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The Comparison and Limit Comparison Tests

Calculus 2 · Chapter 7: Convergence Tests for SeriesSection 7.2 · lesson 33 of 549 min · free

← The Integral Test and p-SeriesNext lesson: The Alternating Series Test →

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In this lesson

Measuring an unknown series against one you know.

  • Aim a direct comparison in the direction the conclusion needs
  • Recognise when the natural inequality points the wrong way
  • Use the ratio's limit to sidestep the inequality entirely
  • Choose b_{n} from the dominant powers of a_{n}

Transcript

Everything said in the video, section by section (1,392 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 7.2

Comparison and Limit Comparison

Section seven point one gave you the p-series benchmark. The sum of one over n to the p converges exactly when p exceeds one. But most series you meet aren't p-series. They're messier. One over n squared plus n plus one. Two n squared plus three n, over root of n to the seventh plus n plus one. The idea of this section is simple and powerful. If you can measure your messy series against a benchmark whose behavior you already know, you inherit that behavior. There are two ways to do that measurement. And the second exists because the first sometimes fails.

Direct Comparison

Two Directions, Not Interchangeable

Start with direct comparison. Suppose the terms of your series are trapped between zero and the terms of another series, for all n beyond some index. Then two conclusions follow. Part one. If the bigger series converges, the smaller one converges. Part two. If the smaller series diverges, the bigger one diverges. That structure should feel familiar. It mirrors exactly the Comparison Theorem for improper integrals, from section two point four. Bound above by a convergent series to prove convergence. Or bound below by a divergent series to prove divergence. And as before, the logic runs one way only. Bounding a series above by a divergent series proves nothing at all. And bounding below by a convergent one proves nothing either. The comparison must point in the direction that matches the conclusion you want.

Figure 7.2.1

The Two Directions, Drawn

The two pictures make the asymmetry concrete. On the left, every term a n sits at or below the terms of a convergent series. So the sum of a n converges. On the right, every term sits at or above the terms of a divergent series. So the sum diverges. Let's do one of each. First. Show that the sum of one over n squared plus n plus one converges. For every n at least one, the denominator n squared plus n plus one is greater than n squared. A bigger denominator means a smaller fraction. So the term is smaller than one over n squared. And the benchmark, sum of one over n squared, is a p-series with p equal to two. Which converges. By Part one, the smaller series converges. Second. Determine whether the sum of one over n minus one half converges. For large n this behaves like one over n. Which suggests divergence. So to use Part two, bound the term below by a divergent benchmark. Since n minus one half is less than n, taking reciprocals reverses the inequality. One over n minus one half is greater than one over n. The harmonic benchmark diverges. So by Part two, the series diverges.

The Weakness

When Direct Comparison Stalls

Now the practical weakness of direct comparison. It requires a clean term-by-term inequality. And that inequality must point the right way. Sometimes the natural benchmark points the wrong way entirely. Consider the sum of one over n squared minus n plus one. For large n this again behaves like one over n squared. So you'd expect convergence. But look at the denominator. n squared minus n plus one is less than n squared. A smaller denominator means a larger fraction. So the term is larger than one over n squared. Larger than the convergent benchmark. And bounding above by one over n squared fails, because the inequality is backward. A convergent lower bound proves nothing. So direct comparison stalls — even though the series clearly should converge. What's needed is a test that captures the idea of behaves like, without demanding a perfect inequality. That's precisely what the Limit Comparison Test provides.

Limit Comparison

Compare the Ratio Instead

Instead of comparing terms by an inequality, the Limit Comparison Test compares them by taking the limit of their ratio. If two positive series have terms that stay roughly proportional as n goes to infinity, they must share the same fate. Formally: let a n and b n both be positive, and let L be the limit of a n over b n. If L is a finite, positive number — strictly between zero and infinity — then both series converge, or both diverge. The intuition is worth holding onto. A finite, positive limit means that for large n, a n and b n are essentially constant multiples of each other. And multiplying a series by a nonzero constant cannot change whether it converges. So the test sidesteps the awkward inequality entirely. Only the ratio's limit matters. Now, the real skill in applying it is picking a good b n. And the reliable strategy is this. Keep only the dominant behavior of a n for large n. The highest-power terms in numerator and denominator. Discard everything else. That produces a p-series or geometric series whose behavior you know. For a ratio of polynomials, b n is the ratio of the leading terms. For terms with roots, keep the dominant power. Root of n squared plus one behaves like n.

