5.2 The Comparison Test

What might have become clear is that determining convergence of series is more complicated than for sequences because the terms themselves are more complicated. Because of this,we want to create tools which will allow us to more easily deduce the properties of a series. One of these is the comparison test, which allows us to determine if a series is convergent by comparing it to a known series such as $\sum_{n=1}^{\infty}\frac{1}{n}$ or $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ .

Theorem 5.3 (Comparison Test)

Let $(x_{n})_{n}$ and $(a_{n})_{n}$ be sequence with no negative terms. Then

(i)

If $\sum_{n=1}^{\infty}a_{n}$ is convergent and $\exists N\in\mathbb{N}$ such that $x_{n}\leq a_{n}\ \forall n\geq N$ , then $\sum_{n=1}^{\infty}x_{n}$ also converges.
(ii)

If $\sum_{n=1}^{\infty}a_{n}$ is divergent and $\exists N\in\mathbb{N}$ such that $x_{n}\geq a_{n}\ \forall n\geq N$ , then $\sum_{n=1}^{\infty}x_{n}$ also diverges.

Proof.

(i)

Let $A=\sum_{n=1}^{\infty}a_{n}$ and note that since all the terms $x_{n}$ are positive, $S_{n}=\sum_{k=1}^{n}x_{k}$ is an increasing sequence. Then, since there exists some integer $N$ such that for all $k\geq N$ we have $x_{n}\leq a_{n}$ , we get

$\sum_{k=N}^{n}x_{k}\leq\sum_{k=N}^{n}a_{k}\leq\sum_{k=1}^{\infty}a_{k}=A.$ (5.17)

Therefore, for all $n\geq 1$ we have

$\displaystyle S_{n}$ $\displaystyle=a_{1}+a_{2}+\dots+a_{N-1}+\sum_{k=N}^{n}x_{k}$ (5.18)

$\displaystyle\leq a_{1}+a_{2}+\dots+a_{N-1}+A,$ (5.19)

and hence $S_{n}$ is bounded above. But now $S_{n}$ is both monotone and bounded and so by the MCT (4.5), $S_{n}$ must be convergent and thus the series $\sum_{n=1}^{\infty}x_{n}$ converges.
(ii)

Suppose by way of contradiction that $\sum_{n=1}^{\infty}x_{n}$ is convergent, then following the argument in part (i) implies that $\sum_{n=1}^{\infty}a_{n}$ is convergent, which is a contradiction, meaning $\sum_{n=1}^{\infty}x_{n}$ must be divergent.

∎

Note that we can also multiply the terms of the generating sequence by a positive constant, as this does not affect the convergence of the series, i.e. we can check that $\exists N\in\mathbb{N},c>0$ such that $x_{n}\leq ca_{n}$ or $x_{n}\geq ca_{n}\ \forall n\geq N$ .

We using the comparison test, often the most convenient series to compare to are the so-called ‘ $p$ -series’, that is, those of the form $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ .

Example 5.4.

Let $p\in\mathbb{R}$ . Then the series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converges if and only if $p>1$ .
Suppose $p\geq 2$ . Then $\frac{1}{n^{p}}\leq\frac{1}{n^{2}}$ for all $n\in\mathbb{N}$ , so $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent by the comparison test.
Now suppose $p\leq 1$ . Then $\frac{1}{n^{p}}\geq\frac{1}{n}$ for all $n\in\mathbb{N}$ , so $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is divergent by the comparison test.
In the case of $p\in(1,2)$ , the series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ can be proved to be convergent by the ‘integral test’.

5.2.1 The Ratio Test

Theorem 5.4 (Ratio Test)

Let $(x_{n})_{n}$ be a sequence of positive terms.

(i)

If $\exists N\in\mathbb{N}$ and $r<1$ such that for all $n\geq N,\frac{x_{n+1}}{x_{n}}\leq r$ , then $\sum_{n=1}^{\infty}x_{n}$ is convergent.
(ii)

If $\exists N\in\mathbb{N}$ and $r>1$ such that for all $n\geq N,\frac{x_{n+1}}{x_{n}}\geq r$ , then $\sum_{n=1}^{\infty}x_{n}$ is divergent.

Proof.

(i)

First note that $x_{N+1}\leq rx_{N}$ , $x_{N+2}\leq rx_{N+1}=r^{2}x_{N}$ , etc. and so $x_{N+k}\leq r^{k}x_{N}$ . Now consider the sequence given by

$a_{n}=r^{n-N}x_{N}=(x_{N}r^{-N})\cdot r^{n}.$ (5.20)

This is a geometric series which is convergent by theorem 5.2. Then notice that for all $n\geq N$ we have

$x_{n}=x_{N+(n-N)}\leq r^{n-N}x_{N}=a_{n},$ (5.21)

and hence by the comparison test $\sum_{n=1}^{\infty}x_{n}$ is convergent.

∎

In effect, what the ratio test says is that if $\lim_{n\to\infty}\frac{x_{x+1}}{x_{n}}$ exists and is less than 1, then the series converges and if it is greater than 1 then the series diverges. If the limit is equal to 1, then we cannot determine the behaviour of the series from the ratio test. The ratio test works best for geometric series because of the cancellation that happens when dividing the exponents. It also works well for some series containing factorials of $n$ for the same reason.

Example 5.5.

Show that the series $\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$ converges.
The generating sequence is given by $x_{n}=\frac{2^{n}}{n!}$ , so

\frac{x_{n+1}}{x_{n}}=\frac{\left(\frac{2^{n+1}}{(n+1)!}\right)}{\left(\frac{2% ^{n}}{n!}\right)}=\frac{2^{n+1}n!}{2^{n}(n+1)!}=\frac{2}{n+1}\to 0\text{ as }n% \to\infty.

(5.22)

Hence by the ratio test, the series converges.

	$\displaystyle S_{n}$	$\displaystyle=a_{1}+a_{2}+\dots+a_{N-1}+\sum_{k=N}^{n}x_{k}$		(5.18)
		$\displaystyle\leq a_{1}+a_{2}+\dots+a_{N-1}+A,$		(5.19)