partial sums. Changing the order changes which partial sums you get, and so it can change the limit. Commutativity of addition is a fact about finite sums, and it does not automatically survive a limit.
Section
Part 6
Pattern
Figure (svg): A flow diagram. First, does the limit of the terms equal zero? If not, the series diverges. If so, test the series of absolute values with the positive-term tests; if it converges, the series converges absolutely. If it diverges, check whether the series alternates with decreasing sizes; if so, it converges conditionally, otherwise another tool is needed.
This is the order to work in whenever a series has negative terms. It starts with the cheapest check, the limit of the terms, because a nonzero limit ends the problem immediately.
The absolute-value question comes next because it is the stronger one: a yes gives absolute convergence and settles both convergence and the independence from order. Only if the absolute values diverge do you reach for the alternating series test, which can upgrade the verdict to conditional convergence but never prove divergence.
The last two items are the ones most often broken. Use the next omitted size, with its sign, to estimate the sum. And never read a failed test as a verdict.
Ranking
Put in order
Order the steps of a complete classification of a series with negative terms.
Why: The cheapest test first, then the strongest question (absolute convergence), then the alternating test, which can only upgrade divergent absolute values to conditional convergence. The error estimate comes last, and only if the alternating test's hypotheses hold.
Drag the steps into order before checking. The cheap divergence check goes first, then the strongest question, absolute convergence, then the alternating test as a fallback.
The error estimate comes last because it is only available once you know the series satisfies the alternating test's hypotheses. Estimating a remainder before you know there is a sum is exactly the error you corrected earlier in the lesson.
Check
Check your understanding
An alternating series has sizes bₙ that tend to 0 but do not decrease. What does the alternating series test conclude?
Answer: C
Why: The alternating series test needs both conditions, and its only verdict is convergence. With a condition failing it says nothing; the series might converge (the trap example did) or diverge (Exercise 298 did).
The question is about what the test can conclude, not about what this particular series does. With the decreasing condition failing, the alternating series test simply does not apply.
The lesson had examples on both sides. Exercise 298 fails this way and diverges; the trap-slide series fails the same way and converges absolutely. So no verdict can follow from the failure alone.
Check
Check your understanding
You approximate the sum of (−1)ⁿ⁺¹/√n by S₁₀₀. Which is a guaranteed bound on the error?
Answer: A
Why: The sizes 1/√n decrease to 0, so Theorem 5.14 applies: the error is at most the first size left out, b₁₀₁ = 1/√101, about 0.0995.
First confirm the hypotheses: the sizes one over root n decrease and tend to zero, so Theorem 5.14 applies even though the convergence is only conditional.
The bound is the first size left out, one over the square root of 101. The most tempting wrong answer uses the last size included instead. That happens to be a slightly larger number, so it is still a true bound, but it is not the one the theorem gives, and the habit of using it will cost you when you choose N from a tolerance.
Check
Check your understanding
Which series converges conditionally?
Answer: B
Why: The sizes 1/ln(n + 1) decrease to 0, so the alternating series converges. Without the signs, 1/ln(n + 1) is bigger than 1/(n + 1), whose series diverges, so the absolute values diverge. Convergent but not absolutely: conditional.
Run the flow diagram on each option. For conditional convergence you need two things: the series converges, and its absolute values do not.
One over ln of n plus 1 is the only option that fits. The sizes decrease to zero, so the signed series converges, while the absolute values are larger than one over n plus 1 and so diverge. Among the others, two converge absolutely by comparison with p-series, and one diverges because its terms tend to one half.
Explain it to yourself
Discussion prompt
In two or three sentences, explain why the two conditions of the alternating series test make the partial sums converge, and why the same picture hands you the error bound.
Write your answer before revealing it. If you can explain this in your own words, you can rebuild the test and the error bound without memorising either.
The key is to connect each condition to a feature of the picture. Alternating signs make the steps reverse; decreasing sizes make each step no longer than the last, which is what traps the sum between consecutive partial sums; sizes tending to zero make the trap close. The error bound is then just the width of the trap.
Exit ticket
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt n} \]
Discussion prompt
Classify this series as absolutely convergent, conditionally convergent or divergent, and find how many terms guarantee an error below 0.01.
This combines the whole lesson. Classify the series, name the tools, then use the remainder estimate to choose N.
Without the signs it is a p-series with p equal to one half, which diverges, so the convergence is not absolute. With the signs, the sizes one over root n decrease to zero, so it converges, conditionally.
For the estimate, the next size is one over the root of N plus 1, and you need it below a hundredth, so N plus 1 must exceed ten thousand. Ten thousand terms for two decimal places: conditional convergence is typically slow, which is one more reason to prefer series that converge absolutely.
Recap
| tool | what it needs | what it concludes |
|---|---|---|
| alternating series test | signs alternate; sizes decrease; sizes → 0 | converges (and nothing if a condition fails) |
| remainder estimate | the test's hypotheses | |R_N| ≤ b(N+1), with the sign of that term |
| absolute convergence | Σ |aₙ| converges | Σ aₙ converges, in any order, to the same sum |
| conditional convergence | Σ aₙ converges, Σ |aₙ| diverges | reordering can give any sum, or none |
\[ |S - S_N| \le b_{N+1} \]
Next, Section 5.6 brings the ratio and root tests, which are built to test absolute convergence directly and are the workhorses for the power series of Chapter 6.
Stewart, Calculus: Early Transcendentals 8e, §11.5 Alternating Series §11.5, pp. 732-736 — the same material in Stewart
One test, one estimate, one distinction. The alternating series test needs signs that alternate and sizes that decrease to zero, and it can only ever conclude convergence. The same zig-zag gives the remainder estimate: the error is at most the first size left out, and its sign is the sign of that term.
Absolute convergence is the stronger property. It implies convergence, and it guarantees that the sum does not depend on the order of the terms. Conditional convergence is fragile: the signs are doing essential work, and reordering can produce any sum at all.
Section 5.6 introduces the ratio and root tests. They test absolute convergence directly, which is why they will become the main tools for the power series of Chapter 6.
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