Math Core

You now have two families of benchmark series whose behavior you know instantly: geometric series and p-series. The comparison tests let you decide the fate of a messier series by showing that it behaves like one of those benchmarks.

The idea

Suppose every term of a positive series is smaller than the matching term of a series you know converges. Then its partial sums are trapped below a finite total, so it must converge too. In the other direction, if every term is larger than the matching term of a divergent positive series, its partial sums are pushed up to infinity.

Direct comparison test

Suppose $0 \le a_n \le b_n$ for all $n$ (or for all $n$ past some point).

If $\sum b_n$ converges, then $\sum a_n$ converges. (Smaller than finite is finite.)
If $\sum a_n$ diverges, then $\sum b_n$ diverges. (Bigger than infinite is infinite.)

Common mistake

The other two combinations tell you nothing. Being smaller than a divergent series, or bigger than a convergent series, gives no conclusion. For example, $\frac{1}{n^2} \le \frac{1}{n}$ and $\sum \frac1n$ diverges, but $\sum \frac{1}{n^2}$ converges. Always check that your inequality points the useful way.

Choosing the comparison series

Look at the terms for large $n$ and keep only the dominant parts of the numerator and denominator. That tells you which benchmark to compare with.

Series	Dominant behavior	Benchmark
$\sum \frac{1}{n^2 + 5}$	$\frac{1}{n^2}$	p-series, $p = 2$
$\sum \frac{1}{3^n + 1}$	$\frac{1}{3^n}$	geometric, $r = \frac13$
$\sum \frac{\sqrt n}{n - 1}$	$\frac{1}{\sqrt n}$	p-series, $p = \frac12$

For $\sum \frac{1}{n^2 + 5}$ , the direct comparison is easy: $n^2 + 5 > n^2$ , so $\frac{1}{n^2 + 5} < \frac{1}{n^2}$ , and $\sum \frac{1}{n^2}$ converges. Therefore $\sum \frac{1}{n^2 + 5}$ converges.

But direct comparison can be awkward. For $\sum \frac{1}{n^2 - 5}$ , the terms are slightly bigger than $\frac{1}{n^2}$ , which is the useless direction, even though the series obviously behaves like $\sum \frac{1}{n^2}$ . The limit comparison test handles these cases cleanly.

The limit comparison test

Limit comparison test

Suppose $a_n > 0$ and $b_n > 0$ , and

\lim_{n\to\infty} \frac{a_n}{b_n} = L, \qquad \text{where } 0 < L < \infty.

Then $\sum a_n$ and $\sum b_n$ both converge or both diverge.

Before using either test, confirm that the terms are positive (at least eventually). The comparison tests are about positive series; for series with negative terms, you will use the alternating series test or absolute convergence, coming up in the next lessons.

The reason: if $\frac{a_n}{b_n} \to L$ , then for large $n$ , $a_n$ is roughly $L\,b_n$ , a constant multiple of $b_n$ , and constant multiples don't change convergence. The limit must be a positive, finite number for the test to apply.

For large x, 1/(x² − 0.5) and 1/x² are nearly identical, so their series behave the same way. That is exactly what limit comparison captures.Open in grapher →

Which comparison test to use

Both tests reach the same conclusions, so pick whichever is less work.

Use direct comparison when an inequality is obvious, typically when you are adding something positive to the denominator of a convergent benchmark (making terms smaller) or to the numerator of a divergent one (making terms larger).
Use limit comparison when the inequality points the wrong way or is messy to prove, which happens with differences like $n^2 - 5$ or with sums that have several competing terms.

On a free-response question, state the benchmark series, state whether it converges or diverges and why (for example, "p-series with $p = 2 > 1$ "), and then state the inequality or the limit.

Worked examples

Worked example: Direct comparison with a geometric series

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n + n}$ converges.

Solution. Since $2^n + n > 2^n$ , we have $0 < \dfrac{1}{2^n + n} < \dfrac{1}{2^n}$ . The series $\sum \left(\frac12\right)^n$ is geometric with $|r| = \frac12 < 1$ , so it converges. By the direct comparison test, $\sum \dfrac{1}{2^n + n}$ converges.

Worked example: Direct comparison for divergence

Determine whether $\displaystyle\sum_{n=3}^{\infty} \frac{\ln n}{n}$ converges.

Solution. For $n \ge 3$ , $\ln n > 1$ , so $\dfrac{\ln n}{n} > \dfrac{1}{n}$ . The harmonic series $\sum \frac1n$ diverges, and our series is bigger term by term, so $\sum \dfrac{\ln n}{n}$ diverges.

Worked example: Limit comparison with a rational series

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{2n^2 + 3}{n^4 - n + 5}$ converges.

Solution. For large $n$ , the terms behave like $\dfrac{2n^2}{n^4} = \dfrac{2}{n^2}$ , so compare with $b_n = \dfrac{1}{n^2}$ :

\lim_{n\to\infty} \frac{\dfrac{2n^2+3}{n^4 - n + 5}}{\dfrac{1}{n^2}} = \lim_{n\to\infty} \frac{2n^4 + 3n^2}{n^4 - n + 5} = 2.

Since $0 < 2 < \infty$ and $\sum \frac{1}{n^2}$ converges ( $p = 2$ ), the series converges.

Worked example: Limit comparison with a trig term

Determine whether $\displaystyle\sum_{n=1}^{\infty} \sin\!\left(\frac{1}{n}\right)$ converges.

Solution. For small angles $\sin\theta \approx \theta$ , so compare with $b_n = \frac1n$ . Using $t = \frac1n \to 0^+$ :

\lim_{n\to\infty} \frac{\sin(1/n)}{1/n} = \lim_{t\to0^+} \frac{\sin t}{t} = 1.

The limit is positive and finite, and $\sum \frac1n$ diverges, so $\sum \sin\!\left(\frac1n\right)$ diverges.

Tip

When you set up a limit comparison, put the "messy" series on top and the benchmark on the bottom. If you get $L = 0$ or $L = \infty$ , you probably picked the wrong benchmark: go back and find the dominant terms again.

Practice

Practice 1

Which statement correctly uses the direct comparison test to analyze $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 5}$ ?

Practice 2

To apply the limit comparison test to $\displaystyle\sum_{n=1}^{\infty} \frac{3n^2 + 1}{n^4 + 2n}$ with $b_n = \dfrac{1}{n^2}$ , compute $\displaystyle\lim_{n\to\infty} \frac{a_n}{b_n}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{3^n + 7}$ converges or diverges.

Practice 4

Which benchmark series is best for a limit comparison with $\displaystyle\sum_{n=1}^{\infty} \frac{2n + 1}{n^2\sqrt{n}}$ , and what is the conclusion?

Practice 5

Suppose $0 \le a_n \le b_n$ for all $n$ . Which of the following must be true?

Practice 6

To test $\displaystyle\sum_{n=1}^{\infty} \left(1 - \cos\frac{1}{n}\right)$ , you compare it with $\displaystyle\sum \frac{1}{n^2}$ . Compute $\displaystyle\lim_{n\to\infty} \frac{1 - \cos(1/n)}{1/n^2}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Determine whether $\displaystyle\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n - 1}$ converges or diverges.