#10.6 Comparison Tests for Convergence

Welcome to AP Calc 10.6! In this lesson, you’ll how to test for convergence by comparing your function to an easier one!

#➕ Comparison Test Theorems

In calculus, we use comparison tests when we are dealing with a series that is too complicated to determine the convergence of directly. There are two types of comparison tests, so we’ll break them both down here!

#➡️ Direct Comparison Test

The Direct Comparison Test states that for two series, $\sum a_n$ and $\sum b_n$ where $a_n,\ b_n\geq0$ and $a_n\leq b_n$ ,

(1) $\sum a_n$ converges if $\sum b_n$ converges and

(2) $\sum b_n$ diverges if $\sum a_n$ .

Let’s think this through for a second and put it in plain English! We have two series, both of which have to be positive ( $a_n,\ b_n\geq0$ is our first condition). Our first function must be smaller than our second function ( $a_n\leq b_n$ is our second condition). If our larger series converges, then a smaller one must also converge. Likewise, if our smaller series diverges, a larger series must also diverge.

We’ll get into when this test is appropriate to use in the examples!

#🔁 Limit Comparison Test

Sometimes we can’t directly compare a function to another one. For these scenarios, we use an indirect method—the Limit Comparison Test, which states that for two series, $\sum a_n$ and $\sum b_n$ where $a_n,\ b_n\geq0$ both series either **converge or diverge if $\lim_{n\to\infty}\frac{a_n}{b_n}$ is positive and finite.

Let’s break this down! Similarly to the first comparison test, we have two series that are positive. But, we’re not directly comparing in this case. Instead, we’re comparing end behavior of our two functions. If our limit of $\frac{a_n}{b_n}$ was zero, it would mean that $b_n$ grew much fast than $a_n$ (it would have to grow to $\infty$ ). If the limit was $\infty$ , then $a_n$ would’ve had to grow much faster to outpace $b_n$ . But, if we have a positive, finite value (like $4$ , for example), then the two functions behave similarly and we can compare them—basically, if one converges, so does the other and if one diverges, so does the second one.

Let’s dig in to some examples so we can understand these tests better!

#🧱 Breaking Down the Theorems

Let’s try a few examples.

#🚶‍♀️Practice Walkthrough 1

(1) Determine if the following series converges.

$\sum_{n=1}^\infty\frac{5}{2n^2+4n+3}$

Without a comparison test, we wouldn’t have the tools to solve this.

If you said direct, you’d be correct! We can compare this series to

$\sum_{n=1}^\infty\frac{5}{n^2}$

We can make this comparison because both functions from the series are positive, and $\frac{5}{2n^2+4n+3}\leq \frac{5}{n^2}$ (this is because $2n^2+4n+3\geq n^2$ ). Using the p-series test, we know that $\sum\frac{5}{n^2}$ converges, so we know that our series also converges!

#🚶‍♀️Practice Walkthrough 2

Let’s try one more to practice the limit comparison test!

(2) Determine if the following series converges.

$\sum_{n=1}^\infty\frac{1}{3^n-n}$

This means that

$a_n=\frac{1}{3^n-n}$

We will try comparing to

$b_n=\frac{1}{3^n}$

We need to find the limit of the ratio between the two functions, like so:

$\lim_{n\to\infty}\frac{\frac{1}{3^n-n}}{\frac{1}{3^n}}\rightarrow \lim_{n\to\infty}\frac{3^n}{3^n-n}\rightarrow \lim_{n\to\infty}\frac{1}{1-\frac{n}{3^n}}\rightarrow \frac{\lim_{n\to\infty}1}{\lim_{n\to\infty}(1-\frac{n}{3^n})}=\frac{1}{1}=1$

Since 1 is positive and finite, we can compare these two series. And, since we know $\sum_{n=1}^\infty \frac{1}{3^n}$ converges using the geometric series test, then our series also converges!

#📝 Practice Problems

Now that you know the basics, try some out on your own!

#❓Problems

Determine whether the following series converge or diverge.

$1. \sum_{n=1}^\infty \frac{2n^2+3n}{\sqrt{5+n^5}}$

$2.\sum_{n=1}^\infty \frac{n^2+3}{n^3+1}$

$3.\sum_{n=1}^\infty \frac{\text{sin}(n)}{n^3}$

#💡 Problem 1 Solution

For this, we’ll apply the limit comparison test, comparing to

$b_n=\frac{2n^2}{\sqrt{n^5}}=\frac{2n^2}{n^{5/2}}=\frac{2}{n^{1/2}}=\frac{2}{\sqrt{n}}$

Then, we take the limit:

$\lim_{n\to\infty}\frac{\frac{2n^2+3n}{\sqrt{5+n^5}}}{\frac{2}{\sqrt{n}}}\rightarrow\lim_{n\to\infty}\frac{(2n^2+3n)\cdot \sqrt{n}}{2\sqrt{5+n^5}}$

Applying L’Hopital’s Rule, we get that the limit is equal to 1. Thus, we can compare the two series. We can apply the p-series test to our second series to find that it diverges. Thus, our original series also diverges.

#💡 Problem 2 Solution

For this, we will use a limit comparison test. We can compare it to

$b_n=\frac{1}{n}$

Then, we take the limit:

$\lim_{n\to\infty}\frac{\frac{n^2+3}{n^3+1}}{\frac{1}{n}}\rightarrow \lim_{n\to\infty}\frac{n^3+3n}{n^3+1}$

Applying L’Hopital’s again, we find that the limit is equal to 1. Thus, we can compare these two series. Since our comparison series is the harmonic series, which we know diverges, we conclude that our original series also diverges.

#💡 Problem 3 Solution

For this problem, we can use a direct comparison test with $b_n=\frac{1}{n^3}$ . This is true because $\text{sin}(n)\leq1$ for all values of $n$ , so

$\frac{\text{sin}(n)}{n}\leq\frac{1}{n^3}$

for all values of $n$ . We use the p-series test to find that $\sum b_n$ converges, therefore, our original series also converges.

#💫 Closing

Great work! Make sure to keep practicing picking between these two tests and build a toolkit of series to compare to. With all of that done, you’ll ace any comparison tests you come across 💯