10.6 Comparison Tests for Convergence

Comparison tests let you decide if a tricky positive-term series converges or diverges by comparing it with a simpler series you already understand, like a p-series or a geometric series. The direct comparison test uses an inequality between terms, while the limit comparison test checks the limit of the ratio of terms to see if two series behave the same way. For AP Calculus BC, name the comparison series and explain why the test applies.

How Do Comparison Tests Work?

Comparison tests decide convergence by matching a difficult positive-term series with a known benchmark series. Use direct comparison when you can prove the right inequality, and use limit comparison when the dominant terms show the two series have the same long-run behavior.

Comparison tests are BC-only, and multiple-choice questions often expect you to pick the right convergence test quickly when a series looks like a messier p-series or geometric series, while free-response credit depends on choosing an appropriate test and justifying it. Spotting which simple series controls a complicated one is the same instinct you will use later with the ratio test, alternating series, and power series convergence.

Direct and Limit Comparison Test Conditions

• Both comparison tests require terms that are nonnegative (at least for large $n$).
• Direct comparison test: if $0 \leq a_n \leq b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges; if $\sum a_n$ diverges, then $\sum b_n$ diverges.
• Limit comparison test: if $\lim_{n\to\infty}\frac{a_n}{b_n}$ is a positive finite number, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.
• Pick your comparison series by looking at the dominant term, usually the highest power of $n$ or the fastest-growing exponential.
• Common comparison series are p-series ($\sum 1/n^p$, converges for $p>1$) and geometric series ($\sum ar^n$, converges for $|r|<1$).
• Use direct comparison when a clean inequality is easy to set up, and limit comparison when the inequality is awkward but the end behavior is clear.

Comparison Test Theorems

Comparison tests are useful when a series is too messy to test directly but looks similar to a series you already know. There are two versions.

Direct Comparison Test

For two series $\sum a_n$ and $\sum b_n$ with $a_n,\ b_n \geq 0$ and $a_n \leq b_n$:

1. $\sum a_n$ converges if $\sum b_n$ converges.
2. $\sum b_n$ diverges if $\sum a_n$ diverges.

In plain English, both series must have nonnegative terms, and your first series must be term-by-term smaller than the second. If the larger series converges, the smaller one must also converge. If the smaller series diverges, the larger one must also diverge.

Limit Comparison Test

Sometimes a clean inequality is hard to set up. In that case, use the limit comparison test. For two series $\sum a_n$ and $\sum b_n$ with $a_n,\ b_n \geq 0$, both series either converge or both diverge if

$$\lim_{n\to\infty}\frac{a_n}{b_n}$$

is positive and finite.

Here you compare end behavior instead of individual terms. If the limit were 0, then $a_n$ is much smaller than $b_n$. If the limit were $\infty$, then $a_n$ is much larger than $b_n$. But if you get a positive, finite value (like 4), the two series behave the same way, so if one converges the other converges, and if one diverges the other diverges.

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Breaking Down the Theorems

Practice Walkthrough 1

Determine if the following series converges.

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

This is hard to test directly, so compare it to

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

The direct comparison test works here because both series have positive terms and $\frac{5}{2n^2+4n+3}\leq \frac{5}{n^2}$ (since $2n^2+4n+3\geq n^2$). By the p-series test, $\sum\frac{5}{n^2}$ converges, so the original series converges too.

Practice Walkthrough 2

Determine if the following series converges.

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

Here the inequality is awkward, so use the limit comparison test. Set

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

and compare to

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

Take the limit of the ratio:

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

Since 1 is positive and finite, the comparison is valid. The series $\sum_{n=1}^\infty \frac{1}{3^n}$ converges by the geometric series test, so the original series converges too.

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Choosing a Comparison Series and Justifying It

Problem Solving:

• Start by identifying the dominant term in the numerator and denominator. That tells you which p-series or geometric series to compare against.
• Confirm the terms are nonnegative (at least for large $n$) before applying either test.
• Choose direct comparison when the inequality direction is clean and matches what you need. Choose limit comparison when the algebra of an inequality is messy but the end behavior is clear.

Common Trap: When you pick a comparison series, make sure the inequality actually points the right way for direct comparison. If you want to prove convergence, your terms must be less than or equal to a convergent series. If you want to prove divergence, your terms must be greater than or equal to a divergent series. An inequality in the wrong direction proves nothing.

Showing Your Work: State the comparison series, confirm nonnegative terms, show the inequality or the limit of the ratio, and then name the test you use on the comparison series (p-series, geometric, harmonic). Writing each step clearly is important for clear exam work on series questions.

Practice 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

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 take the limit:

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

This limit equals 1, which is positive and finite, so the comparison is valid. The comparison series $\sum \frac{2}{\sqrt{n}}$ is a p-series with $p=\frac{1}{2}\leq 1$, so it diverges. Therefore the original series diverges.

Problem 2 Solution

Use the limit comparison test, comparing to

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

Then take the limit:

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

This limit equals 1, which is positive and finite, so the comparison is valid. The comparison series is the harmonic series $\sum \frac{1}{n}$, which diverges, so the original series diverges.

Problem 3 Solution

Use absolute comparison with $b_n=\frac{1}{n^3}$. Since $|\sin(n)|\leq 1$ for all $n$,

$$\left|\frac{\sin(n)}{n^3}\right|\leq\frac{1}{n^3}$$

The comparison series $\sum \frac{1}{n^3}$ is a p-series with $p=3>1$, so it converges. Therefore the original series converges absolutely, so it converges.

(Note: since $\sin(n)$ can be negative, compare absolute values instead of applying direct comparison to the original signed terms.)

Common Misconceptions

• Misconception: Comparison tests apply directly to series with positive and negative terms.
  Actually: Comparison tests apply to series with nonnegative terms. You cannot apply them directly to a series that swings between positive and negative without adjusting, such as using absolute values.
• Misconception: The direction of the inequality does not matter in direct comparison.
  Actually: The direction of the inequality matters in direct comparison. Being smaller than a divergent series tells you nothing, and being larger than a convergent series tells you nothing.
• Misconception: A limit comparison value of 0 or $\infty$ automatically settles the question.
  Actually: A limit comparison value of 0 or $\infty$ does not automatically end things. A positive finite limit is what guarantees the two series share the same behavior.
• Misconception: The choice of comparison series does not matter.
  Actually: Picking the wrong comparison series leads nowhere. Always match the dominant term, so for a rational function compare to the ratio of leading powers of $n$.
• Misconception: If the terms of a series go to 0, the series converges.
  Actually: Converging terms going to 0 is not the same as the series converging. The terms going to 0 is necessary but not enough, which is why you still need a comparison or another test.

Related AP Calculus Guides

• Unit 10 Review: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.4 Integral Test for Convergence