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 such that n   0 < u n K v n and if n = 1 v n converges then n = 1 u n converges.

(b) If there is a positive K such that n   0 < K v n u n 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 . 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 there is 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 such that n N   | u n v n - l | < ε and hence n N   0 < K < u n v n < M . 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.