The direct comparison test and which of its four arrangements are informative, the limit comparison test and why a finite positive limit forces agreement, and the practical skill of reading a term's dominant behaviour to choose a comparison.
Subject: Calculus II · 63 slides · symbolic lesson
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Title
Calculus II · Section 5.4
Deciding a series by measuring it against one you already know
Objectives
The integral test decides a series only when you can integrate its terms. This lesson decides a series by comparing it, term by term, with a geometric series or a p-series whose fate you already know.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, pp. 485-495 — learning objectives 5.4.1 and 5.4.2
Section 5.3 gave you the integral test, and it is powerful, but it has a price: you must be able to integrate the terms, and the function must be decreasing. Plenty of series fail one of those. Try integrating one over two to the x plus one, or sine squared of x over x squared, and you will see the problem.
This lesson replaces integration with comparison. You already know the fate of two whole families, the geometric series and the p-series. If a new series is, term by term, smaller than a convergent one, or bigger than a divergent one, its fate is decided. When the inequality is awkward, a limit of a ratio does the same job.
The one piece of judgement the section asks of you is choosing what to compare with. By the end you will do that by reading the dominant pieces of a formula, which takes a few seconds once you have practised it.
Warm-up
Discussion prompt
Every comparison in this lesson ends by quoting the fate of a known series. Without looking back, state exactly when a p-series converges and exactly when a geometric series converges.
Write both rules down before revealing. Every argument in this lesson finishes by quoting one of them, so if either is shaky, the rest of the lesson will be shaky too.
The p-series rule came from the integral test: convergence exactly when the power is strictly bigger than one. The geometric rule came from the formula for the partial sums in Section 5.2: convergence exactly when the ratio lies strictly between minus one and one, and then the sum is the first term over one minus the ratio.
Pay attention to the boundaries. The harmonic series, with power exactly one, diverges, and a geometric series with ratio exactly one adds the same term forever. In a comparison argument you must state which side of the boundary your comparison series is on, and a wrong boundary sends the whole argument the wrong way.
Section
Part 1
Concept
\[ \sum_{n=1}^{\infty}\frac{1}{n^2+1} \quad\text{beside}\quad \sum_{n=1}^{\infty}\frac{1}{n^2} \]
Adding one to the denominator makes every fraction a little smaller, so each term sits under the matching term of a series you know converges.
\[ 0 < \frac{1}{n^2+1} < \frac{1}{n^2} \quad \text{for every } n \ge 1 \]
Figure (svg): Two rows of dots for k from 1 to 20: the partial sums of one over n squared plus one, levelling off near 1.08, sit below the partial sums of one over n squared, which level off under a dashed line at 1.645.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, pp. 485-486 — Figure 5.16(a) and Table 5.1
Here is the idea of the whole lesson in its simplest form. The series of one over n squared plus one has no telescoping and no common ratio, and integrating it gives an arctangent. But it clearly resembles the reciprocal squares, whose fate you know.
Adding one to a positive denominator makes the fraction smaller, so every term of the new series is below the matching reciprocal square. Now look at the picture. It shows partial sums, the running totals. The orange dots are the running totals of the reciprocal squares, and they never pass one point six four five. The blue dots are the running totals of the new series, and because each term is smaller, each blue dot is below the orange dot above it.
So the blue dots are trapped under a ceiling. The next slide turns that picture into a proof.
Worked example
Turn the inequality between terms into a proof about partial sums.
The partial sums increase
Why: Every term is positive, so each partial sum adds something to the last.
\[ S_{k+1} = S_k + \frac{1}{(k+1)^2+1} > S_k \]
Compare term by term
Why: Add the inequality for n from 1 to k.
\[ S_k = \sum_{n=1}^{k}\frac{1}{n^2+1} < \sum_{n=1}^{k}\frac{1}{n^2} \]
Bound by the whole convergent series
Why: A partial sum of positive terms is less than the full sum.
\[ \sum_{n=1}^{k}\frac{1}{n^2} < \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} \]
Apply the Monotone Convergence Theorem
Why: Increasing and bounded above means convergent.
\[ S_k \uparrow, \quad S_k < \frac{\pi^2}{6} \;\Longrightarrow\; \sum\frac{1}{n^2+1} \text{ converges} \]
Check with Table 5.1
Why: At k equal to 8 the book lists 0.9597 against 1.5274, and far out the sum settles near 1.0767, below the ceiling.
\[ S_8 = 0.9597 < 1.5274, \qquad S_{10^6} \approx 1.0767 < 1.6449 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 485 — the opening argument and Table 5.1
This argument uses the one theorem that proves convergence without finding a sum: the Monotone Convergence Theorem from Section 5.1. It needs two facts about the partial sums, that they increase and that they are bounded above.
Increasing is automatic, because every term is positive, so each partial sum adds something to the one before. Bounded is where the comparison enters. Add up the term-by-term inequality and the partial sum is less than the matching partial sum of the reciprocal squares, which in turn is less than the complete sum of the reciprocal squares, the number pi squared over six.
The check reads the book's Table 5.1 and then goes further than the book. After a million terms the sum has settled near 1.0767, comfortably below the ceiling of 1.6449. Notice that the ceiling is not the sum. Comparison proves the sum exists and gives a bound; it does not give the value.
Picture it
Figure (svg): Two rows of dots for k from 1 to 20: the partial sums of one over n minus one half climb from 2 to about 4.96, always above the harmonic partial sums, which climb from 1 to about 3.60.
Now subtract one half from each denominator. A smaller denominator makes a bigger fraction, so each term sits above the matching term of the harmonic series.
\[ \frac{1}{n - \frac12} > \frac{1}{n} > 0 \quad \text{for every } n \ge 1 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 486 — Figure 5.16(b) and Table 5.2
Now turn the idea around. Subtracting one half from each denominator makes each denominator smaller and each fraction bigger, so every term of this series sits above the matching term of the harmonic series.
The picture shows the two sets of running totals. The orange dots are the harmonic partial sums, which you know grow without bound, even though they grow slowly. The red dots are the new series, and each one is above the orange dot beneath it.
A quantity that is always bigger than something unbounded is itself unbounded. That is all there is to proving divergence by comparison, and the next slide writes it down.
Concept
The same argument with every inequality reversed proves divergence.
\[ S_k = \sum_{n=1}^{k}\frac{1}{n-\frac12} > \sum_{n=1}^{k}\frac{1}{n} = H_k \]
\[ H_k \to \infty \;\Longrightarrow\; S_k \to \infty \;\Longrightarrow\; \sum\frac{1}{n-\frac12} \text{ diverges} \]
Table 5.2 shows it: after eight terms, 4.0436 against 2.7179. One entry there is misprinted: the harmonic sum of four terms is 2.0833, not 2.0933.
\[ H_4 = 1 + \tfrac12 + \tfrac13 + \tfrac14 = \tfrac{25}{12} \approx 2.0833 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, pp. 486-487 — Table 5.2
The divergence argument is shorter than the convergence one, because it does not need the Monotone Convergence Theorem. You only need the partial sums of the new series to be larger than the harmonic partial sums, which follows by adding up the term-by-term inequality.
The harmonic partial sums grow without bound, so the larger partial sums do too, and a series whose partial sums are unbounded cannot converge.
Table 5.2 in the book shows the effect numerically, but one of its entries is misprinted. The harmonic sum of the first four terms is twenty-five twelfths, which is 2.0833, and the table prints 2.0933. It changes nothing in the argument, but it is a good reminder to recompute any table you rely on.
Section
Part 2
Concept
Comparison test (Theorem 5.11) — For series with nonnegative terms: a series whose terms sit, from some point on, below the terms of a convergent series converges; a series whose terms sit above the terms of a divergent series diverges.
\[ \text{(i) } 0 \le a_n \le b_n \text{ for } n \ge N, \;\sum b_n \text{ converges} \;\Longrightarrow\; \sum a_n \text{ converges} \]
\[ \text{(ii) } a_n \ge b_n \ge 0 \text{ for } n \ge N, \;\sum b_n \text{ diverges} \;\Longrightarrow\; \sum a_n \text{ diverges} \]
The two opening examples were one of each. Geometric series and p-series are the usual choices for the comparison series.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, pp. 487-488 — Theorem 5.11
Theorem 5.11 packages the two arguments you have just seen. Part one is the convergence argument: terms smaller than those of a convergent series. Part two is the divergence argument: terms larger than those of a divergent series.
Two details deserve attention. First, the terms must be nonnegative. The whole argument rests on partial sums that only go up, and a series with negative terms could be smaller than a convergent series by being hugely negative. Second, the inequality only needs to hold from some point N onward. The first few terms add a fixed finite amount and never affect whether the series converges.