Worked Examples 3 and 4

Succeeding Where Direct Comparison Failed

Let's return to the series that defeated direct comparison. The sum of one over n squared minus n plus one. Choose b n. The dominant denominator term is n squared. So take b n equal to one over n squared. A convergent p-series. Compute the limit of the ratio. a n over b n is one over n squared minus n plus one, divided by one over n squared. Which simplifies to n squared, over n squared minus n plus one. Divide numerator and denominator by n squared. You get one, over one minus one over n, plus one over n squared. As n goes to infinity, that's one over one minus zero plus zero. Which is one. L equals one. Finite and positive. So the two series share the same behavior. The benchmark converges, so the given series converges. The Limit Comparison Test succeeds exactly where direct comparison stalled. One more, with roots. Determine whether the sum of two n squared plus three n, over root of n to the seventh plus n plus one, converges. Keep dominant powers. The numerator behaves like two n squared. The denominator like root of n to the seventh, which is n to the seven halves. So a n behaves like two n squared over n to the seven halves. Which is two over n to the three halves. Drop the constant and take b n equal to one over n to the three halves. A p-series with p equal to three halves, greater than one. So convergent. Compute the limit. After dividing top and bottom by n to the seven halves, you get two plus three over n, over root of one plus small terms. Which goes to two over one. Two. L equals two. Finite and positive. And the benchmark converges. So the given series converges.

Key Takeaways

What to Carry Forward

Here's a question worth thinking through. The Limit Comparison Test asks for the limit of a n over b n to be finite and positive. Why must it be both? Suppose the limit were zero. Then a n is much smaller than b n. Knowing that the sum of b n converges would tell you something useful about the sum of a n. But knowing that it diverges would tell you nothing — the proportionality has broken. And if the limit were infinity, the reverse problem arises. Only a finite, positive limit guarantees that the two series are genuinely the same size in the long run. Locking their fates together. Which is exactly why choosing b n to match the dominant behavior of a n matters so much. A good choice forces the ratio toward a nonzero constant. Five takeaways. Both comparison tests apply to positive-term series, and work by comparing against a benchmark of known behavior. Usually a p-series or a geometric series. Direct Comparison: bound above by a convergent series to prove convergence. Bound below by a divergent one to prove divergence. Direct comparison requires the inequality to point the right way, and it stalls when the natural benchmark gives a backward inequality. Limit Comparison: if the limit of a n over b n is finite and positive, the two series both converge or both diverge. And choose b n by keeping only the dominant powers of a n. A good choice drives the ratio to a nonzero constant. Next up is section seven point three — the Alternating Series Test. The first test for series whose terms change sign.

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FAQ

Questions students ask us

About The Comparison and Limit Comparison Tests

What is the direct comparison test for series?
The direct comparison test says that if 0 \le a_n \le b_n for all n beyond some index, then convergence of \sum b_n implies convergence of \sum a_n, and divergence of \sum a_n implies divergence of \sum b_n. So you bound above by a convergent series to prove convergence, or below by a divergent series to prove divergence. The logic runs one way only. A divergent upper bound or a convergent lower bound proves nothing.
Why does direct comparison fail for \sum \frac{1}{n^2-n+1}?
Direct comparison fails because the natural inequality points the wrong way. For n \ge 2, n^2-n+1 < n^2, so \frac{1}{n^2-n+1} > \frac{1}{n^2}. At n=1 the two terms are equal. Being at or above the convergent series \sum \frac{1}{n^2} proves nothing. The limit comparison test fixes this. With b_n = \frac{1}{n^2}, the ratio's limit is 1, so the series converges.
How do you choose b_n for the limit comparison test?
Choose b_n by keeping only the dominant, highest-power terms of a_n in the numerator and denominator, then dropping constants. For example, \frac{2n^2+3n}{\sqrt{n^7+n+1}} behaves like \frac{2n^2}{n^{7/2}} = \frac{2}{n^{3/2}}, so take b_n = \frac{1}{n^{3/2}}. The limit of a_n/b_n is 2, which is finite and positive. Since p = \frac{3}{2} > 1, the benchmark converges, so the original series converges too.
Why must the limit in the limit comparison test be finite and positive?
The limit must be finite and positive because only then are a_n and b_n essentially constant multiples of each other for large n, which locks their fates together. If the limit is 0, a_n is much smaller than b_n. Convergence of \sum b_n then still implies convergence of \sum a_n, but divergence of \sum b_n tells you nothing. If the limit is infinite, the reverse problem occurs.

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