The divergence test as the contrapositive of the necessary condition, the integral test comparing a series against an improper integral, the p-series family and the location of its boundary, and estimating a sum by bounding its remainder.
Subject: Calculus II · 62 slides · symbolic lesson
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Title
Calculus II · Section 5.3
Deciding convergence without ever finding the sum
Objectives
Section 5.2 could only decide convergence when it could compute the sum: geometric series and telescoping series. Most series are neither. This lesson builds the first two tools that give a verdict without a value.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, pp. 471-484 — learning objectives 5.3.1 to 5.3.3
Up to now, deciding whether a series converges has meant finding its sum. That worked for geometric series, where there is a formula, and for telescoping series, where almost everything cancels. It fails for nearly everything else, including series as innocent-looking as the sum of one over n squared.
This lesson changes the question. Instead of asking what the sum is, you ask only whether there is one. That turns out to be a much easier question, and the two tools here answer it for a large family of series. The first is a quick screen that can only ever say diverges. The second trades the series for an integral, which you already know how to evaluate from Chapter 3.
The last part brings the value back: once you know a series converges, you will be able to say how close a partial sum is to the true total, and how many terms you need for a given accuracy.
Warm-up
Discussion prompt
A series is an infinite sum. What precisely has to happen for it to converge? Write the definition in terms of the partial sums, without looking back at Section 5.2.
Commit to a written answer before revealing it. If you wrote something like the terms get small, that is the single most common misconception this lesson exists to correct, so it is worth catching now.
A series is really a sequence in disguise: the sequence of partial sums, each one the total of the first k terms. Convergence of the series means convergence of that sequence to a finite limit. The terms themselves are only the step sizes between one partial sum and the next.
Keep that picture of a running total in mind throughout. Every test in this chapter is a clever way of learning something about the running total's limit without ever writing a formula for the running total.
Section
Part 1
Concept
Figure (svg): Dots showing the partial sums of one over n squared for n from 1 to 30, rising quickly then flattening just under a dashed line at 1.645.
\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac14 + \frac19 + \frac1{16} + \cdots \]
Nothing telescopes here, and there is no common ratio. The partial sums visibly flatten, but a picture is not a proof, and there is no formula for the partial sums to take a limit of.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 471 — introduction
Look at the dots. They rise quickly at first, because the early terms are large, and then flatten out as the terms shrink. The eye says there is a ceiling somewhere around 1.64. The eye is right, but you cannot prove a limit exists by looking at thirty points.
Why not just find a formula for the partial sums, as with a geometric series? Because there is not one. No expression in k gives the sum of the first k reciprocal squares in closed form. The exact value of the infinite sum, pi squared over six, was a famous open problem until Euler solved it in 1734, and his method is far beyond this course.
Here is the good news. Proving that the sum exists takes one improper integral, and you will do it in about ten minutes. Knowing it exists is what licenses you to approximate it, which is how every calculator actually computes such numbers.
Intuition
Converges or diverges is a yes-or-no question, and it can often be answered by comparing the series with something simpler whose behaviour you already know.
Once you know a series converges, you can add up enough terms to approximate its sum as closely as you like, and the last part of this lesson shows how to know when you have enough.
If it diverges, no amount of adding will ever give a number, and knowing that early saves you from computing garbage.
It can feel unsatisfying to learn that a series converges without learning to what. But think about how you use the answer. An engineer who needs a sum to four decimal places does not need a closed form; they need to know that adding more terms will home in on something, and they need a bound on how far off they are.
Divergence is just as useful to know. A divergent series has no sum, so any number you compute from its partial sums means nothing. Deciding divergence early stops you from trusting a calculation that is quietly meaningless.
So the plan for the rest of Chapter 5 is: build a toolkit of tests, each with hypotheses you must check and a conclusion it is entitled to draw. This lesson supplies the first two.
Section
Part 2
Concept
Each term is the difference between two consecutive partial sums. That one observation is the whole proof.
\[ a_k = S_k - S_{k-1} \]
\[ \lim_{k\to\infty} a_k = \lim_{k\to\infty} S_k - \lim_{k\to\infty} S_{k-1} \]
\[ \lim_{k\to\infty} a_k = S - S = 0 \]
Both partial-sum limits equal the same number S when the series converges, because the second sequence is just the first one shifted by one place.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 471 — derivation before Theorem 5.8
This is a short argument worth being able to reproduce from memory. The kth term is exactly what you add to get from the partial sum before it to the partial sum after it, so it equals the difference of two consecutive partial sums.
Now suppose the series converges to some number S. Then the partial sums approach S, and so do the partial sums shifted by one place, because a shifted sequence has the same tail. The limit of a difference is the difference of the limits, so the terms approach S minus S, which is zero.
Notice what the argument used: only the existence of the limit S. It says nothing about how fast the terms shrink, and that gap is exactly where the harmonic series will slip through.
Concept
\[ \sum a_n \text{ converges} \;\Longrightarrow\; a_n \to 0 \]
The contrapositive of a true statement is true. Negate both sides and swap them:
\[ a_n \not\to 0 \;\Longrightarrow\; \sum a_n \text{ diverges} \]
Divergence test (Theorem 5.8) — If the limit of the terms is a nonzero number, or does not exist, then the series diverges.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 471 — Theorem 5.8
The previous slide proved an implication: if the series converges, the terms go to zero. An implication and its contrapositive are logically the same statement, so you get the divergence test for free: if the terms do not go to zero, the series cannot converge.
The phrase do not go to zero covers two different situations, and Theorem 5.8 names both. The terms might settle at some other number, or they might never settle at all. Either way, zero is not their limit.
What you must not do is flip the implication the other way. Terms tending to zero is the hypothesis of nothing. It is a necessary condition for convergence, a door every convergent series has walked through, but plenty of divergent series walk through it too.
Notation
Annotate
On: \( \lim_{n\to\infty} a_n = c \ne 0 \quad\text{or}\quad \lim_{n\to\infty} a_n \text{ DNE} \;\Longrightarrow\; \sum_{n=1}^{\infty} a_n \text{ diverges} \)
Step through the annotations one at a time and connect each to the running-total picture. If the terms level off at a nonzero height c, then far out in the series you are adding roughly c at every step, so the total grows by about c each time and cannot level off.
The does-not-exist case is easy to overlook. The terms of the sum of cos n bounce around between minus one and one forever. They do not approach zero, because they do not approach anything, and so the test applies and the series diverges.
The last annotation is the one that costs marks. There is no convergence conclusion anywhere in this theorem. When the limit is zero, write that the test is inconclusive and move on to another test.
Picture it
Figure (svg): Left: the terms n over 3n minus 1 plotted as dots, levelling off at a dashed line one third. Right: the partial sums, rising along a straight line with no ceiling.
On the left, the terms of Example 5.13(a) flatten out at one third. On the right, their running total: every new term adds about a third, so the partial sums climb like a straight line and never level off.
The two panels show the same series from two points of view. On the left, each dot is one term. They start at one half and slide down toward the dashed line at one third, but they never go below it.
On the right, each dot is a running total. Because every term is at least a third, each step adds at least a third to the total, so the dots climb at least as steeply as the dashed line. A straight line with positive slope has no ceiling.
This is the whole content of the divergence test, drawn. Terms that refuse to shrink to zero force the running total to march off to infinity.