In practice the comparison series is almost always a p-series or a geometric series, because those are the families whose fates you can state instantly. The skill is in finding the right one, which Part 4 of this lesson is about.
Notation
Annotate
On: \( 0 \le a_n \le b_n \;(n \ge N), \quad \sum b_n < \infty \;\Longrightarrow\; \sum a_n < \infty \)
Step through the notes one at a time. The first note is the hypothesis students most often forget to state. Write it in every solution: the terms are positive, so the test applies.
The second note is the one that decides whether the argument works at all. Smaller goes with convergent; larger goes with divergent. If you find yourself writing smaller than divergent, or larger than convergent, stop: you have an arrangement that proves nothing.
The third note is a freedom you will use constantly. Many inequalities only become true after a few terms, for instance when a logarithm finally exceeds one. The theorem does not mind. And the last note is a reminder that a comparison transfers a verdict you already have; you must state that verdict and its reason.
Concept
Call the partial sums of the smaller series S sub k, and the sum of the bigger convergent series L.
\[ S_{k+1} = S_k + a_{k+1} \ge S_k \]
\[ \sum_{n=N}^{k} a_n \le \sum_{n=N}^{k} b_n \le \sum_{n=1}^{\infty} b_n = L \]
\[ S_k = (a_1 + \cdots + a_{N-1}) + \sum_{n=N}^{k} a_n \le (a_1 + \cdots + a_{N-1}) + L \]
The right side is one fixed number. The partial sums increase and never pass it, so the Monotone Convergence Theorem makes them converge.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 487 — proof of Theorem 5.11
This is the book's proof, and it is worth following line by line because it is the Monotone Convergence Theorem at work again. The first line says the partial sums increase, since each new term is nonnegative.
The second line handles the tail. From N onward each term is at most the matching comparison term, so a stretch of the series from N to k is at most the same stretch of the comparison series, which is at most the comparison's full sum L.
The third line puts the early terms back. They are a fixed finite amount, whatever they are, so every partial sum is less than that fixed amount plus L. An increasing sequence with a fixed upper bound converges. Notice that the proof never uses the actual value of L, only that it is a finite number.
Concept
Suppose the larger series converged, while the smaller one diverges.
\[ 0 \le b_n \le a_n, \quad \sum a_n \text{ converges} \;\Longrightarrow\; \sum b_n \text{ converges} \]
That is part (i) with the letters swapped, and it contradicts the known divergence of the smaller series. So the larger series cannot converge.
\[ \sum b_n \text{ diverges} \;\Longrightarrow\; \sum a_n \text{ diverges} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 487 — the proof of part ii is the contrapositive
You do not need a second proof. Part two follows from part one by a short argument by contradiction, which is the same as proving the contrapositive.
Suppose the larger series converged. Then the smaller series would be below a convergent series, and part one would force it to converge. But you were told the smaller series diverges. That contradiction means the larger series cannot converge, so it diverges.
This pattern, one theorem and its contrapositive, appeared in Section 5.3 with the divergence test. It is worth recognising, because it means you only have to understand one direction deeply and the other comes along.
Picture it
Figure (svg): Two panels of partial sums. Left: the partial sums of one over n squared in orange, with two series above them, two over n squared (green, levelling off) and one over n (red, climbing). Right: the harmonic partial sums in orange, with two series below them, one over n plus 1 (red, still climbing) and one over n squared (green, levelling off).
Four arrangements are possible and only two conclude anything. Larger than a convergent series, and smaller than a divergent series, are satisfied by series of both fates, so they settle nothing.
There are four ways to line a series up against a known one, and only two conclude anything. This picture shows why the other two cannot.
In the left panel the orange dots are the running totals of the reciprocal squares, which converge. Above them are two other series. The green one, two over n squared, levels off; the red one, the harmonic series, climbs forever. Both are above a convergent series, so being above a convergent series is consistent with either fate, and tells you nothing.
The right panel is the mirror image. Below the harmonic series sit one series that levels off and one that keeps climbing. Being below a divergent series is consistent with either fate too. Whenever you set up a comparison, ask which side of the reference your series is on, and whether that side is one of the two that decides anything.
Sorting
Sort into buckets
Each line is a true inequality and a true fact about the comparison series. Does it prove anything about the first series?
Every inequality on the cards is true, and every fact about the comparison series is true. The only question is whether the arrangement is one that decides anything. Sort each card by the direction of the inequality and the fate of the comparison series.
Three cards use a working arrangement: two are below a convergent series, and one is above the divergent harmonic series. The other three are below a divergent series or above a convergent one, and those settle nothing.
It is worth noticing that the silent cards include series of both fates. The first silent card and the one with two over n squared both converge, while one over ln n diverges. That is exactly why those arrangements must be silent: they are true for series that do opposite things.
Worked example
\[ \sum_{n=1}^{\infty}\frac{1}{n^3+3n+1} \]
Pick the comparison
Why: For large n the denominator is dominated by n cubed.
\[ b_n = \frac{1}{n^3}, \quad p = 3 > 1 \;\Longrightarrow\; \sum b_n \text{ converges} \]
Compare the denominators
Why: The extra 3n plus 1 is positive for every n from 1.
\[ n^3 + 3n + 1 > n^3 \]
Take reciprocals
Why: A bigger positive denominator gives a smaller fraction.
\[ 0 < \frac{1}{n^3+3n+1} < \frac{1}{n^3} \]
Apply Theorem 5.11(i)
Why: Smaller than a convergent series.
\[ \sum_{n=1}^{\infty}\frac{1}{n^3+3n+1} \text{ converges} \]
Check numerically
Why: Ten terms give 0.3249 and the sum settles near 0.3294, well under the sum of the reciprocal cubes.
\[ S_{10} \approx 0.3249, \quad \sum \approx 0.3294 < 1.2021 \approx \sum\frac{1}{n^3} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 488 — Example 5.17a
The comparison series comes from the dominant piece of the denominator. For large n, the n cubed dwarfs the three n plus one, so the terms behave like one over n cubed, a convergent p-series.
Because the comparison converges, you need the terms to be smaller than it, and they are: the denominator is n cubed plus something positive, so it is bigger than n cubed, and the reciprocal is smaller. That is the useful arrangement, and the conclusion is convergence.
The numerical check makes a point about what comparison does and does not give you. The series sums to about 0.3294, while the comparison series sums to about 1.2021. The comparison proved the sum exists and is less than 1.2021, which is true but far from sharp.
Worked example
\[ \sum_{n=1}^{\infty}\frac{1}{2^n+1} \]
Pick the comparison
Why: Drop the plus one: what is left is geometric.
\[ b_n = \left(\frac12\right)^n, \quad r = \frac12, \; |r| < 1 \]
Sum the comparison series
Why: First term one half, ratio one half.
\[ \sum_{n=1}^{\infty}\left(\frac12\right)^n = \frac{1/2}{1 - 1/2} = 1 \]
Compare the terms
Why: Adding one to the denominator shrinks the fraction.
\[ 0 < \frac{1}{2^n+1} < \frac{1}{2^n} \]
Figure (svg): Bars for n from 1 to 7: wide pale bars of height one over two to the n, each with a slightly shorter solid bar of height one over two to the n plus one inside it.
Apply Theorem 5.11(i)
Why: Smaller than a convergent series.
\[ \sum_{n=1}^{\infty}\frac{1}{2^n+1} \text{ converges, with sum } < 1 \]
Check numerically
Why: Five terms already give 0.7336, and the full sum is about 0.7645: below 1, as promised.
\[ S_5 \approx 0.7336, \quad \sum_{n=1}^{\infty}\frac{1}{2^n+1} \approx 0.7645 < 1 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 488 — Example 5.17b
Here the comparison series is geometric rather than a p-series. Dropping the plus one leaves one over two to the n, which is a geometric series with ratio one half, and you can even sum it: the total is exactly one.
The bars make the argument visible. Each solid bar is a term of the series you want; each pale bar is the matching term of the geometric series. Every solid bar fits inside its pale bar, and the pale bars add up to exactly one, so the solid bars add up to less than one.
Notice that the integral test would be unpleasant here: integrating one over two to the x plus one requires a substitution and a logarithm. The comparison takes two lines. The numerical check confirms the bound: the sum is about 0.7645, which is less than one, just as the comparison promised.
