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 =1 converges or not we should compare it with one, =1 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 =1 and =1 are series of positive terms , . Then

(a) If there is a positive such that 0< and if =1 converges then =1 converges.

(b) If there is a positive such that 0< and if =1 diverges then =1 diverges.

Proof.

(a) Suppose that the conditions in (a) are satisfied and =1 ₌ . Then as all are positive the sequence of partial sums =1 is increasing, hence =1 for all . We now look at partial sums _=
=1 for the s.

By assumptions on we have, for each ,

0<~~_=₁₊₊~~<( _1++
₎

and therefore by monotonicity (₎ converges as it is bounded. Hence =1 converges.

(b) Suppose that the conditions in (b) are satisfied. Then by monotonicity the sequence of partial sums from is unbounded and so for all positive there is such that _{1++
_>} .

By assumptions on we have, for each ,

~~_=₁₊₊~~ ( _1++
₎ >.

As was arbitrary, this means that the sequence (₎ of partial sums is unbounded. Hence =1 diverges.

Examples are given in another page.

3. The limit comparison test

The calculations involved in finding constants 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 =1 and =1 are series of positive terms , . Suppose also that

>0

as . Then the series =1 and =1 are either both convergent or both divergent.

Proof.

Suppose that >0 as . Then let =/2 , =/2 , =3/2 . Then there is such that - < and hence 0<< < . Thus for all _< and _< .

Now if =1 converges then so does = and so does = by the comparison test as _< for . Hence =1 converges.

Similarly, if =1 diverges then so does = and so does = by the comparison test as _< for . Hence =1 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.