Comparison Test

Learning Outcomes

Use the comparison test to test a series for convergence

In the preceding two sections, we discussed two large classes of series: geometric series and p-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test. We begin by recalling a principle for comparing variable quantities in the denominator of a fraction.

Recall: Algebra of inequalities

If $0 < a_n \le b_n$ , then $\frac{1}{a_n} \ge \frac{1}{b_n}$ In other words, a larger denominator corresponds to a smaller fraction.

For example, consider the series

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

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}$ .

This series looks similar to the convergent series

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

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ .

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing.

Furthermore, since

0 < \frac{1}{n^{2} + 1} < \frac{1}{n^{2}}

$0<\frac{1}{{n}^{2}+1}<\frac{1}{{n}^{2}}$

for all positive integers $n$ , the $k\text{th}$ partial sum ${S}_{k}$ of $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}$ satisfies

S_{k} = \sum_{n = 1}^{k} \frac{1}{n^{2} + 1} < \sum_{n = 1}^{k} \frac{1}{n^{2}} < \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

${S}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}+1}<\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}}<\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ .

(See Figure 1 (a) and the table below.) Since the series on the right converges, the sequence $\left\{{S}_{k}\right\}$ is bounded above. We conclude that $\left\{{S}_{k}\right\}$ is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, $\left\{{S}_{k}\right\}$ converges, and thus

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

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}$

converges.

Similarly, consider the series

\sum_{n = 1}^{\infty} \frac{1}{n - \frac{1}{2}}

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{n - \frac{1}{2}}$ .

This series looks similar to the divergent series

\sum_{n = 1}^{\infty} \frac{1}{n}

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ .

The sequence of partial sums for each series is monotone increasing and

\frac{1}{n - \frac{1}{2}} > \frac{1}{n} > 0

$\frac{1}{n - \frac{1}{2}}>\frac{1}{n}>0$

for every positive integer $n$ . Therefore, the $k\text{th}$ partial sum ${S}_{k}$ of $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n - \frac{1}{2}}$ satisfies

S_{k} = \sum_{n = 1}^{k} \frac{1}{n - \frac{1}{2}} > \sum_{n = 1}^{k} \frac{1}{n}

${S}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{n - \frac{1}{2}}>\displaystyle\sum _{n=1}^{k}\frac{1}{n}$ .

(See Figure 1 (b) and the following table.) Since the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ diverges to infinity, the sequence of partial sums $\displaystyle\sum _{n=1}^{k}\frac{1}{n}$ is unbounded. Consequently, $\left\{{S}_{k}\right\}$ is an unbounded sequence, and therefore diverges. We conclude that

\sum_{n = 1}^{\infty} \frac{1}{n - \frac{1}{2}}

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{n - \frac{1}{2}}$

diverges.

This shows two graphs side by side. The first shows plotted points for the partial sums for the sum of 1/n^2 and the sum 1/(n^2 + 1). Each of the partial sums for the latter is less than the corresponding partial sum for the former. The second shows plotted points for the partial sums for the sum of 1/(n - 0.5) and the sum 1/n. Each of the partial sums for the latter is less than the corresponding partial sum for the former.

Comparing a series with a p-series (p = 2)
$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}+1}$	$0.5$	$0.7$	$0.8$	$0.8588$	$0.8973$	$0.9243$	$0.9443$	$0.9597$
$\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}}$	$1$	$\begin{array}{l}1.25\hfill \end{array}$	$1.3611$	$1.4236$	$1.4636$	$1.4914$	$1.5118$	$1.5274$

Comparing a series with the harmonic series
$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\displaystyle\sum _{n=1}^{k}\frac{1}{n - \frac{1}{2}}$	$2$	$2.6667$	$3.0667$	$3.3524$	$3.5746$	$3.7564$	$3.9103$	$4.0436$
$\displaystyle\sum _{n=1}^{k}\frac{1}{n}$	$1$	$1.5$	$\begin{array}{l}1.8333\hfill \end{array}$	$2.0933$	$2.2833$	$2.45$	$2.5929$	$2.7179$

Theorem: Comparison Test

Suppose there exists an integer $N$ such that $0\le {a}_{n}\le {b}_{n}$ for all $n\ge N$ . If $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ converges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges.
Suppose there exists an integer $N$ such that ${a}_{n}\ge {b}_{n}\ge 0$ for all $n\ge N$ . If $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ diverges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.