Worked example
\[ \sum_{n=2}^{\infty}\frac{1}{\ln n} \]
Pick the comparison
Why: A logarithm grows more slowly than n, so its reciprocal should be bigger than one over n.
\[ b_n = \frac{1}{n}, \quad \sum_{n=2}^{\infty}\frac1n \text{ diverges} \]
Compare the denominators
Why: For every n from 2 on, the logarithm is smaller than n itself.
\[ 0 < \ln n < n \]
Take reciprocals
Why: The smaller positive denominator gives the bigger fraction.
\[ \frac{1}{\ln n} > \frac{1}{n} > 0 \]
Figure (svg): Two curves for x from 2 to 20 with dots at the whole numbers: y equals one over ln x, falling slowly from about 1.44 to 0.33, stays far above y equals one over x, falling from 0.5 to 0.05.
Apply Theorem 5.11(ii)
Why: Larger than a divergent series.
\[ \sum_{n=2}^{\infty}\frac{1}{\ln n} \text{ diverges} \]
Check with partial sums
Why: From 2 to 100 the series has reached 29.99, while the harmonic terms over the same range total only 4.19.
\[ \sum_{n=2}^{100}\frac{1}{\ln n} \approx 29.99 > 4.19 \approx \sum_{n=2}^{100}\frac1n \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 488 — Example 5.17c
This example goes in the divergence direction. The logarithm grows far more slowly than n, so one over ln n is bigger than one over n, and the harmonic series diverges. Bigger than divergent is the working arrangement.
The key inequality is that ln n is less than n for every whole number from two onward. If you want to see why, compare the curves: the logarithm starts at zero and grows ever more slowly, while n grows steadily. The picture shows the consequence for the reciprocals: the red curve, one over ln n, stays far above the orange curve, one over n.
The series starts at n equal to two because ln one is zero. The partial-sum check is striking. By a hundred terms the series has passed 29, while the harmonic terms over the same range add to barely more than four. The divergence here is not delicate at all.
Step zero
\[ \sum_{n=1}^{\infty}\frac{n}{n^3+n+1} \]
Discussion prompt
Checkpoint 5.16 asks for the comparison test here. Before writing any inequality, decide which known series to compare with, and whether you need the terms to be smaller or larger than it.
Write your answer before revealing. The habit this slide builds is to decide the comparison and the direction before writing a single inequality, because the direction you need depends on the fate of the comparison series.
Keep the biggest power of n on top and the biggest on the bottom: that leaves n over n cubed, which is one over n squared. The reciprocal squares converge, so the only useful arrangement is to show your terms are smaller.
If instead you had guessed a divergent comparison, you would need your terms to be bigger, and here they are not. Deciding the direction first saves you from proving a true inequality that points the useless way.
Worked example
\[ \sum_{n=1}^{\infty}\frac{n}{n^3+n+1} \]
Enlarge the fraction by shrinking its denominator
Why: Removing the positive n plus 1 makes the denominator smaller.
\[ \frac{n}{n^3+n+1} < \frac{n}{n^3} \]
Simplify
Why: Cancel one factor of n.
\[ \frac{n}{n^3} = \frac{1}{n^2} \]
Name the comparison's verdict
Why: A p-series with p equal to 2.
\[ p = 2 > 1 \;\Longrightarrow\; \sum\frac{1}{n^2} \text{ converges} \]
Apply Theorem 5.11(i)
Why: Positive terms, smaller than a convergent series.
\[ \sum_{n=1}^{\infty}\frac{n}{n^3+n+1} \text{ converges} \]
Check numerically
Why: Ten terms: 0.7923 against 1.5498 for the reciprocal squares; the full sum is about 0.8872, under 1.6449.
\[ S_{10} \approx 0.7923 < 1.5498, \quad \sum \approx 0.8872 < 1.6449 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 489 — Checkpoint 5.16
The move here is to make a fraction bigger by making its denominator smaller. Dropping the n plus one from the denominator leaves n cubed, which is smaller, so the fraction grows. Then cancel one n, and the bound is one over n squared.
That is the working arrangement: smaller than a convergent p-series with power two. State the comparison's fate, state the inequality, state the conclusion. Those three sentences are a complete answer.
The numerical check compares ten terms of each series, 0.7923 against 1.5498, and then the full sums, about 0.8872 against 1.6449. The partial sums of your series stay under those of the reciprocal squares at every stage, which is precisely what the proof used.
Trap
A tempting line, since the harmonic series diverges:
\[ \frac{1}{n^2+1} < \frac{1}{n} \]
\[ \Longrightarrow \sum\frac{1}{n^2+1} \text{ diverges?} \]
Wrong. Every step but the last is true.
Being under something infinite puts no limit on a total at all. This very series converges (it is under one over n squared). Smaller only helps when the bigger series is finite.
\[ a_n \le b_n: \text{ only } \sum b_n < \infty \text{ helps} \]
Every line on the left is true until the last. The inequality is correct, and the harmonic series does diverge. The error is believing that being smaller than something infinite forces you to be infinite.
Think of heights. Knowing that you are shorter than a mountain tells nobody how tall you are. Being under an infinite total is no restriction on a total at all, and the series here in fact converges, because it is also under the reciprocal squares. When a comparison fails to conclude anything, the fix is not to force a conclusion but to choose a different comparison.
Counterexample
Discussion prompt
Someone argues: every term of the sum of 2 over n squared is bigger than one over n squared, and the reciprocal squares converge, so the sum of 2 over n squared diverges. Break the claim, and say what the example proves.
Try to find the counterexample yourself before revealing. The claim sounds plausible because bigger feels like it should mean worse, but it confuses bigger with infinite.
Two over n squared is twice the reciprocal squares, and multiplying a convergent series by two just doubles its sum, to pi squared over three, about 3.29. So a series above a convergent series can converge.
The harmonic series is also above the reciprocal squares, and it diverges. Having one example of each fate on the same side of the same reference is the complete proof that the arrangement is silent. That is what a counterexample does: it does not just refute one claim, it shows that no argument of that shape can ever work.
Worked example
\[ \sum_{n=1}^{\infty}\frac{\sin^2 n}{n^2} \]
Bound the numerator
Why: Sine squared is never negative and never more than 1.
\[ 0 \le \sin^2 n \le 1 \]
Divide by n squared
Why: Dividing by a positive number keeps the inequalities.
\[ 0 \le \frac{\sin^2 n}{n^2} \le \frac{1}{n^2} \]
Figure (svg): Stems and dots for sine squared of n over n squared, n from 1 to 14, bouncing irregularly, every one on or under the curve y equals one over x squared.
Apply Theorem 5.11(i)
Why: The integral test could not be used: these terms are not decreasing. The comparison needs no such hypothesis.
\[ \sum\frac{1}{n^2} \text{ converges} \;\Longrightarrow\; \sum\frac{\sin^2 n}{n^2} \text{ converges} \]
Check numerically
Why: Ten terms give 1.0209 and a million give 1.0708, which matches the known exact value, pi minus 1 over 2, and sits below the ceiling.
\[ \sum_{n=1}^{\infty}\frac{\sin^2 n}{n^2} = \frac{\pi - 1}{2} \approx 1.0708 < 1.6449 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 493 — Exercise 202
This example shows what comparison can do that the integral test cannot. The terms jump up and down as sine squared of n wanders between zero and one, so they are not decreasing, and the integral test's hypothesis fails.
The comparison test has no decreasing hypothesis. It only needs a ceiling, and sine squared is never more than one, so each term is at most one over n squared. The picture shows the terms bouncing irregularly, every one under the orange curve. Nonnegative and below a convergent series: done.
The check uses a known exact value. This sum is pi minus one, over two, about 1.0708, and adding a million terms numerically agrees to four decimal places. It is less than the bound of 1.6449, as it must be.
Section
Part 3
Concept
\[ \sum_{n=2}^{\infty}\frac{1}{n^2-1} \quad\text{beside}\quad \sum_{n=2}^{\infty}\frac{1}{n^2} \]
The obvious comparison is the reciprocal squares, but subtracting one makes the denominator smaller and the fraction bigger.
\[ \frac{1}{n^2-1} > \frac{1}{n^2} \quad \text{for every } n \ge 2 \]
Figure (svg): Dots for n from 2 to 10: one over n squared minus one in red sits above one over n squared in orange at every n, the gap shrinking quickly.
Larger than a convergent series: a silent arrangement. Yet the two sequences are nearly equal. What is needed is a test that compares sizes by a ratio instead of an inequality.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 489 — the series 1/(n² − 1)
Now a series where the direct test gets stuck. The natural comparison for one over n squared minus one is the reciprocal squares, because the minus one hardly matters for large n. But subtracting one makes each denominator smaller and each fraction bigger.
So the terms are above a convergent series, and that is a silent arrangement. The picture shows how frustrating this is: the two sequences of terms are almost identical after the first few, yet the direct comparison is on the wrong side and cannot say anything.
You could hunt for a cleverer comparison, and one exists, but there is a better way. When two sequences are almost the same size, measure that by their ratio. If the ratio settles down to a fixed positive number, the two must share a fate. That is the limit comparison test.
Concept
Suppose the ratio of the terms tends to a positive number L. Then from some N on the ratio lies within one half of L of its limit.
\[ \lim_{n\to\infty}\frac{a_n}{b_n} = L > 0 \;\Longrightarrow\; \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2} \quad (n \ge N) \]
\[ \frac{L}{2}\,b_n < a_n < \frac{3L}{2}\,b_n \quad (n \ge N) \]
Figure (svg): Dots for the ratio n squared over n squared minus one, n from 2 to 20: it starts at 1.33 and drops toward a solid line at 1, staying inside a shaded band between one half and three halves.
The right inequality puts the series below a multiple of a convergent series when the comparison converges; the left puts it above a multiple of a divergent one when it diverges. Either way the fates match.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 489 — the idea behind the limit comparison test
This argument is the heart of the lesson, and it builds the new test entirely out of the old one. Suppose the ratio of your terms to the comparison terms tends to a positive number L. Then eventually the ratio is within half of L of its limit, so it lies between half of L and one and a half times L.
Multiply through by the comparison term and you get two inequalities at once. The picture shows the ratio for the series on the last slide: it enters the green band between one half and three halves and never leaves it.
Now each inequality does one job. If the comparison converges, the right inequality puts your series below a multiple of a convergent series, so it converges. If the comparison diverges, the left inequality puts it above a multiple of a divergent series, so it diverges. Either way the two series share a fate.
Intuition
The squeeze used multiples of the comparison series, not the series itself. That costs nothing.
\[ \sum_{n=1}^{k} c\,b_n = c\sum_{n=1}^{k} b_n \quad (c > 0) \]
Multiplying every partial sum by a fixed positive number keeps a finite limit finite and an unbounded sequence unbounded. So the actual value of L never matters: 3, one half or a thousand, only that it is positive and finite.
The argument on the last slide compared your series with multiples of the comparison series, half of L or one and a half times L of it. That is legitimate because multiplying a series by a fixed positive constant multiplies every partial sum by that constant.
If the partial sums had a finite limit, they still have one, just scaled. If they grew without bound, they still grow without bound. So a constant multiple never changes convergence or divergence.
This is why the value of the limit in the limit comparison test never matters. A limit of three, of one half or of a thousand gives the same conclusion. What matters is only that the limit is a genuine positive number, neither zero nor infinite.
Worked example
\[ \sum_{n=2}^{\infty}\frac{1}{n^2-1}, \qquad b_n = \frac{1}{n^2} \]
Form the ratio
Why: Dividing by a fraction is multiplying by its reciprocal.
\[ \frac{a_n}{b_n} = \frac{1/(n^2-1)}{1/n^2} = \frac{n^2}{n^2-1} \]
Divide the top and bottom by n squared
Why: That exposes the limit.
\[ \frac{n^2}{n^2-1} = \frac{1}{1 - 1/n^2} \to \frac{1}{1-0} = 1 \]
Read the case
Why: One is positive and finite, and the reciprocal squares converge.
\[ L = 1, \quad \sum\frac{1}{n^2} \text{ converges} \;\Longrightarrow\; \sum_{n=2}^{\infty}\frac{1}{n^2-1} \text{ converges} \]
Check by summing it exactly
Why: Partial fractions make it telescope: one half of one over n minus 1, minus one over n plus 1.
\[ \frac{1}{n^2-1} = \frac12\left(\frac{1}{n-1} - \frac{1}{n+1}\right) \]
Verify the total
Why: Only 1 and one half survive the cancelling, so the sum is finite, as the test said. Summing a million terms numerically gives 0.749999.
\[ \sum_{n=2}^{\infty}\frac{1}{n^2-1} = \frac12\left(1 + \frac12\right) = \frac34 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 489 — the limit comparison for 1/(n² − 1)
This is the series where the direct comparison failed, now settled in three moves. Form the ratio, which turns into n squared over n squared minus one. Divide the top and bottom by n squared, and the limit is one. One is positive and finite, the reciprocal squares converge, so this series converges.
The check is unusually strong, because this series can be summed exactly. Partial fractions split each term into half of one over n minus one, minus one over n plus one. Writing out the first few terms, almost everything cancels, and only half of one plus one half survives. The total is three quarters.
A finite sum confirms convergence, and adding a million terms by computer gives 0.749999. You will rarely be able to check a comparison so completely, which makes this a good example to remember.
Concept
If the ratio tends to zero, it is eventually below 1, which gives only the upper inequality.
\[ \frac{a_n}{b_n} \to 0 \;\Longrightarrow\; a_n \le b_n \text{ eventually} \;\Longrightarrow\; \text{only a convergent } \sum b_n \text{ helps} \]
If the ratio grows without bound, it is eventually above 1, which gives only the lower inequality.
\[ \frac{a_n}{b_n} \to \infty \;\Longrightarrow\; a_n \ge b_n \text{ eventually} \;\Longrightarrow\; \text{only a divergent } \sum b_n \text{ helps} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, pp. 489-490 — the cases L = 0 and L = ∞
The limit of the ratio does not have to be a positive number. It can be zero, or it can grow without bound, and each of these gives you half of the squeeze.
If the ratio tends to zero, it is eventually below one, so your terms are eventually smaller than the comparison terms. Smaller is useful only against a convergent series. So a limit of zero lets you copy convergence, and nothing else.
If the ratio grows without bound, it is eventually above one, so your terms are eventually bigger. Bigger is useful only against a divergent series. So an infinite limit lets you copy divergence, and nothing else. You do not need to memorise these cases if you remember that zero means smaller and infinity means bigger.
Notation
Annotate
On: \( a_n, b_n \ge 0: \quad \lim_{n\to\infty}\frac{a_n}{b_n} = L \;\begin{cases} 0 < L < \infty & \text{same fate} \\ L = 0 & \sum b_n < \infty \Rightarrow \sum a_n < \infty \\ L = \infty & \sum b_n = \infty \Rightarrow \sum a_n = \infty \end{cases} \)
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 490 — Theorem 5.12
Step through the notes and connect each case to the idea behind it. Case one is the two-sided squeeze: both inequalities, so the fates match in both directions.
Cases two and three are the one-sided squeezes. A limit of zero makes your terms eventually smaller, so they can inherit convergence. An infinite limit makes them eventually bigger, so they can inherit divergence.
The last note is the one that costs marks. A limit of zero with a divergent comparison, or an infinite limit with a convergent one, is not a case of the theorem. It is the limit version of the silent arrangements. The right response is to write that the test gives no information with this comparison, and to choose another.
Picture it
Figure (svg): Three sequences of ratios for n from 1 to 40: root n over root n plus one rises toward 1; ln n over n falls toward 0; ln n climbs past 3.6 with no ceiling.
Green settles at 1, so its two series share a fate. Orange sinks to zero and red climbs forever: each ties its pair together in one direction only, and which direction is fixed by the case, not by you.
The three sequences in the picture are the three cases, drawn. The green dots are the ratio from Example 5.18(a), which you will meet in a moment. They climb toward one, so the two series in that pair are tied together in both directions.
The orange dots sink toward zero, which means the top series is becoming negligible compared with the bottom one. The red dots climb without bound, which means the top series is becoming overwhelmingly bigger. Each of those is a one-directional tie.
The shape of the ratio tells you how well matched your comparison is. A ratio that settles at a positive number means you have found the right size. A ratio going to zero or infinity means your comparison was too big or too small, and you may need to adjust it.
Worked example
Compare two series whose fates you already know with the convergent reciprocal cubes.
The ratio for one over root n
Why: Dividing by one over n cubed multiplies by n cubed.
\[ \frac{1/\sqrt n}{1/n^3} = \frac{n^3}{\sqrt n} = n^{5/2} \to \infty \]
The ratio for one over n squared
Why: The same move.
\[ \frac{1/n^2}{1/n^3} = n \to \infty \]
Read the case
Why: An infinite limit with a convergent comparison is not one of the three cases.
\[ L = \infty, \quad \sum\frac{1}{n^3} < \infty \;\Longrightarrow\; \text{no information} \]
Check against the known fates
Why: The same limit, infinity, came from a divergent series and a convergent one, so no conclusion could be right. The book prints the two p-series inequalities backwards on p. 490; the correct ones are below.
\[ p = \tfrac12 < 1: \text{ diverges}, \qquad p = 2 > 1: \text{ converges} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 490 — the note after Theorem 5.12
The book uses this pair to show that the silent cases are genuinely silent. Compare one over root n and one over n squared with the reciprocal cubes. Both ratios grow without bound, n to the five halves and n.