Worked example
Apply the divergence test to the series below.
\[ \sum_{n=1}^{\infty} \frac{n}{3n-1} \]
Divide the top and bottom by n
Why: The highest power of n in the denominator is n itself; dividing by it exposes the limit.
\[ \frac{n}{3n-1} = \frac{1}{3 - \frac{1}{n}} \]
Let n go to infinity
Why: The one over n in the denominator vanishes.
\[ \lim_{n\to\infty} \frac{1}{3 - \frac1n} = \frac{1}{3 - 0} = \frac13 \]
Compare with zero and conclude
Why: One third is not zero, so Theorem 5.8 applies.
\[ \frac13 \ne 0 \;\Longrightarrow\; \sum \frac{n}{3n-1} \text{ diverges} \]
Check against the partial sums
Why: Every term is at least one third, so after k terms the total is at least a third of k, which grows without bound.
\[ S_k \ge \frac{k}{3} \to \infty \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 472 — Example 5.13a
The only real work is the limit, and the standard move for a rational function is to divide numerator and denominator by the highest power of n in the denominator. Here that is just n, which turns the fraction into one over three minus one over n.
As n grows, one over n goes to zero and the limit is one third. That single number settles the question: one third is not zero, so the series diverges. Notice how little you had to do, and that you never touched a partial sum.
The last step is a habit worth building: check the verdict against the definition when it is cheap to do so. Every term here is bigger than one third, so after k terms the total is more than k over three. That grows without bound, which is what diverges means, so the test and the definition agree.
Worked example
Apply the divergence test to this series.
\[ \sum_{n=1}^{\infty} e^{1/n^2} \]
Find the limit of the exponent
Why: Squaring n makes the denominator grow, so the fraction shrinks.
\[ \frac{1}{n^2} \to 0 \]
Pass the limit through the exponential
Why: The exponential function is continuous, so the limit may move inside it.
\[ e^{1/n^2} \to e^{0} = 1 \]
Conclude
Why: The terms approach one, not zero.
\[ 1 \ne 0 \;\Longrightarrow\; \sum e^{1/n^2} \text{ diverges} \]
Check with a single term
Why: Every term is bigger than one, since the exponent is positive, so the partial sums exceed k.
\[ e^{1/n^2} > 1 \;\Longrightarrow\; S_k > k \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 472 — Example 5.13c
This one tempts people to stop too early. You see a fraction with n squared in the denominator, the fraction goes to zero, and it is easy to write zero and declare the test inconclusive. But the fraction is an exponent. What goes to zero is the exponent, so the term goes to e to the zero, which is one.
The step that passes the limit inside the exponential relies on continuity of the exponential function. You will use this move constantly: whenever a continuous function wraps a sequence, the limit can go inside.
The check is almost free. A positive exponent makes the exponential bigger than one, so every term exceeds one and the first k terms add up to more than k. Divergence confirmed without any theorem at all.
Worked example
Apply the divergence test to the reciprocal cubes.
\[ \sum_{n=1}^{\infty} \frac{1}{n^3} \]
Take the limit of the terms
Why: The denominator grows without bound.
\[ \lim_{n\to\infty} \frac{1}{n^3} = 0 \]
Read the theorem's hypothesis
Why: Theorem 5.8 needs a nonzero limit or no limit. Neither holds, so the theorem says nothing.
\[ a_n \to 0 \;\Longrightarrow\; \text{inconclusive} \]
Figure (svg): Two sequences of partial sums on the same axes: the harmonic partial sums keep climbing past 4, while the partial sums of one over n squared flatten below 1.7.
Check why nothing could be said
Why: The harmonic series also has terms tending to zero, and it diverges; reciprocal squares have terms tending to zero, and they converge. Same test result, opposite answers.
\[ \frac1n \to 0 \text{ and diverges}, \qquad \frac{1}{n^2} \to 0 \text{ and converges} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 472 — Example 5.13b
The terms of the reciprocal cubes do go to zero, so the hypothesis of Theorem 5.8 fails and the theorem simply does not speak. The correct write-up is one line: the terms tend to zero, so the divergence test is inconclusive.
The picture shows why nothing more could possibly be said. Both sequences of partial sums come from series whose terms tend to zero. One of them, the harmonic series, keeps climbing forever, tracking a logarithm. The other, the reciprocal squares, levels off. A test that looks only at the limit of the terms sees the same thing in both cases, so it cannot tell them apart.
The reciprocal cubes do in fact converge, as you will prove with the integral test in a few slides. For now, the lesson is that the divergence test is a screen, not a verdict machine.
Trap
A classic wrong line on a test:
\[ \lim_{n\to\infty} \frac1n = 0 \]
\[ \Longrightarrow\; \sum \frac1n \text{ converges?} \]
Wrong. The second line does not follow from the first.
The divergence test has one verdict, diverges. A limit of zero means the test is out of the game, not that the series won. The harmonic series is the permanent counterexample.
\[ a_n \to 0 \text{: necessary, not sufficient} \]
This mistake is so common that it is worth saying out loud why it is tempting. The phrase the terms go to zero sounds like the series settles down, and settling down sounds like converging. But the terms settling is not the total settling.
Keep the harmonic series as your permanent counterexample. Its terms go to zero, its partial sums pass ten after about twelve thousand terms and one hundred only after about ten to the forty-third terms, and they never stop growing. Whenever you are tempted to conclude convergence from shrinking terms, test your reasoning on the harmonic series and watch it fail.
Prediction
\[ \sum_{n=1}^{\infty} \cos\left(\frac{1}{n^2}\right) \]
Predict first
What does the divergence test say about this series?
Correct: It diverges
Why: One over n squared tends to zero, and cosine is continuous, so the terms tend to cos 0, which is 1. A nonzero limit means the series diverges. It is easy to see the fraction shrinking and stop there, but the fraction is inside a cosine.
Commit to one of the options before revealing. This checkpoint is built to catch exactly the error from Example 5.13(c): seeing a fraction shrink and forgetting what the fraction is inside.
The inner fraction goes to zero, cosine is continuous, and cosine of zero is one. So the terms approach one, the test applies, and the series diverges. The running total grows by nearly one at every step.
If you picked inconclusive, find the exact moment in your reasoning where you took the limit of the inside and reported it as the limit of the whole term. That is the slip to watch for from now on.
Sorting
Sort into buckets
For each series, does the divergence test settle it, or is it silent?
For each item, find the limit of the terms first, then decide which bucket it belongs in. You are sorting by what the test can say, not by what the series actually does.
Three of these have terms that do not approach zero. Two of them approach one: n over n plus 2, and e to the minus two over n, where the exponent shrinks to zero. The third, cos n, has no limit at all. All three diverge by Theorem 5.8.
The other three all have terms that shrink to zero, so the test is silent about them. As it happens, the one over root n series and the ln n over n series both diverge, and the last one converges. None of that can be learned from this test, which is the point of the next part of the lesson.
Counterexample
Discussion prompt
Someone claims: if the terms tend to zero, the series converges. Give a counterexample, and say in one sentence what the counterexample proves about the divergence test.
This is a good habit for any theorem: ask whether the reverse is true, and if not, find the example that breaks it. For the divergence test the reverse would be a convergence test, and the harmonic series destroys it.
The second half of the question matters more. A counterexample does not just refute one claim; it tells you something structural. Here it tells you that the limit of the terms, on its own, does not carry enough information to decide convergence. Any test that could certify convergence must look at more than that one number, for instance at how fast the terms shrink. The integral test does exactly that.
Section
Part 3
Concept
Figure (svg): Rectangles of width one and heights 1, 1/2, 1/3 and so on, each drawn from n to n plus 1, sitting above the curve y equals one over x; the curve's area from 1 to infinity is infinite.
Draw each term as a rectangle of width one: the nth rectangle runs from n to n plus 1 with height one over n. Its area is exactly the term.
\[ S_k = 1 + \frac12 + \cdots + \frac1k > \int_1^{k+1} \frac{1}{x}\,dx \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, pp. 472-473 — Figure 5.12
This picture is the key idea of the whole section, so take a moment with it. Each term of the harmonic series becomes a rectangle one unit wide, so its area equals the term. The first rectangle has area one, the second one half, and so on.
The heights are taken at the left end of each unit interval, where the curve one over x is highest. So every rectangle sticks up above the curve, and the total area of the rectangles is more than the area under the curve over the same stretch.
The animation adds the rectangles one at a time, the way the partial sums add terms. Each partial sum is the area of the rectangles so far, and it always exceeds the area under the curve from 1 to one step past the last rectangle.
Worked example
Turn the picture into a proof that the harmonic series diverges.
Compare the rectangles with the region
Why: The rectangles cover the region under one over x from 1 to k plus 1, with room to spare.
\[ S_k > \int_1^{k+1} \frac{1}{x}\,dx \]
Evaluate the integral
Why: The antiderivative of one over x is the natural logarithm.
\[ \int_1^{k+1} \frac1x\,dx = \ln x \Big|_1^{k+1} = \ln(k+1) - \ln 1 = \ln(k+1) \]
Let k grow
Why: The logarithm is unbounded, and the partial sums sit above it.
\[ S_k > \ln(k+1) \to \infty \]
Conclude
Why: Unbounded partial sums cannot converge.
\[ \sum_{n=1}^{\infty}\frac1n \text{ diverges} \]
Check with numbers
Why: At k equal to 10 the harmonic sum is 2.929 and the logarithm of 11 is 2.398; at 100 they are 5.187 and 4.615. The inequality holds, and both keep growing.
\[ S_{10} = 2.929 > 2.398 = \ln 11, \qquad S_{100} = 5.187 > 4.615 = \ln 101 \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 473 — the harmonic series revisited
Now the picture becomes a proof in three lines. The rectangles out-area the region, the region's area is a natural logarithm, and the logarithm grows without bound. Something bigger than an unbounded quantity is unbounded too, and an unbounded sequence of partial sums cannot converge.
Look at the check with numbers. At ten terms the harmonic sum is about 2.93 and the logarithm of 11 is about 2.40; at one hundred terms, 5.19 against 4.62. The gap between them stays roughly constant, a little over half, which is a hint about something you will meet on the real-world slide later.
This is a different proof from the grouping argument in Section 5.2, and it is more useful because it generalises: replace one over x by any positive decreasing function and the same three lines go through.
Picture it
Figure (svg): Rectangles of heights 1/4, 1/9, 1/16 and so on, drawn from n minus 1 to n, tucked under the curve y equals one over x squared from x equals 1 onward; the first term, a square of height 1, is drawn separately on the left.
This time draw the nth rectangle from n minus 1 to n. Its height, one over n squared, is the curve's value at the RIGHT end, the lowest point on that stretch, so every rectangle from the second on fits under the curve.
To prove convergence you need the opposite inequality: the partial sums must be smaller than something finite. So this time the rectangles have to fit under the curve.
The trick is to shift each rectangle one unit to the left. The rectangle for the nth term now runs from n minus 1 to n, and its height, one over n squared, is the curve's value at the right-hand end, which is the lowest point of the curve on that interval. So from the second rectangle onwards, every rectangle sits entirely under the curve.
The first term, the square of height one, has nowhere to go: it would have to sit under the curve between zero and one, where the curve blows up. So it is simply counted separately. That extra one is why the bound will be two rather than one.
Worked example
Turn the second picture into a convergence proof.
Separate the first term, then compare
Why: Terms two through k fit under the curve from 1 to k.
\[ S_k = 1 + \sum_{n=2}^{k}\frac{1}{n^2} < 1 + \int_1^{k}\frac{1}{x^2}\,dx \]
Evaluate the integral
Why: Power rule with exponent minus two.
\[ \int_1^{k}x^{-2}\,dx = \left[-\frac1x\right]_1^{k} = 1 - \frac1k \]
Combine
Why: The partial sums are trapped below two for every k.
\[ S_k < 1 + 1 - \frac1k = 2 - \frac1k < 2 \]
Invoke the Monotone Convergence Theorem
Why: Every term is positive, so the partial sums only increase; increasing and bounded above means convergent.
\[ S_{k} = S_{k-1} + \frac{1}{k^2} > S_{k-1}, \quad S_k < 2 \;\Longrightarrow\; S_k \to S \le 2 \]
Check with numbers
Why: The partial sums at 10 and 100 terms are 1.550 and 1.635, both under two, and the true sum is 1.645.
\[ S_{10} = 1.550 < 1.9, \qquad S_{100} = 1.635 < 1.99 \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, pp. 473-474 — Figure 5.13
Follow the logic in order. The rectangles from the second on fit under the curve from 1 to k, so their total is less than that integral. The integral works out to one minus one over k. Add the separate first term and the partial sums are always less than two.
Being bounded is not enough on its own: a sequence can stay below two and still oscillate forever. What finishes the proof is that every term is positive, so each partial sum is bigger than the one before. An increasing sequence that is bounded above must converge, and that is the Monotone Convergence Theorem from Section 5.1.
The numbers confirm it: 1.550 at ten terms, 1.635 at one hundred, creeping toward 1.645 and never near two. The bound of two is crude, and the last part of this lesson will show how to trap the sum much more tightly.
Concept
Integral test (Theorem 5.9) — Suppose the terms are positive and a function f is continuous and decreasing with f(n) equal to the nth term for every n from some N onward. Then the series and the improper integral of f from N to infinity either both converge or both diverge.
\[ a_n = f(n) \text{ for } n \ge N \quad\Longrightarrow\quad \sum_{n=1}^{\infty} a_n \text{ and } \int_N^{\infty} f(x)\,dx \text{ share a fate} \]
| hypothesis | what goes wrong without it |
|---|---|
| terms positive | negative terms can cancel; the rectangles' areas no longer add up to the partial sums |
| f decreasing | a bumpy f can dodge between the rectangles, so neither inequality holds |
| f continuous | the improper integral may not even be defined |
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 475 — Theorem 5.9 and Figure 5.14
Everything in the last four slides is packaged into Theorem 5.9. The two rectangle arguments go through for any function that is positive, continuous and decreasing and that hits the terms at the whole numbers. If the integral is finite, the rectangles-below argument bounds the partial sums. If the integral is infinite, the rectangles-above argument makes them unbounded.
The table is there so you know why each hypothesis is present, which makes them much easier to remember. Positive terms make the partial sums increase, which the Monotone Convergence Theorem needs. Decreasing is what guarantees the rectangles sit on the correct side of the curve. Continuity makes the integral meaningful.
The hypotheses only need to hold from some N onwards. A few awkward early terms change the value of the sum, but never whether it exists.
Notation
Annotate
On: \( a_n = f(n),\; n \ge N \quad\Longrightarrow\quad \sum_{n=1}^{\infty} a_n \;\sim\; \int_N^{\infty} f(x)\,dx \)
The first annotation is the practical one: to find f, you usually just replace n by x in the formula for the terms. The theorem needs a function of a real variable because only such a function can be integrated.
The fourth annotation is the one to take seriously. The squiggle means the two objects share a fate, both finite or both infinite. It does not mean they are equal. The next slide shows a case where both are known exactly and they come out different.
Starting the integral at N rather than at 1 is not a technicality. Series such as one over n ln n start at n equal to 2, and some functions only become decreasing after a few terms. The theorem lets you begin wherever the hypotheses begin to hold.
Worked example
Compare a series with its own integral when both can be computed exactly.
\[ \sum_{n=1}^{\infty} \left(\frac1e\right)^n \quad\text{vs}\quad \int_1^{\infty} e^{-x}\,dx \]
Sum the geometric series
Why: First term one over e, ratio one over e, which is less than one.
\[ \sum_{n=1}^{\infty} \left(\frac1e\right)^n = \frac{1/e}{1 - 1/e} = \frac{1}{e-1} \approx 0.582 \]
Evaluate the improper integral
Why: The antiderivative of e to the minus x is minus e to the minus x.
\[ \int_1^{\infty} e^{-x}\,dx = \lim_{b\to\infty}\left[-e^{-x}\right]_1^b = \lim_{b\to\infty}\left(e^{-1} - e^{-b}\right) = \frac1e \approx 0.368 \]
Figure (svg): Rectangles of heights one over e, one over e squared, and so on, drawn from n minus 1 to n, beside the curve y equals e to the minus x shaded from 1 to infinity; the rectangles' total 0.582 is not the shaded area 0.368.
Check what the test promised
Why: Both are finite, exactly as Theorem 5.9 predicts, and they are different numbers. The test transmits the verdict, not the value.
\[ 0.582 \ne 0.368, \quad \text{both finite} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 475 — the geometric example after Theorem 5.9
This example is chosen because both sides can be computed exactly. The series is geometric with first term and ratio both equal to one over e, so the Section 5.2 formula gives one over e minus 1, about 0.582. The integral of e to the minus x from 1 to infinity is one over e, about 0.368.
Both are finite, exactly as the integral test says they must be together. But they are different numbers, and the picture shows why: the rectangles and the region under the curve are different shapes. The rectangles here were drawn to the left, under the curve's left-hand part, so they even cover area the integral from 1 does not count.
Carry this away: the integral test is a messenger that carries one bit of information, finite or infinite. It never carries the value.
Worked example
\[ \sum_{n=1}^{\infty} \frac{1}{n^3} \]
Choose f and check the hypotheses
Why: For x at least 1, one over x cubed is positive, continuous and decreasing.
\[ f(x) = x^{-3}, \quad f'(x) = -3x^{-4} < 0 \]
Write the improper integral as a limit
Why: Replace infinity by b and let b grow.
\[ \int_1^{\infty} x^{-3}\,dx = \lim_{b\to\infty}\int_1^{b} x^{-3}\,dx \]
Integrate
Why: Power rule: raise the exponent to minus two and divide by minus two.
\[ \int_1^{b} x^{-3}\,dx = \left[-\frac{1}{2x^2}\right]_1^{b} = -\frac{1}{2b^2} + \frac12 \]
Take the limit
Why: The first term vanishes.
\[ \lim_{b\to\infty}\left(\frac12 - \frac{1}{2b^2}\right) = \frac12 \]
Figure (svg): The curve y equals one over x cubed, shaded from x equals 1 to the right edge, with the label: total area one half.
Check and conclude
Why: The integral is finite, so the series converges; its first term alone is 1, larger than one half, which confirms again that the integral is not the sum.
\[ \int_1^{\infty} x^{-3}\,dx = \frac12 \;\Longrightarrow\; \sum \frac{1}{n^3} \text{ converges} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 476 — Example 5.14a
This is the model integral-test solution, and every line in it earns its place. First name f and check the hypotheses; the derivative settles decreasing in one line. Then write the improper integral as a limit, because infinity is not a number you can substitute.
The integration itself is the power rule: add one to the exponent, making it minus two, and divide by minus two. Evaluating from 1 to b and letting b grow leaves exactly one half.
The check at the end is worth copying. The integral came out as one half, but the first term of the series is already one, so the series sums to more than one half. If you ever find yourself reporting the integral as the sum, a quick look at the first term will usually expose it.
Worked example
\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{2n-1}} \]
Choose f and check it decreases
Why: The denominator increases for x at least 1, so the fraction decreases; it is positive and continuous there.
\[ f(x) = (2x-1)^{-1/2}, \quad f'(x) = -(2x-1)^{-3/2} < 0 \]
Find an antiderivative
Why: Substitute u equal to 2x minus 1, so du is 2 dx.
\[ \int (2x-1)^{-1/2}\,dx = \frac12\int u^{-1/2}\,du = u^{1/2} = \sqrt{2x-1} \]
Evaluate from 1 to b
Why: At the lower limit the root is one.
\[ \int_1^{b}(2x-1)^{-1/2}\,dx = \sqrt{2b-1} - 1 \]
Take the limit
Why: The root grows without bound.
\[ \lim_{b\to\infty}\left(\sqrt{2b-1} - 1\right) = \infty \]
Figure (svg): Two curves: y equals one over the square root of 2x minus 1, falling slowly, and the accumulated area square root of 2b minus 1 minus 1, climbing past 3 with no ceiling.
Check and conclude
Why: An infinite integral means the series diverges, even though its terms shrink to zero; the divergence test could not have said this.
\[ \int_1^{\infty}\frac{dx}{\sqrt{2x-1}} = \infty \;\Longrightarrow\; \sum \frac{1}{\sqrt{2n-1}} \text{ diverges} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 476 — Example 5.14b
The terms here shrink to zero, so the divergence test is silent, and the series looks as if it might converge. The integral settles it.
The antiderivative comes from a substitution: set u equal to 2x minus 1, so du equals 2 dx, and the integral becomes one half of the integral of u to the minus one half, which is u to the one half. Undo the substitution to get the square root of 2x minus 1.
Evaluated from 1 to b this is the root of 2b minus 1, minus 1, and that grows without bound. The picture makes the point visually: the heights shrink, but only like one over a square root, which is too slow, and the accumulated area climbs forever. Terms that shrink is not enough; they must shrink fast enough.
Step zero
\[ \sum_{n=1}^{\infty} \frac{n}{3n^2+1} \]
Discussion prompt
Checkpoint 5.13 asks for the integral test here. Before you integrate, what three things must you check about f(x) = x/(3x² + 1), and which one needs actual work?
Write your answer before revealing. The instinct is to jump straight to integrating, and this slide is here to slow that instinct down by about thirty seconds.
Positivity and continuity are free here: for x at least one, both the top and the bottom are positive, and the bottom is never zero. Decreasing is the hypothesis that needs work, because the numerator and the denominator both grow, and it is not obvious which wins.
The quotient rule answers it. The derivative's numerator is one minus three x squared, which is negative for every x from one onward. So f is decreasing and the integral test may be used. On an exam this is the line that separates a complete answer from a lucky one.
Worked example
\[ \sum_{n=1}^{\infty} \frac{n}{3n^2+1} \]
Record the hypotheses
Why: Positive, continuous, and decreasing for x at least 1 by the derivative on the previous slide.
\[ f(x) = \frac{x}{3x^2+1}, \quad f'(x) = \frac{1-3x^2}{(3x^2+1)^2} < 0 \]
Substitute
Why: The numerator is one sixth of the derivative of the denominator, so let u equal 3x squared plus 1.
\[ u = 3x^2 + 1, \quad du = 6x\,dx, \quad x\,dx = \frac{du}{6} \]
Integrate
Why: The integral of one over u is a logarithm.
\[ \int \frac{x}{3x^2+1}\,dx = \frac16\int\frac{du}{u} = \frac16\ln(3x^2+1) \]
Evaluate and take the limit
Why: At x equal to 1 the argument is 4.
\[ \int_1^{b}\frac{x}{3x^2+1}\,dx = \frac16\left[\ln(3b^2+1) - \ln 4\right] \to \infty \]
Check and conclude
Why: For large n the terms behave like one third of one over n, a multiple of the harmonic series, so divergence is exactly what should happen.
\[ \frac{n}{3n^2+1} \approx \frac{1}{3n} \quad\Longrightarrow\quad \sum \frac{n}{3n^2+1} \text{ diverges} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 476 — Checkpoint 5.13
With the hypotheses already checked, the integral is a standard substitution. The numerator x is one sixth of the derivative of the denominator, so let u be the denominator, and the integral becomes one sixth of the integral of du over u, a logarithm.
A logarithm grows without bound, however slowly, so the integral is infinite and the series diverges.
