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**:

- Volume 2, Section 5.4
*Comparison Tests*(link to textbook section)

**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).

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:

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

- 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.
- 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 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.

**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.

## 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):

- 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.
- 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.
- 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 $\lim_{n \to \infty} \frac{a_n}{b_n} = L$.

- 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.
- If $L = 0$ and if $\sum_{n=1}b_n$ converges, then $\sum_{n=1}a_n$ also converges.
- 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.

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

**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.

## 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.

### Additional Video Resources

#### Exit ticket

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

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