Both limits are infinite and the comparison converges, which is not one of the three cases. And the known fates confirm that no conclusion would be safe: one over root n diverges and one over n squared converges, yet they produced the same kind of limit.
A note on the book itself. On page 490 it states the two p-series facts with the inequalities printed the wrong way round, saying one half is greater than one and two is less than one. The verdicts it draws are right, and the correct inequalities are on the slide.
Matching
Match the pairs
Why: A positive finite limit copies the verdict whichever it is. A limit of zero copies only convergence, and a limit of infinity copies only divergence. A zero limit against a divergent comparison says the terms are smaller than something infinite, which is no information.
Each card gives you two pieces of information, the limit of the ratio and the fate of the comparison series. You need both to reach a verdict, which is the point of the exercise.
A limit of four is positive and finite, so the fates match, and the comparison diverges. A limit of zero with a convergent comparison is case two. An infinite limit with a divergent comparison is case three.
The last card is the trap: a limit of zero with a divergent comparison. Your terms are eventually smaller than terms of a divergent series, which is the silent arrangement again, so there is no information. If you matched it to a verdict, go back to the one-sided cases slide.
Worked example
\[ \sum_{n=1}^{\infty}\frac{1}{\sqrt n + 1}, \qquad b_n = \frac{1}{\sqrt n} \]
Form the ratio
Why: Multiply by the reciprocal of the comparison term.
\[ \frac{1/(\sqrt n + 1)}{1/\sqrt n} = \frac{\sqrt n}{\sqrt n + 1} \]
Divide the top and bottom by root n
Why: The one becomes one over root n, which vanishes.
\[ \frac{\sqrt n}{\sqrt n + 1} = \frac{1}{1 + 1/\sqrt n} \to 1 \]
Read the case
Why: L equal to 1 is positive and finite; the comparison is a p-series with p one half.
\[ L = 1, \quad p = \tfrac12 < 1 \;\Longrightarrow\; \sum\frac{1}{\sqrt n + 1} \text{ diverges} \]
Check with a direct comparison
Why: For n at least 1, one is at most root n, so the denominator is at most twice root n, and the terms beat half a divergent series.
\[ \sqrt n + 1 \le 2\sqrt n \;\Longrightarrow\; \frac{1}{\sqrt n + 1} \ge \frac12 \cdot \frac{1}{\sqrt n} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 491 — Example 5.18a
The plus one in the denominator is negligible next to root n for large n, so the natural comparison is one over root n, a p-series with power one half, which diverges.
Notice that the direct test is awkward here: your terms are smaller than one over root n, which is the silent arrangement. The limit test sidesteps that. The ratio is root n over root n plus one; divide by root n and the limit is one. Positive and finite, and the comparison diverges, so your series diverges.
The check shows that a direct comparison was possible after all, with a little ingenuity. Because one is at most root n, the denominator is at most twice root n, so each term is at least half of one over root n. Half of a divergent series diverges. The limit test simply spared you from finding that trick.
Worked example
\[ \sum_{n=1}^{\infty}\frac{2^n+1}{3^n}, \qquad b_n = \left(\frac23\right)^n \]
Form the ratio
Why: Divide by two to the n over three to the n; the three to the n cancels.
\[ \frac{(2^n+1)/3^n}{2^n/3^n} = \frac{2^n+1}{2^n} \]
Split the fraction
Why: Each piece has an obvious limit.
\[ \frac{2^n+1}{2^n} = 1 + \left(\frac12\right)^n \to 1 \]
Read the case
Why: A geometric comparison with ratio two thirds converges.
\[ L = 1, \quad r = \tfrac23 < 1 \;\Longrightarrow\; \sum\frac{2^n+1}{3^n} \text{ converges} \]
Figure (svg): Bars for n from 1 to 8: pale bars of height two thirds to the n, and solid bars of height two to the n plus one over three to the n, which start taller (1 at n equals 1) but shrink to match the pale bars.
Check by summing it exactly
Why: The series splits into two geometric series, sums 2 and one half, so it converges to 2.5. Summing eighty terms numerically gives 2.500000.
\[ \sum_{n=1}^{\infty}\left(\frac23\right)^n + \sum_{n=1}^{\infty}\left(\frac13\right)^n = 2 + \frac12 = \frac52 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 491 — Example 5.18b
For large n, the two to the n in the numerator dwarfs the one, so the terms behave like two to the n over three to the n, a geometric series with ratio two thirds, which converges.
The terms are bigger than the comparison terms, the silent arrangement for a convergent comparison, so the direct test does not apply as it stands. The ratio fixes that: it simplifies to one plus one half to the n, which tends to one. The picture shows the pairs of bars drawing level with each other as n grows.
The check is exact. Split the fraction into two to the n over three to the n plus one over three to the n, and the series becomes the sum of two geometric series, with sums two and one half. The total is two and a half, and summing eighty terms numerically gives the same.
Worked example
\[ \sum_{n=1}^{\infty}\frac{\ln n}{n^2} \]
Try the harmonic series
Why: Since ln n is less than n, the terms are smaller than one over n.
\[ \frac{\ln n/n^2}{1/n} = \frac{\ln n}{n} \]
Evaluate with L'Hôpital's rule
Why: Pass to the real variable x; both top and bottom grow without bound.
\[ \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = 0 \]
Read the case
Why: A zero limit against a divergent comparison.
\[ L = 0, \quad \sum\frac1n = \infty \;\Longrightarrow\; \text{no information} \]
Try the reciprocal squares instead
Why: The n squared cancels, leaving ln n.
\[ \frac{\ln n/n^2}{1/n^2} = \ln n \to \infty \]
Check the case again
Why: Infinity against a convergent comparison: silent a second time. One comparison was too big and the other too small.
\[ L = \infty, \quad \sum\frac{1}{n^2} < \infty \;\Longrightarrow\; \text{no information} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, pp. 491-492 — Example 5.18c
This example is valuable because the first two tries fail, and the book shows them. Logarithms sit awkwardly between the powers, and that makes choosing the comparison harder.
Against the harmonic series, the ratio is ln n over n, which L'Hôpital's rule sends to zero. A zero limit against a divergent comparison is silent. The comparison was too big: your terms are negligible next to one over n, but that does not stop them from diverging.
Against the reciprocal squares, the ratio is ln n, which grows without bound. An infinite limit against a convergent comparison is silent too. This comparison was too small. The two failures together tell you where to look next: somewhere between one over n and one over n squared.
Worked example
Choose a p-series between the two failures: p strictly between 1 and 2, say three halves.
Form the ratio
Why: Multiply by n to the three halves.
\[ \frac{\ln n/n^2}{1/n^{3/2}} = \frac{\ln n}{n^{1/2}} \]
Evaluate with L'Hôpital's rule
Why: The derivative of root x is one over twice root x.
\[ \lim_{x\to\infty}\frac{\ln x}{\sqrt x} = \lim_{x\to\infty}\frac{1/x}{1/(2\sqrt x)} = \lim_{x\to\infty}\frac{2}{\sqrt x} = 0 \]
Figure (svg): The curve ln x over root x for x from 1 to one million on a logarithmic horizontal axis: it rises from 0 to a peak of about 0.736 at x equal to e squared, about 7.39, then sinks toward zero, reaching 0.014 at one million.
Read the case
Why: A zero limit against a convergent comparison: case (ii).
\[ L = 0, \quad p = \tfrac32 > 1 \;\Longrightarrow\; \sum\frac{\ln n}{n^2} \text{ converges} \]
Check with partial sums
Why: They level off: 0.8817 after 100 terms, 0.9296 after 1000, 0.9375 after a million.
\[ S_{100} \approx 0.8817, \quad S_{1000} \approx 0.9296, \quad S_{10^6} \approx 0.9375 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 492 — Example 5.18c
The failures pointed to a power strictly between one and two, and three halves is the natural choice. It is above one, so the comparison converges, and that means a zero limit will be good enough.
The ratio is ln n over root n, and L'Hôpital's rule turns it into two over root n, which tends to zero. So the case is zero with a convergent comparison, case two, and the series converges.
The picture shows the ratio on a logarithmic horizontal axis. It rises at first, peaking at two over e near x equal to e squared, because the logarithm starts out growing faster than the square root. Then the square root wins, and the ratio sinks slowly toward zero. The partial sums confirm the verdict: they level off near 0.9375.
