Hi everyone! In this lesson we’ll add another two more convergence tests to our series convergence/divergence toolbox. We’ll continue adding to this toolbox over the next few lessons.

Lesson 16: Comparison Tests

Learning goals:

  • Use the comparison test to determine whether a series converges or diverges.
  • Use the limit comparison test to determine whether a series converges or diverges.

Topic:

WeBWorK:

  • Series – Comparison Tests

Motivating question

If we know about the convergence or divergence of one series, how can we use that knowledge to determine the convergence or divergence of another series?

Warmup exercise 1

Determine whether the series converges or diverges:

    \[\sum_{n=1}^\infty \frac{1}{n^2}\]

Show answer 1

This is an example of a p-series \sum_{n=1}^\infty \frac{1}{n^p}. Since p=2 is greater than 1, the series converges.

Warmup exercise 2

Determine whether the series converges or diverges:

    \[\sum_{n=1}^\infty \frac{1}{n}\]

Show answer 2

This is an example of a p-series \sum_{n=1}^\infty \frac{1}{n^p}. Since p=1 is not greater than 1, the series diverges.

(Direct) comparison test

The comparison test is sometimes called the direct comparison test to contrast it with the limit comparison test, which we’ll investigate later in today’s lesson. Before we give the formal definition of the test, let’s return to the blobs from Lesson 10 (link here).

Figure 1: Blobs

Remember that the blobs helped us when we were comparing improper integrals to see if they converged or diverged. With respect to the blobs, we said:

All we care about is whether the blobs are big or small. We don’t have a scale, so all we have to go on is how the blobs compare to one another. Here are two claims we can make:

  1. Let’s assume that blob A is big. Well, we know that blob B is even bigger than blob A. So we can conclude that blob B must also be big.
  2. Let’s assume that blob B is small. Well, we know that blob A is even smaller than blob B. So we can conclude that blob A must also be small.

Just like the comparison theorem for improper integrals, the comparison test for infinite series uses formalizations of these comparisons as well.

Theorem (Comparison test): Let \sum_{n=1}^\infty a_n and \sum_{n=1}^\infty b_n be two series. Assume 0 \leq a_n \leq b_n for all n (this means that a_n is playing the role of blob A and b_n is playing the role of blob B).

  1. Assume that \sum_{n=1}^\infty a_n diverges. Well, we know that b_n \geq a_n. So we can conclude that the series \sum_{n=1}^\infty b_n also diverges.
  2. Assume that \sum_{n=1}^\infty b_n converges. Well, we know that a_n \leq b_n. So we can conclude that the series \sum_{n=1}^\infty a_n also converges.

Remember that if the series \sum_{n=1}^\infty a_n converges, this can’t tell us anything about whether the series \sum_{n=1}^\infty b_n converges or diverges. Likewise, if \sum_{n=1}^\infty b_n diverges, this can’t tell us anything about \sum_{n=1}^\infty a_n.

Examples

Often when applying the direct comparison test, the trickiest part is choosing the second series to compare our series to. It’s good if we can start with a guess about whether our series converges or diverges, and then use that guess to choose the second series. Video 1 below takes us through the application of the direct comparison test and three examples.

Video 1

Video 2 shows the application of the direct comparison test to another example. Notice that the divergence test from Lesson 15 (link here) is used first; it doesn’t end up being conclusive, so another test has to be used. It’s always a good idea to try the divergence test first because if it is conclusive, it is usually ends up being faster than other tests.

Video 2

Videos 3, 4, and 5 show three more examples of the direct comparison test. Keep in mind, if the direct comparison test is inconclusive, that just means that the series we chose to compare our original series to is not helpful. It doesn’t mean that there’s no helpful comparison that we could make. But often if the first or second comparisons don’t yield a conclusion with the direct comparison test, it’s time to try out another test from the series convergence/divergence toolbox.

Video 3
Video 4
Video 5

Limit comparison test

The limit comparison test might be thought of as the “indirect” comparison test. To apply this test, we’ll still need to choose a second series whose convergence properties we know to compare our original series. It might be helpful to think of the limit comparison test as the test we can use when the direct comparison test doesn’t quite work out. The example in Video 5 is a good candidate for the limit comparison test. The limit comparison test is more sensitive than the direct comparison test and it applies in more examples.

Intuition

Before we state the theorem, let’s imagine a simpler situation to build some intuition. Again, we’ll have two series \sum_{n=1}^\infty a_n and \sum_{n=1}b_n. Again, we’ll assume that all terms are positive. For now, let’s also assume that \frac{a_n}{b_n} = L for all n. We can rewrite this relationship as a_n = L b_n. In particular, a_n is a constant multiple of b_n (and vice versa).

Let’s examine three situations, depending on L (this won’t be 100% precise, but it will help us understand the precise statement later):

  1. Assume that L is some finite, nonzero number. Then \sum_{n=1}a_n = \sum_{n=1}Lb_n = L \sum_{n=1}a_n. That is, the whole series \sum_{n=1}a_n is a nonzero constant multiple of the series \sum_{n=1}b_n (and vice versa). This means that if \sum_{n=1}a_n converges, then \sum_{n=1}a_n also converges. It also means that if \sum_{n=1}b_n, then \sum_{n=1}a_n also converges. So they either both converge or both diverge.
  2. Now assume that L is arbitrarily close to 0. This means that a_n is waaaaay smaller than b_n. This also means that \sum_{n=1}a_n is waaaay smaller than \sum_{n=1}b_n. Then by the direct comparison test, if \sum_{n=1}b_n converges, \sum_{n=1}a_n also converges.
  3. Finally, assume that L is an arbitrarily large number. This means that a_n is waaaaay bigger than b_n, so \sum_{n=1}a_n is waaaay bigger than \sum_{n=1}b_n. Then by the direct comparison test, if \sum_{n=1}b_n diverges, \sum_{n=1}a_n also diverges.

Theorem statement

For the limit comparison test, we don’t assume that \frac{a_b}{b_n} = L but we do assume that \lim_{n \to \infinity} \frac{a_b}{b_n} = L. You can think of this assumption as saying that as n gets large, \frac{a_b}{b_n} is eventually, practically L. This is good enough to draw the conclusions in the theorem.

Theorem (limit comparison test): Let a_n and b_n be positive for all natural numbers n. Let \lim_{n \to \infty} \frac{a_n}{b_n} = L.

  1. If L \neq 0, L \neq \infty, then the series \sum_{n=1}a_n and \sum_{n=1}b_n either both converge or they both diverge.
  2. If L = 0 and if \sum_{n=1}b_n converges, then \sum_{n=1}a_n also converges.
  3. If L = \infty and if \sum_{n=1}b_n diverges, then \sum_{n=1}a_n also diverges.

Remark about notation: Saying \lim_{n \to \infty} \frac{a_n}{b_n} = L might imply that the limit actually exists, so then it wouldn’t make sense to then say that L = \infty, as we did in case 3 of the statement. If you’re worried about this and want to make this more precise, you can just assume that \lim_{n \to \infty} \frac{a_n}{b_n} = \infty without mentioning L.

Examples

Again, a good candidate for an example where the limit comparison test applies is one where we tried to apply the direct comparison test and it didn’t quite work for us.

Video 6 below shows us an example of a series where both the limit comparison test and the direct comparison test apply.

Video 6

Recall that the direct comparison in Video 5 above was inconclusive. Video 7 below revisits this example with the limit comparison test.

Video 7

Videos 8, 9, 10, and 11 show still more examples of the limit comparison test. For these, pause the video after you see the series in question and see if you can apply the limit test yourself before you watch the rest of the video.

Video 8
Video 9
Video 10
Video 11

Summary

For both the direct comparison test and the limit comparison test, we have to choose a second series to compare our series to. We have to know whether that series converges or diverges. Depending on whether the comparison goes in the right direction for the direct comparison test, we may be able to use this test or we may have to try the limit comparison test.

Exit ticket

Determine whether the series converge or diverge. State which convergence test you are using.

  1.     \[\sum_{n=1}^\infty \frac{9n^3}{3n^5+5}\]

  2.     \[\sum_{n=1}^\infty \frac{9n^3}{3n^5-5}\]