Comparison Tests — AP® Calculus BC Notes

Most series cannot be summed exactly, and many are neither geometric nor $p$ -series. The comparison tests let you borrow the answer from a series you already know: set the new series beside a geometric series or a $p$ -series and read its behavior from the comparison.

Why positive terms make comparison work

When every term $a_n$ is positive, each partial sum $S_n$ is larger than the one before. An increasing list of numbers can do only two things: level off at a limit, or grow past every bound. So a positive series converges exactly when its partial sums stay under some ceiling. A comparison gives you that ceiling, or shows there is none.

Partial sums S sub N for N = 1 to 12 of two series, plotted as dots. The upper dots, for the series of 1 over n squared, rise from 1 and level off just under a dashed line at pi squared over 6, about 1.645. The lower dots, for the series of 1 over (n squared + 1), rise from 0.5 and level off near 1.08, always below the upper dots. Smaller positive terms give smaller partial sums, trapped under the same ceiling. — **Figure 1.** Smaller positive terms give smaller partial sums under the same ceiling. LevlPrep original diagram.

The figure shows the idea with $\sum \frac{1}{n^2 + 1}$ and $\sum \frac{1}{n^2}$ . Each term $\frac{1}{n^2 + 1}$ is below $\frac{1}{n^2}$ , so each partial sum of the first series is below the matching partial sum of the second, and those level off below $\frac{\pi^2}{6}$ .

The direct comparison test

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

If $\sum b_n$ converges, then $\sum a_n$ converges. Smaller than convergent: converges.
If $\sum a_n$ diverges, then $\sum b_n$ diverges. Larger than divergent: diverges.

The other two cases prove nothing. A series smaller than a divergent one may converge ( $\frac{1}{n^2} < \frac{1}{n}$ ) or diverge ( $\frac{1}{2n} < \frac{1}{n}$ ). A series larger than a convergent one may do either too.

Worked example 1. Does $\sum_{n=1}^{\infty} \frac{1}{4^n + 3}$ converge?

The terms are positive. Since $4^n + 3 > 4^n$ , the fractions satisfy $0 < \frac{1}{4^n + 3} < \frac{1}{4^n}$ for every $n$ .

$\sum \frac{1}{4^n}$ is geometric with ratio $r = \frac{1}{4}$ , and $|r| < 1$ , so it converges.

By the direct comparison test, $\sum \frac{1}{4^n + 3}$ converges.

Worked example 2. Does $\sum_{n=1}^{\infty} \frac{n + 1}{n^2}$ converge?

For every $n \ge 1$ , $\frac{n + 1}{n^2} > \frac{n}{n^2}$ $= \frac{1}{n} > 0$ .

The harmonic series $\sum \frac{1}{n}$ diverges, and our terms are larger, so $\sum \frac{n + 1}{n^2}$ diverges by the direct comparison test.

The limit comparison test

Often the obvious comparison points the wrong way. The terms of $\sum \frac{1}{n^2 - 3}$ (from $n = 2$ ) are a little larger than $\frac{1}{n^2}$ , so a direct comparison with a convergent series says nothing. Yet for large $n$ the two terms are almost equal. The limit comparison test turns "almost equal" into a proof.

Suppose $a_n > 0$ and $b_n > 0$ , and $\lim_{n \to \infty} \frac{a_n}{b_n}$ $= L$ where $L$ is a finite positive number. Then $\sum a_n$ and $\sum b_n$ both converge or both diverge.

Worked example 3. Does $\sum_{n=1}^{\infty} \frac{2n + 5}{n^3 + 1}$ converge?

Choose $b_n$ by keeping the highest power on top and bottom: $\frac{2n}{n^3} = \frac{2}{n^2}$ , so use $b_n = \frac{1}{n^2}$ .

$\lim_{n \to \infty} \frac{a_n}{b_n}$ $= \lim_{n \to \infty} \frac{(2n + 5)n^2}{n^3 + 1}$ $= \lim_{n \to \infty} \frac{2n^3 + 5n^2}{n^3 + 1}$ $= 2.$

The limit is finite and positive, and $\sum \frac{1}{n^2}$ converges because it is a $p$ -series with $p = 2 > 1$ . By the limit comparison test, the series converges.

Choosing the series to compare with

Term looks like	Compare with	Known behavior
a ratio of polynomials in $n$	$\frac{1}{n^p}$ with $p$ = (bottom degree) $-$ (top degree)	converges if $p > 1$
roots, such as $\frac{1}{\sqrt{n} + 2}$	$\frac{1}{n^{1/2}}$	$p = \frac{1}{2}$ : diverges
exponentials, such as $\frac{1}{3^n + n}$	$\frac{1}{3^n}$	geometric, $r = \frac{1}{3}$ : converges
a bounded factor, such as $\frac{\cos^2 n}{n^3}$	$\frac{1}{n^3}$ , using $0 \le \cos^2 n \le 1$	converges

Writing it the way graders want

Name the comparison series and say why its behavior is known: "geometric with $r = \frac{1}{4}$ " or "a $p$ -series with $p = 2 > 1$ ". Do not call a series harmonic unless it is $\sum \frac{1}{n}$ . For direct comparison, state the inequality and that the terms are positive. For limit comparison, write the limit in $n$ , compute it, and say it is finite and positive. Finally, say what the series does: "the series converges", not "the terms converge", since the terms approach 0 for every series that converges and for many that do not.

Both tests are for series with positive terms. A series whose terms change sign needs a different test, which comes next.