The Comparison Test For Series

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Sometimes we’re given a series that is very close to something we can deal with. For example, the series

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

is very close to the $p$ -series

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

In cases like this, we can use the comparison test to decide whether the series converges. First, remember that the second series above converges since it’s a p-series with $p=2$ . Second, notice that

$n^2+1 > n^2,$

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

(if you don’t understand or don’t believe it, plug in some numbers for $n$ .)

Since each term of the first series is smaller than each term of the second series, we expect that the first series converges since the second one does. That’s exactly what the comparison test tells us.

Details of The Comparison Test

Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms. If $a_n\leq b_n$ and $\sum b_n$ converges, so does $\sum a_n$ . On the other hand, if $a_n\leq b_n$ and $\sum a_n$ diverges, then so does $\sum b_n$ .

Also, notice that not every of the first series needs to be less than the corresponding series. It just needs to be true eventually. For example, the series

$1+ \frac{1}{2}+ \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots$

converges, and we can use this to show that

$3+2+1+\frac{1}{9}+\frac{1}{17} + \frac{1}{33} +\dots$

does, too, even though the first few terms are actually bigger than the corresponding terms of the first series.

The comparison test is useful especially in certain situations. Take a look at these problems using the comparison test:

Problem: Show that

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

converges.

Solution: Remember that $\sin x$ is never bigger than 1 and never less than -1. That means that

$0\leq |\sin n|\leq 1$

for every integer $n$ . Dividing each of these terms by $n^2$ , we get the inequality

$0\leq\frac{|\sin n|}{n^2}\leq \frac{1}{n^2}.$

We know that $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges because it’s a p-series with $p=2$ . Therefore the given series converges by the comparison test.

In the above problem, there was a $\sin n$ that was making things complicated. We used the comparison test to get rid of it and simplify the series we were looking at. That’s the general philosophy you should use when you’re thinking about the comparison test. Always compare to something simpler than what you’re looking at. Here’s another problem where we can use the comparison test to simply the series we’re thinking about: