# The comparison tests

## 1. Introduction

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 $∑ n = 1 ∞ u n$ converges or not we should compare it with one, $∑ n = 1 ∞ v n$ 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.

## 2. The comparison test

Comparison Test.

Suppose $∑ n = 1 ∞ u n$ and $∑ n = 1 ∞ v n$ are series of positive terms $u n$, $v n$. Then

(a) If there is a positive $K ∈ ℝ$ such that and if $∑ n = 1 ∞ v n$ converges then $∑ n = 1 ∞ u n$ converges.

(b) If there is a positive $K ∈ ℝ$ such that and if $∑ n = 1 ∞ v n$ diverges then $∑ n = 1 ∞ u n$ diverges.

Proof.

(a) Suppose that the conditions in (a) are satisfied and $∑ n = 1 ∞ v n = l$. Then as all $v n$ are positive the sequence of partial sums $∑ k = 1 n v k$ is increasing, hence $∑ k = 1 n v k ⩽ l$ for all $k ∈ ℕ$. We now look at partial sums $s n = ∑ k = 1 n u k$ for the $u n$s.

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

$0 < s n = u 1 + ⋯ + u n < K ( v 1 + ⋯ + v n ) ⩽ K l$

and therefore by monotonicity $( s n )$ converges as it is bounded. Hence $∑ n = 1 ∞ u n$ 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 $B ∈ ℝ$ there is $n ∈ ℕ$ such that $v 1 + ⋯ + v n > B K$.

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

$s n = u 1 + ⋯ + u n ⩾ K ( v 1 + ⋯ + v n ) > B ⁢$

As $B$ was arbitrary, this means that the sequence $( s n )$ of partial sums is unbounded. Hence $∑ n = 1 ∞ u n$ diverges.

Examples are given in another page.

## 3. The limit comparison test

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 $∑ n = 1 ∞ u n$ and $∑ n = 1 ∞ v n$ are series of positive terms $u n$, $v n$. Suppose also that

$u n v n → l > 0$

as $n → ∞$. Then the series $∑ n = 1 ∞ u n$ and $∑ n = 1 ∞ v n$ are either both convergent or both divergent.

Proof.

Suppose that $u n v n → l > 0$ as $n → ∞$. Then let $ε = l / 2$, $K = l / 2$, $M = 3 l / 2$. Then there is $N ∈ ℕ$ such that and hence . Thus for all $n ⩾ N$ $u n < M v n$ and $K v n < u n$.

Now if $∑ n = 1 ∞ v n$ converges then so does $∑ n = N ∞ v n$ and so does $∑ n = N ∞ u n$ by the comparison test as $u n < M v n$ for $n ⩾ N$. Hence $∑ n = 1 ∞ u n$ converges.

Similarly, if $∑ n = 1 ∞ v n$ diverges then so does $∑ n = N ∞ v n$ and so does $∑ n = N ∞ u n$ by the comparison test as $K v n < u n$ for $n ⩾ N$. Hence $∑ n = 1 ∞ u n$ diverges.

Examples are given in another page.

## 4. Conclusion

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.