This web page discusses one of the most powerful tests
for convergence of series of positive terms: the comparison test.
The main idea is, to determine if a series
$\sum _{n=1}^{\infty}{u}_{n}$
converges or not we should compare it with one,
$\sum _{n=1}^{\infty}{v}_{n}$
that we know the behaviour of. There are two forms
of the comparison test: the comparison test

itself
and the limit comparison test

which is is often slightly
easier to use but is just a slightly different form.

To apply ideas of monotonicity
from a previous web page
we shall assume all series here are series of *positive*
terms except where stated otherwise.

Comparison Test.

Suppose $\sum _{n=1}^{\infty}{u}_{n}$ and $\sum _{n=1}^{\infty}{v}_{n}$ are series of positive terms ${u}_{n}$, ${v}_{n}$. Then

(a) If there is a positive $K\in \mathbb{R}$ such that $\forall n\in \mathbb{N}0{u}_{n}\u2a7dK{v}_{n}$ and if $\sum _{n=1}^{\infty}{v}_{n}$ converges then $\sum _{n=1}^{\infty}{u}_{n}$ converges.

(b) If there is a positive $K\in \mathbb{R}$ such that $\forall n\in \mathbb{N}0K{v}_{n}\u2a7d{u}_{n}$ and if $\sum _{n=1}^{\infty}{v}_{n}$ diverges then $\sum _{n=1}^{\infty}{u}_{n}$ diverges.

**Proof.**

(a) Suppose that the conditions in (a) are satisfied and $\sum _{n=1}^{\infty}{v}_{n}=l$. Then as all ${v}_{n}$ are positive the sequence of partial sums $\sum _{k=1}^{n}{v}_{k}$ is increasing, hence $\sum _{k=1}^{n}{v}_{k}\u2a7dl$ for all $k\in \mathbb{N}$. We now look at partial sums ${s}_{n}=\sum _{k=1}^{n}{u}_{k}$ for the ${u}_{n}$s.

By assumptions on $K$ we have, for each $n$,

and therefore by monotonicity $\left({s}_{n}\right)$ converges as it is bounded. Hence $\sum _{n=1}^{\infty}{u}_{n}$ converges.

(b) Suppose that the conditions in (b) are satisfied. Then by monotonicity the sequence of partial sums from ${v}_{n}$ is unbounded and so for all positive $B\in \mathbb{R}$ there is $n\in \mathbb{N}$ such that ${v}_{1}+\cdots +{v}_{n}>\frac{B}{K}$.

By assumptions on $K$ we have, for each $n$,

As $B$ was arbitrary, this means that the sequence $\left({s}_{n}\right)$ of partial sums is unbounded. Hence $\sum _{n=1}^{\infty}{u}_{n}$ diverges.

Examples are given in another page.

The calculations involved in finding constants $K$ in the comparison test can be awkward. It is often easier to use the following form of the comparison test instead.

Limit Comparison Test.

Suppose $\sum _{n=1}^{\infty}{u}_{n}$ and $\sum _{n=1}^{\infty}{v}_{n}$ are series of positive terms ${u}_{n}$, ${v}_{n}$. Suppose also that

as $n\to \infty $. Then the series $\sum _{n=1}^{\infty}{u}_{n}$ and $\sum _{n=1}^{\infty}{v}_{n}$ are either both convergent or both divergent.

**Proof.**

Suppose that $\frac{{u}_{n}}{{v}_{n}}\to l>0$ as $n\to \infty $. Then let $\epsilon =l/2$, $K=l/2$, $M=3l/2$. Then there is $N\in \mathbb{N}$ such that $\forall n\u2a7eN\left|\frac{{u}_{n}}{{v}_{n}}-l\right|\epsilon $ and hence $\forall n\u2a7eN0K\frac{{u}_{n}}{{v}_{n}}M$. Thus for all $n\u2a7eN$ ${u}_{n}<M{v}_{n}$ and $K{v}_{n}<{u}_{n}$.

Now if $\sum _{n=1}^{\infty}{v}_{n}$ converges then so does $\sum _{n=N}^{\infty}{v}_{n}$ and so does $\sum _{n=N}^{\infty}{u}_{n}$ by the comparison test as ${u}_{n}<M{v}_{n}$ for $n\u2a7eN$. Hence $\sum _{n=1}^{\infty}{u}_{n}$ converges.

Similarly, if $\sum _{n=1}^{\infty}{v}_{n}$ diverges then so does $\sum _{n=N}^{\infty}{v}_{n}$ and so does $\sum _{n=N}^{\infty}{u}_{n}$ by the comparison test as $K{v}_{n}<{u}_{n}$ for $n\u2a7eN$. Hence $\sum _{n=1}^{\infty}{u}_{n}$ diverges.

Examples are given in another page.

This web page has introduced you to one of the most powerful techniques for proving (non)convergence of series consisting of positive terms: the comparison test. Strictly speaking you only need to learn the basic comparison test, but the limit comparison test is much more convenient to use in many situations.