The check shows you a shortcut you will formalise in the next section. For large n, the plus one in the denominator is negligible, so the terms behave like n over three n squared, which is one third of one over n. A constant multiple of the harmonic series diverges, so of course this one does. Section 5.4 turns that kind of reasoning into a theorem, and it is often much faster than integrating.
Worked example
Exercise 160 in the section's exercise set. The sum starts at 2 because ln 1 is zero.
\[ \sum_{n=2}^{\infty} \frac{1}{n\ln n} \]
Choose f and check it
Why: For x at least 2, both x and ln x are positive and increasing, so their product increases and its reciprocal decreases.
\[ f(x) = \frac{1}{x\ln x}, \quad x \ge 2 \]
Substitute u equal to ln x
Why: Then du is dx over x, which is exactly what is left.
\[ \int\frac{dx}{x\ln x} = \int\frac{du}{u} = \ln|u| = \ln(\ln x) \]
Evaluate from 2 to b
Why: Start the integral at N equal to 2, where the hypotheses begin to hold.
\[ \int_2^{b}\frac{dx}{x\ln x} = \ln(\ln b) - \ln(\ln 2) \]
Take the limit
Why: A logarithm of a logarithm is still unbounded.
\[ \lim_{b\to\infty}\ln(\ln b) = \infty \;\Longrightarrow\; \text{diverges} \]
Figure (svg): The accumulated area under one over x ln x from 2 to b, which equals ln ln b minus ln ln 2, plotted for b from 2 to 1000 on a log scale: it creeps up to about 2.3 and is still rising.
Check the scale of it
Why: The integral first reaches 3 around b equal to 530 million: this series diverges, but more slowly than any computer could ever show.
\[ \ln(\ln b) = 3 \iff b = e^{e^3} \approx 5.3\times 10^{8} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 482 — Exercise 160
This series is famous because it sits right on the edge. It diverges, but more slowly than anything you have seen. It starts at n equal to 2, because at n equal to 1 the logarithm is zero and the term would be undefined.
The substitution u equal to ln x is exactly suited to it: du is dx over x, which is the rest of the integrand, leaving the integral of one over u. The result is the log of the log of x, and a log of a log is still unbounded, so the series diverges.
The graph's horizontal axis is logarithmic. Multiply b by a thousand and the area goes up by only about two. The accumulated area does not reach three until b is about 530 million. No computer adding terms could ever convince you this series diverges; only the integral can.
Error analysis
Annotate
On: \( \int_1^{\infty}\frac{\cos x}{x^2}\,dx < \infty \;\Longrightarrow\; \sum_{n=1}^{\infty}\frac{\cos n}{n^2} \text{ converges} \)
Read the argument first and see whether you can spot the problem before revealing the annotations. The integral written down is in fact finite, so the calculation is not where the mistake lives.
The error is that the integral test was applied to a series it does not cover. Cosine is negative at many whole numbers, so the terms are not all positive, and the function oscillates rather than decreasing. The rectangle arguments behind the theorem both collapse.
The uncomfortable part is that the conclusion is true. This series does converge, by tools from the next two sections. But a true conclusion reached by an invalid argument is still an invalid argument, and on an exam it earns nothing. Always check the hypotheses first, even when you think you know the answer.
Trap
Having just shown the integral equals one half:
\[ \sum_{n=1}^{\infty}\frac{1}{n^3} = \int_1^{\infty}\frac{dx}{x^3} = \frac12 \]
Wrong. Equal fates, not equal values.
The very first term of the series is 1, already bigger than one half. The integral test gives converges, and nothing more. The actual sum is 1.2020569…, which the last part of this lesson will pin down.
\[ \sum_{n=1}^{\infty}\frac{1}{n^3} \approx 1.202 \ne \frac12 \]
This is the most common misuse of the integral test, and the quickest defence is to look at the first term. The series of reciprocal cubes starts with one, so its sum is more than one. It cannot possibly equal one half.
The integral and the series are different areas: one is the region under a smooth curve, the other a staircase of rectangles. They are linked tightly enough that one is finite exactly when the other is, and no more tightly than that. The true sum here, 1.2020569, is known as Apery's constant, and there is no known simple formula for it. The next part of the lesson shows how to trap it between two numbers anyway.
Tweak it
Parameter explorer
The curve shows the area under 1/xᵖ from 1 to x, as x grows. Slide p. For which p does the area level off, and for which does it climb forever?
\[ \int_1^{x} t^{-{p}}\,dt \]
Start with p at two and move it slowly toward one. For p bigger than one, the curve rises and then flattens toward a horizontal level, which is the finite value one over p minus one. As p approaches one that level climbs higher and higher, and the curve takes longer and longer to flatten.
At p equal to one the curve becomes the natural logarithm, which never flattens. Push p below one and the growth becomes a power of x, faster still.
Keep this image in mind for the next part. There is no gradual transition from converging to diverging: at every p above one the area is finite, and at p equal to one it is not. The finite values simply get larger without bound as you approach the boundary.
Section
Part 4
Concept
p-series — For any real number p, the series whose nth term is one over n to the power p.
\[ \sum_{n=1}^{\infty}\frac{1}{n^{p}} = 1 + \frac{1}{2^{p}} + \frac{1}{3^{p}} + \cdots \]
Figure (svg): Three curves on the same axes: y equals one over square root x, one over x, and one over x squared, from x equals 1 to 10. The first two stay high; the last collapses toward zero.
You already know two members: p equal to 1 is the harmonic series, which diverges, and p equal to 2 converges. The question is where the dividing line sits.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 477 — definition of the p-series
The p-series are the most useful reference series in the whole chapter, because in Section 5.4 you will decide the fate of messy series by comparing them with a p-series. So it is worth having their behaviour at your fingertips.
The picture shows three members side by side. The curve for p equal to one half stays high for a long time. The curve for p equal to one falls, but slowly. The curve for p equal to two collapses toward zero almost at once. Only the last encloses finite area.
The question for the next slide is exactly where between one half and two the switch happens, and which side the boundary itself belongs to.
Worked example
Settle every p at once.
Dispose of p at most zero
Why: Then the terms do not shrink: they grow when p is negative and equal 1 when p is zero.
\[ p < 0: \frac{1}{n^p} = n^{|p|} \to \infty, \qquad p = 0: \frac{1}{n^0} = 1 \]
Apply the divergence test to that case
Why: Neither limit is zero.
\[ p \le 0 \;\Longrightarrow\; \text{diverges} \]
For p positive, integrate
Why: Now one over x to the p is positive, continuous and decreasing, and for p not equal to 1 the power rule applies.
\[ \int_1^{b}x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^{b} = \frac{b^{1-p} - 1}{1-p} \]
Let b grow when p is greater than one
Why: The exponent one minus p is negative, so b to that power dies away.
\[ p > 1: \quad b^{1-p} \to 0, \quad \int_1^{\infty}x^{-p}\,dx = \frac{1}{p-1} \]
Let b grow when p is less than one
Why: The exponent is positive, so the power grows without bound.
\[ 0 < p < 1: \quad b^{1-p} \to \infty, \quad \int_1^{\infty}x^{-p}\,dx = \infty \]
Check against the two known members
Why: At p equal to 2 the integral is 1, finite, so convergence, as proved earlier; p equal to 1 was the logarithm, divergence.
\[ \sum_{n=1}^{\infty}\frac{1}{n^{p}} \;\begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \le 1 \end{cases} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 477 — equation 5.9
The derivation splits into cases, and the order matters. If p is zero or negative, the terms do not even shrink, so the cheap divergence test disposes of them immediately; there is no need to integrate at all.
For positive p, the function one over x to the p is positive, continuous and decreasing, so the integral test applies. For p not equal to one, the power rule gives b to the one minus p, minus one, all over one minus p. Everything depends on the sign of the exponent one minus p. If p is bigger than one, the exponent is negative and b to that power shrinks to zero, leaving a finite value. If p is less than one, the exponent is positive and the power grows without bound.
The case p equal to one needs the logarithm instead of the power rule, and you did it at the start: the harmonic series diverges. Put the cases together and you get a rule simple enough to state in one line.
Intuition
The harmonic series diverges with almost nothing to spare: its partial sums grow like a logarithm, the slowest-growing function you know. Any power bigger than one, even 1.001, makes the terms shrink just enough faster to tip the total over into convergence.
\[ \int_1^{\infty}\frac{dx}{x^{1.001}} = \frac{1}{0.001} = 1000 \]
Finite, but huge. Series just past the boundary converge, but painfully slowly, and that will matter when you try to estimate their sums.
It can seem strange that 1.001 converges and 1 does not, when the terms of the two series are almost identical for the first million terms. The resolution is that convergence is a statement about the infinitely long tail, not about any finite stretch.
The integral shows what just-barely converging looks like: at p equal to 1.001 the area is one thousand. Finite, but enormous, and the partial sums take an astronomically long time to approach it.
This matters practically. A series that converges because its exponent is only slightly above one will need a huge number of terms before a partial sum is a good approximation. The remainder estimate at the end of the lesson puts a number on that.
Worked example
Decide each without integrating, using the rule just derived.
\[ \text{(a) } \sum_{n=1}^{\infty}\frac{1}{n^{4}} \qquad \text{(b) } \sum_{n=1}^{\infty}\frac{1}{n^{2/3}} \]
Identify p in (a)
Why: The power in the denominator is four.
\[ p = 4 > 1 \;\Longrightarrow\; \text{(a) converges} \]
Identify p in (b)
Why: A cube root of n squared is n to the two thirds.
\[ p = \tfrac23 < 1 \;\Longrightarrow\; \text{(b) diverges} \]
Checkpoint 5.14 the same way
Why: Five quarters is bigger than one.
\[ \sum\frac{1}{n^{5/4}}: \quad p = \tfrac54 > 1 \;\Longrightarrow\; \text{converges} \]
Check (b) with the integral
Why: One minus p is one third, so the integral from 1 to b is 3 times the cube root of b, minus 3, which grows without bound.
\[ \int_1^{b}x^{-2/3}\,dx = 3b^{1/3} - 3 \to \infty \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 478 — Example 5.15 and Checkpoint 5.14
Once the rule is established, a p-series is decided by reading off the exponent. There is no integral to do and no hypothesis to check, because the rule already did that work for every p at once.
The only care needed is in identifying p when the power is written as a root. The cube root of n squared is n to the two thirds, and two thirds is less than one, so that series diverges. Five quarters is more than one, so the checkpoint series converges.
The final line re-derives part (b) from the integral as a check. Seeing the cube root of b grow without bound makes the abstract rule concrete: the integral is infinite, so the series is too.
Worked example
Exercise 165: the terms do not look like powers of n at all.
\[ \sum_{n=1}^{\infty} 2^{-\ln n} \]
Rewrite the base as an exponential
Why: Any positive number a equals e to the ln a.
\[ 2^{-\ln n} = e^{-(\ln 2)(\ln n)} \]
Regroup the exponent
Why: The same product, now read as ln n times ln 2.
\[ e^{-(\ln 2)(\ln n)} = \left(e^{\ln n}\right)^{-\ln 2} = n^{-\ln 2} \]
Read off p
Why: The natural log of 2 is about 0.693.
\[ \sum \frac{1}{n^{\ln 2}}, \quad p = \ln 2 \approx 0.693 < 1 \;\Longrightarrow\; \text{diverges} \]
Check with the sister series
Why: With base 3 instead, p becomes ln 3, about 1.099, just over the line, so that one converges.
\[ 3^{-\ln n} = n^{-\ln 3}, \quad \ln 3 \approx 1.099 > 1 \;\Longrightarrow\; \text{converges} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 482 — Exercises 165 and 166
Some series are p-series without looking like one, and the exponential rewrite is the tool that unmasks them. Any positive base can be written as e to its own logarithm, so two to the minus ln n becomes e to the minus ln 2 times ln n.
Now read that exponent the other way round, as ln n times ln 2, and it becomes n raised to the minus ln 2. So the series is a p-series with p equal to ln 2, about 0.693, which is less than one. It diverges.
The sister series with base three is a lovely contrast: ln 3 is about 1.099, just over the line, so it converges. Changing the base from two to three moves the exponent across the boundary.
Comparison
Comparison matrix
| series | p | verdict |
|---|---|---|
| sum of 1/n | 1 | diverges |
| sum of 1/√n | 1/2 | diverges |
| sum of 1/(n√n) | 3/2 | converges |
| sum of n^(−1.01) | 1.01 | converges |
| sum of n | −1 | diverges |
Fill in each blank before checking. The main skill being practised is rewriting a term as n to some power: the square root of n is n to the one half, n times root n is n to the three halves, and n on its own is one over n to the minus one.
Two rows deserve a second look. The boundary row, p equal to one, is the harmonic series and diverges. The row with p equal to 1.01 converges, even though it will do so painfully slowly. And the last row, with a negative p, is caught by the divergence test before the p-series rule is even needed: its terms grow.
Matching
Match the pairs
Why: Three different reasons to diverge. The divergence test catches n/(n+1); the p-series rule catches 1/n; and 1/(n ln n) is caught only by integrating, since it is not a p-series and its terms tend to zero.
Each series pairs with a verdict and a reason, and the reason is the point of the exercise. Three of the four diverge, but for three different reasons, and the reason tells you which tool to reach for.
The divergence test catches n over n plus one, because its terms tend to one. The p-series rule catches one over n. The series one over n ln n is caught by neither: its terms tend to zero and it is not a power of n, so only the integral test, with the log of the log, decides it. The remaining series has p equal to one and a half, comfortably above one, so it converges.
Trap
The rule gets remembered as a strict inequality that points the wrong way:
\[ \sum\frac{1}{n^{p}} \text{ converges for } p \ge 1 \]
Wrong at exactly one point: the boundary.
The boundary case p equal to 1 is the harmonic series, and it diverges. Convergence needs p strictly bigger than one. If in doubt, test the rule on the one member you are sure of.
\[ \sum\frac{1}{n^{p}} \text{ converges} \iff p > 1 \]
Under exam pressure the p-series rule gets remembered as a greater-than-or-equal, and the one case you know best gets filed on the wrong side. The protection is to always test your remembered rule on the harmonic series before using it.
The boundary is a strict inequality because at p equal to one the integral is a logarithm, which is unbounded. Everything at or below one diverges; everything strictly above one converges.
Section
Part 5
Concept
Knowing a series converges, approximate its sum by a partial sum. The error is everything you did not add.
\[ R_N = \sum_{n=1}^{\infty} a_n - S_N = a_{N+1} + a_{N+2} + a_{N+3} + \cdots \]
remainder — The tail of the series after the first N terms: the exact error made by stopping at the Nth partial sum.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 478 — the remainder R_N
Now the value comes back. When you know a series converges, the natural way to approximate its sum is to add up the first N terms. The remainder is the error you make by doing that: the entire infinite tail that you did not add.
Because every term is positive in this setting, the remainder is positive too, so the partial sum always underestimates the true sum. The question is by how much. The remainder is itself an infinite series, so you cannot compute it directly, but the same rectangle pictures that proved convergence can bound it.
Picture it
Figure (svg): Two panels with N equal to 3. Left: the tail rectangles a4, a5, … drawn from n minus 1 to n fit under the curve from x equals 3, so the tail is less than the integral from N. Right: the same rectangles drawn from n to n plus 1 stick out above the curve from x equals 4, so the tail is more than the integral from N plus 1.
Shift the tail rectangles left and they fit under the curve from N; shift them right and they cover the curve from N plus 1. The tail is caught between the two areas.
Both panels show the same tail of the reciprocal cubes after three terms: the rectangles for the fourth, fifth, sixth terms and onward. Only their horizontal position differs.
In the left panel each rectangle is shifted left so it ends where its height is measured. Every rectangle sits under the curve, so the whole tail is less than the area under the curve from N onwards. In the right panel each rectangle starts where its height is measured, so every rectangle pokes above the curve, and the tail is more than the area from N plus 1 onwards.
Two areas you can compute, one on each side of the number you want. That is the remainder estimate, and the next slide writes it down.
Notation
Annotate
On: \( \int_{N+1}^{\infty} f(x)\,dx < R_N < \int_{N}^{\infty} f(x)\,dx \)
\[ S_N + \int_{N+1}^{\infty} f(x)\,dx < \sum_{n=1}^{\infty} a_n < S_N + \int_{N}^{\infty} f(x)\,dx \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, pp. 478-479 — Theorem 5.10 and equation 5.10
The upper bound is the workhorse. The integral from N to infinity is always bigger than the error, so if you make that integral small enough, you have guaranteed the accuracy of your partial sum. That is how you choose N in practice.
The lower bound earns its keep when you add both to the partial sum. The second line of the slide traps the true sum of the series in an interval whose width is the difference of the two integrals. For well-behaved series that interval is very short, much shorter than either bound alone would suggest.
The hypotheses are exactly those of the integral test. If you could not use the integral test on a series, you cannot use this estimate either.
Worked example
Compute the tenth partial sum and bound its error.
Add the first ten terms
Why: A calculator does this; the digits matter, so keep six places.
\[ S_{10} = 1 + \frac18 + \frac1{27} + \cdots + \frac{1}{1000} \approx 1.197532 \]
Find the tail integral from a general N
Why: The same power rule as Example 5.14(a), now from N.
\[ \int_N^{\infty}\frac{dx}{x^3} = \lim_{b\to\infty}\left[-\frac{1}{2x^2}\right]_N^{b} = \frac{1}{2N^2} \]
Upper bound at N equal to 10
Why: Put N equal to 10.
\[ R_{10} < \frac{1}{2(10)^2} = 0.005 \]
Lower bound, from N plus 1
Why: Put 11 in the same formula.
\[ R_{10} > \frac{1}{2(11)^2} = \frac{1}{242} \approx 0.004132 \]
Trap the sum
Why: Add both bounds to the partial sum.
\[ 1.197532 + 0.004132 < \sum\frac{1}{n^3} < 1.197532 + 0.005 \]
Check against the known value
Why: The true sum is 1.2020569, and it sits inside the interval: the ten-term estimate was already good to about three decimal places.
\[ 1.201664 < 1.202057 < 1.202532 \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, pp. 480-481 — Example 5.16a
The partial sum is pure arithmetic, best done on a calculator, and you should keep more decimal places than you think you need, because the error bounds are in the third decimal place.
The tail integral is worth doing once for a general N, since you get a formula you can reuse: the integral from N to infinity of one over x cubed is one over 2N squared. Put in N equal to 10 for the upper bound and 11 for the lower bound.
Adding both bounds to the partial sum traps the true sum between about 1.2017 and 1.2025, an interval less than a thousandth wide, from only ten terms. The check against the known value, 1.2020569, lands inside the interval, which is exactly what the theorem promised.
Worked example
Choose N so that the Nth partial sum is within one thousandth of the true sum.
Make the guaranteed bound small enough
Why: The error is below one over 2N squared, so it is enough to force that below 0.001.
\[ R_N < \frac{1}{2N^2} \le 0.001 \]
Solve for N squared
Why: Cross-multiply; both sides are positive.
\[ 2N^2 \ge 1000 \;\Longrightarrow\; N^2 \ge 500 \]
Take the square root
Why: The root of 500 is about 22.36.
\[ N \ge \sqrt{500} \approx 22.36 \]
Round up
Why: N counts terms, so it must be a whole number, and rounding down would break the inequality.
\[ N = 23 \]
Check the answer
Why: At 23 the bound is below a thousandth; at 22 it is not.
\[ \frac{1}{2(23)^2} \approx 0.000945 < 0.001, \qquad \frac{1}{2(22)^2} \approx 0.001033 > 0.001 \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 481 — Example 5.16b
This runs the estimate backwards: instead of asking how accurate ten terms are, you ask how many terms buy a given accuracy. The idea is to make the guaranteed upper bound on the error small enough, because then the actual error is smaller still.
The algebra is an inequality, not an equation, so be careful at the end. The square root of 500 is about 22.36, and N must be a whole number. Rounding down to 22 would leave the bound slightly above a thousandth, so you must round up to 23.
The final check shows that the rounding direction really matters: at 22 terms the bound fails, at 23 it holds. Whenever a count comes out of an inequality, check the integer on each side.
Worked example
\[ \sum_{n=1}^{\infty}\frac{1}{n^4}: \quad \text{find } S_5 \text{ and bound } R_5 \]
Add five terms
Why: One, one sixteenth, one eighty-first, and so on.
\[ S_5 = 1 + 0.0625 + 0.012346 + 0.003906 + 0.0016 = 1.080352 \]
Find the tail integral
Why: Power rule with exponent minus four.
\[ \int_N^{\infty}x^{-4}\,dx = \frac{1}{3N^3} \]
Bound the error
Why: Use N equal to 5 for the upper bound and 6 for the lower.
\[ \frac{1}{3(6)^3} < R_5 < \frac{1}{3(5)^3} \;\Longrightarrow\; 0.001543 < R_5 < 0.002667 \]
Check against the exact value
Why: This sum is known exactly: pi to the fourth over 90, about 1.082323, which lands inside the trapped interval.
\[ 1.081895 < \frac{\pi^4}{90} \approx 1.082323 < 1.083019 \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 481 — Checkpoint 5.15
This is the same procedure on a faster-converging series. Five terms already give 1.0804, and the tail integral from N of one over x to the fourth is one over 3N cubed, which is tiny.
The upper bound at N equal to 5 is about 0.0027, and the lower bound, using 6, is about 0.0015. The true sum therefore lies between about 1.0819 and 1.0830.
This series also happens to have a known exact value, pi to the fourth over ninety, found by Euler along with the reciprocal squares. It is about 1.0823, comfortably inside the trapped interval. Compare how many terms it took here with the next slide, where the exponent is only two.
Estimation
\[ \sum_{n=1}^{\infty}\frac{1}{n^2}, \quad \text{error below } 10^{-4} \]
Predict first
Guess first, then check with the bound one over N: roughly how many terms guarantee four-decimal accuracy?
Correct: About 10,000
Why: The tail integral from N of one over x squared is one over N, so you need one over N at most one ten-thousandth: N equal to 10,000. Compare the reciprocal cubes, where 23 terms bought three decimals. The closer p sits to one, the slower the convergence and the more terms an estimate costs.
\[ R_N < \int_N^{\infty}\frac{dx}{x^2} = \frac1N \le 10^{-4} \iff N \ge 10{,}000 \]
Commit to a guess. Most people guess far too low, because the reciprocal squares seem to converge quickly from the first few partial sums.
The tail integral from N of one over x squared is simply one over N, so four-decimal accuracy needs N to be ten thousand. That is a stark contrast with the reciprocal cubes, where twenty-three terms gave three decimals, and the reciprocal fourth powers, where five terms nearly did.
The pattern is general: the closer the exponent is to one, the slower the convergence. The remainder estimate turns that vague idea into an exact count of terms.
Real world
Figure (svg): The harmonic numbers H sub n plotted as dots for n up to 40, tracking just above the curve ln n with a gap that settles near 0.577.