Tweak it
Parameter explorer
The curve is the ratio of ln x over x² to 1/xᵖ. Slide p. For which p does the ratio sink to zero, and for which p does that zero actually prove convergence?
\[ \frac{\ln x / x^2}{1/x^{{p}}} = x^{{p}-2}\ln x \]
Start with p at three halves and move it slowly. For every p below two, the ratio eventually sinks to zero, although for p close to two it takes a long time and the curve rises for a while first.
Now ask when that zero is useful. A zero limit only proves convergence against a convergent comparison, which needs p bigger than one. So the working window is p strictly between one and two. Below one, the limit is still zero, but the comparison diverges and the test is silent.
At p equal to two and beyond, the ratio grows without bound, and an infinite limit against a convergent comparison is silent again. The slider shows that there is not one right comparison but a whole window of them, and that the book's three halves is just a convenient choice in the middle.
Prediction
\[ \sum_{n=1}^{\infty}\frac{5^n}{3^n+2} \]
Predict first
Compare with the geometric series of ratio 5/3. What will the limit comparison test say?
Correct: Diverges: the ratio tends to 1
Why: Dividing by five thirds to the n leaves three to the n over three to the n plus two, which tends to 1. A positive finite limit copies the verdict of the comparison, and a geometric series with ratio five thirds diverges, because its ratio is bigger than one.
Commit to an option before revealing. The comparison has been chosen for you, the geometric series with ratio five thirds, so the question is what the ratio does and what the comparison's fate is.
The ratio simplifies to three to the n over three to the n plus two, which tends to one. A positive finite limit copies the comparison's fate, and a geometric series with ratio five thirds diverges, because five thirds is bigger than one.
If you chose converges, check whether you read the geometric rule the wrong way round. The ratio must be less than one in absolute value for convergence, and five thirds is not.
Worked example
\[ \sum_{n=1}^{\infty}\frac{5^n}{3^n+2}, \qquad b_n = \left(\frac53\right)^n \]
Form the ratio
Why: The five to the n cancels.
\[ \frac{5^n/(3^n+2)}{5^n/3^n} = \frac{3^n}{3^n+2} \]
Divide the top and bottom by three to the n
Why: Two over three to the n vanishes.
\[ \frac{3^n}{3^n+2} = \frac{1}{1 + 2/3^n} \to 1 \]
Read the case
Why: Positive finite limit, and the comparison ratio exceeds 1.
\[ L = 1, \quad r = \tfrac53 > 1 \;\Longrightarrow\; \sum\frac{5^n}{3^n+2} \text{ diverges} \]
Check with the divergence test
Why: The terms themselves grow: 1, 2.27, 4.31, 7.53, 12.76 for n from 1 to 5. Terms that do not tend to zero confirm divergence without any comparison.
\[ a_1 = 1, \; a_2 \approx 2.27, \; a_5 \approx 12.76 \;\Longrightarrow\; a_n \not\to 0 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 492 — Checkpoint 5.17
Written out, the argument is the same three moves as every limit comparison: form the ratio, take its limit, read the case. The five to the n cancels in the ratio, and dividing by three to the n shows the limit is one.
The comparison series is geometric with ratio five thirds, which diverges, so the series diverges.
The check is a reminder to try the cheapest test first. The terms themselves are one, then about 2.27, 4.31, 7.53 and 12.76, and they keep growing. Terms that do not go to zero mean divergence by the divergence test from Section 5.3, with no comparison needed at all. The limit comparison is correct, but it was more work than this series required.
Error analysis
Annotate
On: \( \lim_{n\to\infty}\frac{1/n^2}{1/n} = 0 \text{ and } \sum\frac1n \text{ diverges} \;\Longrightarrow\; \sum\frac{1}{n^2} \text{ diverges} \)
Look for the error before revealing the notes. The limit is computed correctly, and the harmonic series does diverge, so the mistake is in the conclusion drawn from them.
A limit of zero means the top series is eventually much smaller than the bottom one. Being smaller than a divergent series is the silent arrangement, so case two of Theorem 5.12 does not apply; it needs a convergent comparison.
Here the conclusion is not just unjustified but false: the reciprocal squares converge. That makes this a good example to remember, because it shows the rule is not a technicality. Reading a zero limit the wrong way gives wrong answers.
Trap
Wanting the reciprocal squares as a ceiling:
\[ \frac{1}{n^2-1} < \frac{1}{n^2} \]
\[ \Longrightarrow \text{ converges?} \]
Wrong. The inequality is false.
A smaller denominator gives a bigger fraction, so the terms are above the comparison. The conclusion happens to be true, but the argument is not. Either use the limit comparison test, or a correct bound:
\[ \frac{1}{n^2-1} \le \frac{2}{n^2} \quad (n \ge 2) \]
This is the most common mistake with the direct test: writing the inequality you want instead of the one that is true. Subtracting one from a denominator makes the fraction bigger, not smaller, so the inequality on the left is false.
The conclusion, that the series converges, happens to be true, which makes the mistake easy to miss. But an argument with a false line is not an argument. There are two honest fixes. Use the limit comparison test, as you did a few slides ago, or find a true bound: for n at least two, the denominator is at least half of n squared, so each term is at most two over n squared, and twice a convergent series converges.
Section
Part 4
Concept
For large n, a sum is dominated by its fastest-growing piece. Keep that piece in the numerator and in the denominator, drop the rest, and simplify: the result is the comparison series.
| term | dominant pieces | compare with | verdict |
|---|---|---|---|
| (n + 1)/(n³ + 2) | n / n³ | 1/n² | converges |
| √n/(n² + 1) | n^(1/2) / n² | 1/n^(3/2) | converges |
| (2ⁿ + n)/(3ⁿ − 1) | 2ⁿ / 3ⁿ | (2/3)ⁿ | converges |
| (n² + 1)/(n³ − n + 5) | n² / n³ | 1/n | diverges |
Exponentials beat powers, and powers beat logarithms. Logarithms need the careful treatment of Example 5.18(c).
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 485 — typically compared with geometric series or p-series
This is the skill that makes comparison fast. For large n, a sum of terms behaves like its fastest-growing piece, so you can predict the size of a complicated fraction by keeping only the dominant piece of the numerator and of the denominator.
Read the table row by row. In the first row, n over n cubed simplifies to one over n squared. In the second, root n over n squared is one over n to the three halves. In the third, the exponentials dominate, leaving two thirds to the n. In the last, n squared over n cubed is one over n, so that series diverges.
Remember the growth order: exponentials beat powers, and powers beat logarithms. The dominant-pieces guess tells you what to compare with; the limit comparison test then proves that the guess was right, because the ratio will tend to a positive number.
Worked example
\[ \sum_{n=2}^{\infty}\frac{3n^2+5}{n^4-n} \]
Read the dominant pieces
Why: Top three n squared, bottom n to the fourth. The sum starts at 2 because the denominator is zero at n equal to 1.
\[ \frac{3n^2}{n^4} = \frac{3}{n^2} \;\Longrightarrow\; b_n = \frac{1}{n^2} \]
Form the ratio
Why: Multiply by n squared.
\[ \frac{a_n}{b_n} = \frac{(3n^2+5)\,n^2}{n^4-n} = \frac{3n^4+5n^2}{n^4-n} \]
Divide the top and bottom by n to the fourth
Why: Every term but the leading ones vanishes.
\[ \frac{3 + 5/n^2}{1 - 1/n^3} \to 3 \]
Read the case
Why: Three is positive and finite; the constant is irrelevant.
\[ L = 3, \quad p = 2 > 1 \;\Longrightarrow\; \text{converges} \]
Check the ratio numerically
Why: At n equal to 10, 100 and 1000 the ratio is 3.0531, 3.0005 and 3.000005: it really is heading to 3.
\[ \frac{a_{10}}{b_{10}} \approx 3.0531, \quad \frac{a_{100}}{b_{100}} \approx 3.0005, \quad \frac{a_{1000}}{b_{1000}} \approx 3.000005 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 490 — Theorem 5.12(i), applied
First notice where the series starts. At n equal to one the denominator is zero, so the sum begins at two. Always check a formula's domain before you start comparing.
The dominant pieces are three n squared on top and n to the fourth on the bottom, giving three over n squared. The three is a constant, and constants never matter, so compare with the reciprocal squares. The ratio simplifies, after dividing by n to the fourth, to a limit of three.
The numerical check watches the ratio approach its limit: 3.05 at n equal to ten, 3.0005 at a hundred, 3.000005 at a thousand. Seeing the limit confirmed by numbers is a good habit when the algebra has several steps.