Proof

We prove part i. The proof of part ii. is the contrapositive of part i. Let $\left\{{S}_{k}\right\}$ be the sequence of partial sums associated with $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ , and let $L=\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ . Since the terms ${a}_{n}\ge 0$ ,

S_{k} = a_{1} + a_{2} + \dots + a_{k} \leq a_{1} + a_{2} + \dots + a_{k} + a_{k + 1} = S_{k + 1}

${S}_{k}={a}_{1}+{a}_{2}+\cdots +{a}_{k}\le {a}_{1}+{a}_{2}+\cdots +{a}_{k}+{a}_{k+1}={S}_{k+1}$ .

Therefore, the sequence of partial sums is increasing. Further, since ${a}_{n}\le {b}_{n}$ for all $n\ge N$ , then

\sum_{n = N}^{k} a_{n} \leq \sum_{n = N}^{k} b_{n} \leq \sum_{n = 1}^{\infty} b_{n} = L

$\displaystyle\sum _{n=N}^{k}{a}_{n}\le \displaystyle\sum _{n=N}^{k}{b}_{n}\le \displaystyle\sum _{n=1}^{\infty }{b}_{n}=L$ .

Therefore, for all $k\ge 1$ ,

S_{k} = (a_{1} + a_{2} + \dots + a_{N - 1}) + \sum_{n = N}^{k} a_{n} \leq (a_{1} + a_{2} + \dots + a_{N - 1}) + L

${S}_{k}=\left({a}_{1}+{a}_{2}+\cdots +{a}_{N - 1}\right)+\displaystyle\sum _{n=N}^{k}{a}_{n}\le \left({a}_{1}+{a}_{2}+\cdots +{a}_{N - 1}\right)+L$ .

Since ${a}_{1}+{a}_{2}+\cdots +{a}_{N - 1}$ is a finite number, we conclude that the sequence $\left\{{S}_{k}\right\}$ is bounded above. Therefore, $\left\{{S}_{k}\right\}$ is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that $\left\{{S}_{k}\right\}$ converges, and therefore the series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges.

$_\blacksquare$

To use the comparison test to determine the convergence or divergence of a series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ , it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and p-series, these series are often used. If there exists an integer $N$ such that for all $n\ge N$ , each term ${a}_{n}$ is less than each corresponding term of a known convergent series, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges. Similarly, if there exists an integer $N$ such that for all $n\ge N$ , each term ${a}_{n}$ is greater than each corresponding term of a known divergent series, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.

Try It

Example: Using the Comparison Test

For each of the following series, use the comparison test to determine whether the series converges or diverges.

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}+3n+1}$
$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{n}+1}$
$\displaystyle\sum _{n=2}^{\infty }\frac{1}{\text{ln}\left(n\right)}$

Show Solution

Compare to $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}$ Since $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}$ is a p-series with $p=3$ , it converges. Further,

$\frac{1}{{n}^{3}+3n+1}<\frac{1}{{n}^{3}}$

for every positive integer $n$ . Therefore, we can conclude that $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}+3n+1}$ converges.
Compare to $\displaystyle\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n}$ . Since $\displaystyle\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n}$ is a geometric series with $r=\frac{1}{2}$ and $|\frac{1}{2}|<1$ , it converges. Also,

$\frac{1}{{2}^{n}+1}<\frac{1}{{2}^{n}}$

for every positive integer $n$ . Therefore, we see that $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{n}+1}$ converges.
Compare to $\displaystyle\sum _{n=2}^{\infty }\frac{1}{n}$ . Since

$\frac{1}{\text{ln}\left(n\right)}>\frac{1}{n}$

for every integer $n\ge 2$ and $\displaystyle\sum _{n=2}^{\infty }\frac{1}{n}$ diverges, we have that $\displaystyle\sum _{n=2}^{\infty }\frac{1}{\text{ln}\left(n\right)}$ diverges.

try it

Use the comparison test to determine if the series $\displaystyle\sum _{n=1}^{\infty }\frac{n}{{n}^{3}+n+1}$ converges or diverges.

Hint

Find a value $p$ such that $\frac{n}{{n}^{3}+n+1}\le \frac{1}{{n}^{p}}$ .

Show Solution

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.4.1” here (opens in new window).

Module 5: Sequences and Series