Discussion prompt
A cereal brand hides one of N different cards in each box. The expected number of boxes needed to collect all N cards is N times the harmonic number H(N). Use the picture to estimate how many boxes it takes for N = 50.
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 484 — Exercise 186, the coupon collector
The harmonic partial sums are called the harmonic numbers, and they appear throughout probability and computer science. The picture shows why the natural logarithm is the right way to estimate them: the dots run parallel to the log curve, with a gap that settles near 0.577. That constant is the Euler-Mascheroni constant.
In the card-collecting problem, each new card is harder to find than the last, because more of the cards you open are ones you already have. When you have all but one, you expect to wait N boxes for the last card alone. Adding up those waits gives N times the harmonic number.
For fifty cards the estimate is about 224 boxes. The divergence of the harmonic series is the mathematical reason collections take so long to finish: the expected waiting time grows faster than the number of cards.
Estimation
Predict first
A computer adds a million terms of the harmonic series every second. Roughly how long until the partial sum passes 100?
Correct: Far, far longer than the age of the universe
Why: The partial sums grow like ln N plus 0.577, so reaching 100 needs ln N near 99.4: N is about 1.5 times ten to the 43rd power. At a million per second that is about ten to the 37th seconds, tens of billions of billions of times the age of the universe. The series diverges, and no computer will ever see it.
\[ \ln N + 0.577 \approx 100 \;\Longrightarrow\; N \approx e^{99.4} \approx 1.5\times 10^{43} \]
OpenStax Calculus Volume 2, §5.3 The Divergence and Integral Tests §5.3, p. 484 — Exercise 192
Pick an answer before revealing. The instinct is that a computer adding a million terms a second will get anywhere quickly.
The harmonic numbers grow like the natural logarithm, so reaching one hundred needs a number of terms whose logarithm is about ninety-nine and a half. That is about one and a half times ten to the forty-third. At a million terms per second the wait is about ten to the thirty-seventh seconds, tens of billions of billions of times the age of the universe.
This is the practical meaning of logarithmic divergence. The series really does diverge, yet the partial sums you could ever compute look as though they have settled down. Only the integral test tells you the truth.
Section
Part 6
Pattern
Figure (svg): A flow diagram: first compute the limit of the terms; if it is not zero the series diverges. If it is zero, check positive, decreasing, continuous; if those hold, integrate and let the integral's verdict be the series' verdict.
This is the order to work in whenever the terms are positive. It runs from the cheapest test to the most expensive, because the cheap ones can end the problem in one line.
The divergence test comes first because it costs a single limit. If the terms tend to zero, check whether you are looking at a p-series, possibly in disguise; if so, the exponent decides it immediately. Only then set up an integral, and only after checking the three hypotheses.
The final rule is the one most often broken: the integral gives a verdict, never a value. If you need a value, bound the remainder. If any hypothesis fails, this lesson's tools do not apply, and the comparison, alternating, ratio and root tests of Sections 5.4 to 5.6 take over.
Ranking
Put in order
Order the steps of a complete integral-test argument.
Why: The cheap test goes first because it can end the problem in one line. The hypotheses must be checked before the integral is trusted, and the conclusion states a verdict, never a value.
Drag the steps into order before checking. The only order that is really up for debate is where the divergence test goes, and the answer is first, because if it succeeds you never need an integral at all.
Checking the hypotheses must come before evaluating the integral, not after. If you integrate first, it is very tempting to skip the check once you have a satisfying number.
Check
Check your understanding
For which series does the divergence test prove divergence?
Answer: B
Why: The terms (n + 1)/(2n) tend to one half, which is not zero, so the series diverges by Theorem 5.8. The other three all have terms tending to zero, so the test says nothing about them, even though two of them do in fact diverge.
The question is about what this particular test can prove, not about which series diverge. Two of the options do diverge, but only one of them diverges for the reason the test can detect.
The terms n plus one over 2n tend to one half, which is not zero, so Theorem 5.8 applies. The harmonic series and the one over root n series both diverge too, but their terms go to zero, so their divergence needs the integral test or the p-series rule.
Check
Check your understanding
Which series can the integral test NOT be applied to directly?
Answer: C
Why: Sine takes negative values at many whole numbers, so the terms are not all positive, and the corresponding f is not decreasing. The integral test's hypotheses fail, whatever the integral turns out to be.
Run through the three hypotheses for each option. The series with sine is the one that fails, because sine is negative at many whole numbers.
The other three all satisfy the hypotheses, possibly after checking a derivative or starting at N equal to 2. Starting the integral later is always allowed; the theorem only needs the hypotheses to hold eventually.
Check
Check your understanding
For the sum of 1/n², which is a guaranteed upper bound on the error after 20 terms?
Answer: A
Why: The remainder is below the integral from N of one over x squared, which is one over N. With N equal to 20 that is 0.05. The integral from N plus 1, one over 21, is the LOWER bound.
The upper bound on the error is the integral starting at N itself, which for one over x squared equals one over N. With twenty terms that is five hundredths.
The most tempting wrong answer is one over 21, which is a genuine bound but on the wrong side: it is the integral from N plus 1, and it bounds the error from below. Keep the picture in mind: the tail fits under the curve from N, so the integral from N is the one that is bigger than the error.
Explain it to yourself
Discussion prompt
In the harmonic picture the rectangles sat above the curve; in the reciprocal-squares picture they sat below it. Explain, in two sentences, why the proof of divergence needs them above and the proof of convergence needs them below.
Write your two sentences before revealing. This is the idea that makes the integral test work, and being able to explain it means you can rebuild the theorem without memorising it.
The direction of the inequality you need decides where the rectangles go. To show a sum is infinite, you need it to be bigger than something infinite, so the rectangles must contain the region under the curve. To show a sum is finite, you need it to be smaller than something finite, so the rectangles must fit inside the region. Left-endpoint heights put them above a decreasing curve; right-endpoint heights put them below.
Exit ticket
\[ \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} \]
Discussion prompt
Decide whether this converges, name the test, and give a number N that guarantees the partial sum is within 0.01 of the true sum.
This combines the whole lesson in one problem. Classify the series, name the tool, then use the remainder estimate to choose N.
The exponent is three halves, above one, so it converges as a p-series. The tail integral from N of x to the minus three halves is two over the square root of N. Making that at most a hundredth needs the square root of N to be at least two hundred, so N is forty thousand.
Forty thousand terms for two decimal places: an exponent only half a unit above one converges, but slowly. That is the practical lesson of the boundary at one.
Recap
| tool | what it needs | what it can conclude |
|---|---|---|
| divergence test | only the limit of the terms | diverges (if the limit is not 0); otherwise nothing |
| integral test | positive, continuous, decreasing f with f(n) = aₙ | converges or diverges, the same as the integral |
| p-series rule | terms 1/nᵖ | converges exactly when p > 1 |
| remainder estimate | the integral test's hypotheses | the error is between the integrals from N + 1 and from N |
\[ \int_{N+1}^{\infty} f < R_N < \int_{N}^{\infty} f \]
Next, Section 5.4 compares a new series with one whose fate you already know, so you rarely need to integrate at all.
Stewart, Calculus: Early Transcendentals 8e, §11.3 The Integral Test and Estimates of Sums §11.3, pp. 719-726 — the same material in Stewart
Two tests and one estimate. The divergence test costs one limit and can only ever prove divergence. The integral test needs three hypotheses and then gives the same verdict as an improper integral, never its value. The p-series rule, derived from the integral test, is the one fact from this lesson you should know without thinking: convergence exactly when p is strictly greater than one.
The remainder estimate brings values back: after N terms the error lies between the integrals from N plus 1 and from N. That lets you both judge an approximation and choose how many terms to add.
The integral test has one drawback: you have to be able to integrate. Section 5.4 removes that requirement by comparing a new series with one you already understand, most often a p-series, so the table on this slide becomes your reference list.
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