Estimation
\[ a_n = \ln\left(1 + \frac{1}{n^2}\right), \qquad n = 100 \]
Predict first
Exercise 210 asks about this series. Before any theorem, estimate the 100th term.
Correct: About 0.0001
Why: One over 100 squared is one ten-thousandth, and the logarithm of one plus a tiny number is almost exactly that tiny number. So the term is very close to one over n squared: the ratio at n equal to 100 is 0.99995, which suggests comparing with the reciprocal squares.
\[ \lim_{n\to\infty}\frac{\ln(1 + 1/n^2)}{1/n^2} = \lim_{t\to 0^+}\frac{\ln(1+t)}{t} = 1 \;\Longrightarrow\; \text{converges} \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 493 — Exercise 210
Make an estimate before revealing. One over a hundred squared is one ten-thousandth, and the question is what the logarithm does to one plus that tiny number.
Near one, the logarithm is almost a straight line with slope one, so the natural log of one plus a small number is very nearly that small number. At n equal to a hundred the term is 0.0000999950, which is one ten-thousandth to four significant figures.
That estimate is really the limit comparison in disguise. It says the term behaves like one over n squared, so the ratio tends to one. Substituting t for one over n squared turns the limit into the standard one, the log of one plus t over t, which is one. So Exercise 210 converges.
Worked example
\[ \sum_{n=1}^{\infty}\left(\frac1n - \sin\frac1n\right) \]
Substitute x equal to one over n
Why: Then x tends to zero from above as n grows.
\[ a_n = x - \sin x, \quad x = \frac1n \to 0^+ \]
Guess the size of the gap
Why: Near zero, sine x is x minus x cubed over 6 plus smaller terms.
\[ x - \sin x \approx \frac{x^3}{6} \;\Longrightarrow\; b_n = \frac{1}{n^3} \]
Form the ratio and apply L'Hôpital's rule once
Why: Top and bottom both tend to zero.
\[ \lim_{x\to 0^+}\frac{x - \sin x}{x^3} = \lim_{x\to 0^+}\frac{1 - \cos x}{3x^2} \]
Apply it twice more
Why: Still zero over zero each time.
\[ \lim_{x\to 0^+}\frac{\sin x}{6x} = \lim_{x\to 0^+}\frac{\cos x}{6} = \frac16 \]
Figure (svg): Two curves for x from 0 to 1: y equals x minus sine x and y equals x cubed over 6, almost indistinguishable, both rising from 0 to about 0.16; dots mark x equal to 1, 1/2, 1/3, 1/4 and 1/5 on the first.
Read the case
Why: One sixth is positive and finite, and the reciprocal cubes converge.
\[ L = \tfrac16, \quad p = 3 > 1 \;\Longrightarrow\; \text{converges} \]
Check the ratio numerically
Why: At n equal to 10 and 100 the ratio is already 0.16658 and 0.166666.
\[ 10^3\,a_{10} \approx 0.16658, \quad 100^3\,a_{100} \approx 0.166666 \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 493 — Exercise 217
The terms here are tiny differences, and the first move is to see how tiny. Writing x for one over n turns each term into x minus sine x, with x shrinking to zero.
Sine x is very close to x for small x, so the difference is much smaller than x itself. The picture shows how close: the curve x minus sine x is almost exactly x cubed over six near zero, and the terms of the series are the yellow dots on that curve. That suggests comparing with one over n cubed.
L'Hôpital's rule needs to be applied three times, because each time top and bottom are both still zero. The result is one sixth, positive and finite, and the reciprocal cubes converge. The numerical check confirms the ratio: 0.16658 at n equal to ten, already very close to one sixth.
Discrimination
Sort into buckets
Each series is to be compared with the obvious p-series or geometric series. Does the easy inequality point the useful way, or do you need the limit version?
For each series the comparison is given. The question is only whether the obvious inequality between the two points the useful way, which is below a convergent series or above a divergent one.
Plus signs in a denominator make terms smaller, and minus signs make them bigger. So one over n cubed plus seven is below a convergent series, and one over two to the n plus n is too: direct comparison works at once. One over root n minus a half is above a divergent series, which also works.
The other three have the inequality the wrong way round: above a convergent series or below a divergent one. For each of them the ratio to the comparison tends to one, so the limit comparison test settles them with no extra ingenuity. When you are unsure which test to use, the limit version is the safe default.
Comparison
Comparison matrix
| series | compare with | limit of aₙ/bₙ | verdict |
|---|---|---|---|
| Σ √n/(n² + 1) | 1/n^(3/2) | 1 | converges |
| Σ (n + 1)/(n² + 3) | 1/n | 1 | diverges |
| Σ 1/(4ⁿ − 3ⁿ) | (1/4)ⁿ | 1 | converges |
| Σ (1 − cos(1/n)) | 1/n² | 1/2 | converges |
| Σ 1/(n ln n) | 1/n | 0 | no information |
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 493 — Exercises 211 and 218
Fill in each blank before checking. In each row you are practising the same chain: choose the comparison from the dominant pieces, find the limit of the ratio, then combine that limit with the comparison's fate.
Two rows need a moment's thought. For one minus cosine of one over n, the same trick as the estimation slide applies: near zero, one minus cosine x behaves like half of x squared, so the limit is one half against the reciprocal squares.
The last row is the preview of the next part. For one over n ln n compared with one over n, the limit is zero and the comparison diverges, so the answer is no information. That row is not a failure of your algebra; it is a limit of the method.
Section
Part 5
Concept
A comparison transfers a verdict; it never creates one. So the test is only as strong as your list of series with known fates.
Figure (svg): Three curves for x from 3 to 30: one over x on top (red), one over x ln x in the middle (blue), one over x squared at the bottom (green). The middle curve is below a divergent series and above a convergent one.
\[ \frac{1}{n^2} < \frac{1}{n\ln n} < \frac{1}{n} \quad (n \ge 3) \]
The series of one over n ln n is below a divergent series and above a convergent one: both silent arrangements. You met its true fate in Section 5.3, through an integral.
Comparison is a way of transferring a verdict from a series you know to one you do not. It cannot create a verdict. So its reach is exactly as wide as your stock of series with known fates, which for now means p-series and geometric series.
The picture shows a series that slips through the gaps. One over x ln x lies below one over x, which diverges, and above one over x squared, which converges. Both are silent arrangements.
You already know what happens to this series from Section 5.3: it diverges, because its integral is the log of the log. The next slide shows that no p-series at all can settle it by comparison, which is why the integral test remains in your toolkit.
Worked example
\[ \sum_{n=2}^{\infty}\frac{1}{n\ln n} \]
Try the harmonic series
Why: The n cancels.
\[ \frac{1/(n\ln n)}{1/n} = \frac{1}{\ln n} \to 0 \quad \text{with } \sum\frac1n = \infty: \text{ silent} \]
Try any convergent p-series
Why: Take p bigger than 1, and write the ratio as a power over a logarithm.
\[ \frac{1/(n\ln n)}{1/n^{p}} = \frac{n^{p-1}}{\ln n}, \quad p - 1 > 0 \]
Evaluate with L'Hôpital's rule
Why: Any positive power of x outgrows the logarithm.
\[ \lim_{x\to\infty}\frac{x^{p-1}}{\ln x} = \lim_{x\to\infty}(p-1)x^{p-1} = \infty \quad \text{with } \sum\frac{1}{n^{p}} < \infty: \text{ silent} \]
Check with the integral test
Why: Section 5.3 settled it: the integral is a log of a log, unbounded, so the series diverges. No p-series could have decided it.
\[ \int_2^{b}\frac{dx}{x\ln x} = \ln(\ln b) - \ln(\ln 2) \to \infty \]
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.3, p. 482 — Exercise 160, now seen from §5.4
This is the precise version of the picture. Against the harmonic series, the ratio is one over ln n, which tends to zero, and a zero limit against a divergent comparison is silent.
Against any convergent p-series, with power p bigger than one, the ratio is n to the p minus one, over ln n. Any positive power of n eventually outgrows the logarithm, so this ratio grows without bound, and an infinite limit against a convergent comparison is silent too. Even p equal to 1.1 fails, although the ratio dips as low as about 0.27 before it climbs.
So the series sits between the two families, bigger than every convergent p-series eventually and smaller than the harmonic series, and no p-series can decide it. The integral test can. That is the answer to a question you might have been wondering about: every test in this chapter covers ground the others miss.
Intuition
The stock can be stretched cleverly. Exercise 247 deletes from the harmonic series every term whose n has a digit 9, and asks whether what is left converges.
\[ \text{d-digit numbers with no 9: } 8 \cdot 9^{d-1}, \quad \text{each } \ge 10^{d-1} \]
\[ \sum_{\text{no } 9}\frac1n \le \sum_{d=1}^{\infty}\frac{8\cdot 9^{d-1}}{10^{d-1}} = 8\sum_{d=1}^{\infty}\left(\frac{9}{10}\right)^{d-1} = 80 \]
Grouping by the number of digits turns it into a comparison with a geometric series, so the depleted series converges. Adding every surviving term below ten million gives only 12.21.
OpenStax Calculus Volume 2, §5.4 Comparison Tests §5.4, p. 495 — Exercise 247
This exercise shows how far a clever comparison can reach. Start with the harmonic series and delete every term whose denominator has a nine anywhere in its digits. It sounds like a small deletion, but among large numbers almost every number contains a nine.
Group the surviving terms by the number of digits d. There are eight times nine to the d minus one such numbers with d digits and no nine, since the first digit has eight choices and each other digit has nine. Each of them is at least ten to the d minus one, so each term is at most one over that.
Adding up the groups gives a geometric series with ratio nine tenths, whose sum is eighty. So the depleted series converges, to less than eighty. Adding every surviving term below ten million gives only about 12.21, and the total creeps up extremely slowly from there.
Real world
Figure (svg): Five identical blocks stacked on a table, each shifted right over the one below; the top block's right end sits about 1.14 block lengths beyond the table edge, so it is entirely past the edge.
Discussion prompt
Blocks of length 1 are stacked so that block k from the top sticks out 1/(2k) past the block beneath it. A cautious builder uses only 1/(2k + 3). Does the cautious stack still reach as far out as you like? How many blocks does each builder need to overhang the table by 2 lengths?
This is a real puzzle about stacking identical blocks at the edge of a table. With the best possible stacking, block k from the top overhangs the block under it by one over 2k of a block length, and the overhangs add up. The picture shows five blocks: the top one is already entirely beyond the table.
The total overhang is half a harmonic sum, which diverges, so a tall enough stack reaches as far out as you like. The cautious builder's overhangs, one over 2k plus 3, are smaller, so direct comparison with the harmonic series is silent. The limit comparison is not: the ratio to one over k tends to one half, so the cautious sum diverges too.
Divergence guarantees reach, but it says nothing about cost. Two block lengths of overhang takes 31 blocks for the bold builder and 109 for the cautious one, and each further length costs far more than the last, because the sums grow only like a logarithm.
Section
Part 6
Pattern
Figure (svg): A flow diagram: keep the dominant terms to get a comparison series b sub n; if the inequality points the useful way, use the direct test; otherwise take the limit of a sub n over b sub n. A positive finite limit gives the same fate; zero with a convergent b gives convergence; infinity with a divergent b gives divergence; any other combination means choose a new b.
This is the order to work in. First confirm the terms are nonnegative, because both tests depend on it. Then read the dominant pieces: that gives you the comparison series, and you should immediately state whether it converges or diverges.
Next look at the direction of the easy inequality. If your terms are below a convergent comparison, or above a divergent one, the direct test finishes the job. If the inequality points the other way, take the limit of the ratio instead.
If the limit lands in a silent case, do not force a conclusion. Adjust the comparison, usually by choosing a power between the ones that failed, as in Example 5.18(c), or switch to a different test. The flow diagram is the same procedure as a picture.
Ranking
Put in order
Order the steps of a complete limit-comparison argument.
Why: The hypothesis comes first, the comparison is chosen from the dominant behaviour, and its verdict must be stated before the limit can be read. The conclusion depends on the limit and that verdict together.
Drag the steps into order before checking. The step people most often skip is stating the comparison series' fate before interpreting the limit.
That order matters because the limit alone does not decide anything unless it is positive and finite. A limit of zero means one thing with a convergent comparison and nothing with a divergent one, so you cannot read the case until you have written down the comparison's verdict.
Check
Check your understanding
Which argument proves that the sum of 1/(n² + 4n) converges?
Answer: A
Why: Adding 4n to the denominator makes each term smaller than one over n squared, and the reciprocal squares converge, so Theorem 5.11 part one applies. The sum in fact telescopes to 25 over 48.
Each option is a correct inequality or a correct limit, so the question is only about whether the arrangement concludes anything.
Only the first option is below a convergent series. The second is below a divergent series and the third is above a convergent one, both silent. The fourth is a zero limit against a divergent comparison, which is silent too. As a bonus, this series telescopes, and its exact sum is twenty-five forty-eighths, about 0.52.
Check
Check your understanding
The terms are positive and the ratio aₙ / (1/n^(3/2)) tends to 0. What follows?
Answer: A
Why: The comparison is a p-series with p three halves, which converges. A limit of zero means the terms are eventually smaller than the comparison terms, and smaller than convergent gives convergence: case two of Theorem 5.12.
Decide what the zero limit says about the sizes of the terms, then combine that with the comparison's fate.
A zero limit says your terms are eventually smaller than one over n to the three halves. That comparison is a convergent p-series, so smaller than convergent gives convergence: case two. The same zero limit against a divergent comparison would have been silent, which is why the fate of the comparison has to be part of every answer.
Check
Check your understanding
Which series is the best choice for a limit comparison with the sum of (n + 1)/(n³ + 2)?
Answer: A
Why: The dominant pieces are n on top and n cubed below, which leave one over n squared. The ratio then tends to 1, and since the reciprocal squares converge, so does the series. The ratio at n equal to 100 is already 1.01.
Use the dominant pieces. The numerator is dominated by n and the denominator by n cubed, so the terms behave like one over n squared.
The other options are all legitimate series to compare with, but each leads to a silent case. The harmonic series gives a zero limit with a divergent comparison. The reciprocal cubes and the geometric series give infinite limits with convergent comparisons. The best comparison is the one whose ratio tends to a positive number, and the dominant-pieces rule finds it first time.
Explain it to yourself
Discussion prompt
In two or three sentences, explain why a ratio tending to a positive finite limit L makes the two series share a fate, using only the comparison test.
Write your explanation before revealing. If you can explain this test from the comparison test, you never have to memorise its cases.
The key is that a limit is a statement about eventually. When the ratio tends to L, from some point on it stays between half of L and one and a half times L. That turns a single limit into two inequalities, one on each side, and the comparison test uses whichever one points the useful way for the comparison's fate.
Exit ticket
\[ \sum_{n=1}^{\infty}\frac{2n+1}{\sqrt{n^5+1}} \]
Discussion prompt
Choose a comparison series from the dominant behaviour, compute the limit of the ratio, and give the verdict with the case of Theorem 5.12 you used.
This combines every step of the lesson in one problem. Choose the comparison from the dominant pieces, compute the limit, and name the case.
The dominant pieces are two n on top and the square root of n to the fifth below, which is n to the five halves. That leaves two over n to the three halves, so compare with one over n to the three halves. Dividing the top and bottom of the ratio by n to the five halves shows the limit is two.
Two is positive and finite, and the comparison is a convergent p-series, so the series converges by case one. The numbers agree: the ratio is 2.001 at n equal to a thousand.
Recap
| test | what it needs | what it can conclude |
|---|---|---|
| comparison (5.11) | 0 ≤ aₙ ≤ bₙ, or aₙ ≥ bₙ ≥ 0, eventually | below convergent: converges; above divergent: diverges |
| limit comparison (5.12) | aₙ, bₙ ≥ 0 and the limit of aₙ/bₙ | 0 < L < ∞: same fate; L = 0 copies convergence; L = ∞ copies divergence |
| the silent cases | above convergent, below divergent, 0 with divergent, ∞ with convergent | nothing: choose another bₙ |
\[ \frac{L}{2}\,b_n < a_n < \frac{3L}{2}\,b_n \quad \text{eventually, when } a_n/b_n \to L > 0 \]
Both tests need a known series to compare with, and they cannot handle terms that change sign. Section 5.5 turns to alternating series.
Stewart, Calculus: Early Transcendentals 8e, §11.4 The Comparison Tests §11.4, pp. 727-731 — the same material in Stewart
Two tests, both built on the Monotone Convergence Theorem. The comparison test needs an inequality that points the useful way: below a convergent series, or above a divergent one. The limit comparison test replaces that inequality with a limit, and a positive finite limit makes the two series share a fate, because it supplies both inequalities at once.
Learn the silent cases as carefully as the working ones. Above convergent and below divergent say nothing, and so do a zero limit against a divergent series and an infinite limit against a convergent one.
Both tests have two limits of their own: they need a series with a known fate to compare with, and they need nonnegative terms. Section 5.5 deals with the second, series whose terms alternate in sign.
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