The ratio test

1 Introduction

This web page discusses a further test for the convergence or non-convergence of infinte series of positive terms, the ratio test. In some ways it is not as powerful as the comparison test, but the ratio test is particularly useful to test convergence for power series as we shall see.

2 The ratio test

We are going to look at the ratio test here. Essentially, the ratio test is just a version of the comparison test when the sequence to be compared against is ∑n=1∞rn and says nothing much that is new that you cannot deduce from the comparison test, but it is presented in a very useful form that is easy to compute, and the ratio test is almost always successful in the area of power series—one of the most common type of series in practice. In other words, if you were only going to learn one test, it should be the comparison test, but in this case you would miss out on some easy calculations!

Like the comparison test, the ratio test discusses series of positive terms. This allows us to make some simplifications, via monotonicity for example, as we have already seen.

Theorem 2.1 (Ratio Test)

Suppose ∑n=1∞an is a series of positive terms, and suppose that the limit l=limn→∞an+1an exists. Then:

(a) if l<1 then the series ∑n=1∞an converges;

(b) if l>1 then the series ∑n=1∞an diverges.

Proof

Since each an>0, we have an+1an>0 hence by the theorem on bounds for convergent sequences l≥0. Also, by hypothesis we may assume l≠1.

Let ε=l-12 and let N0∈ℕ be such that

∀n≥N0 an+1an-l<ε.

Then for all n≥N0

max(0,l-ε)<an+1an<l+ε

so

an⋅max(0,l-ε)<an+1<(l+ε)⋅an.

We now look at the two cases when l>1 and l<1.

If l>1 then l-ε>1 by our choice of ε and an+1>(l-ε)an for all n≥N0. Hence

an>(l-ε)(n-N0)aN0=(l-ε)naN0(l-ε)N0

so ∑n=1∞an diverges, by the comparison test, using comparison with ∑n=1∞(l-ε)n.

If l<1 then l+ε<1 by our choice of ε and an+1<(l+ε)an for all n≥N0. Hence

an<(l+ε)(n-N0)aN0=(l+ε)naN0(l+ε)N0

so ∑n=1∞an converges, by the comparison test, using comparison with ∑n=1∞(l+ε)n.

Apart from what the ratio test actually says, the most important thing to notice about the ratio test is that it is silent (i.e., say nothing about convergence or divergence) if l=limn→∞an+1an=1.

3 Examples

Example 3.1

Let x>0 and consider ∑n=1∞nxn. Let an=nxn Then

an+1an=(n+1)xn+1nxn=n+1n⋅x=1+1nx→(1+0)x=x

as n→∞. The series consists of positive terms, so by the ratio test converges if x<1 and diverges if x>1. This result is in agreement with Exercise 3, sheet 8 which produced the same conclusion by more direct means. Note that the ratio test doesn't say what happens if x=1 but in this case an=n so this sequence (an) is unbounded and hence non-convergent (by boundedness of convergent sequences) and hence not null. Therefore the series diverges for x=1 by the null sequence test.

The ratio test generally works well and is the first test to try for power series such as ∑n=1∞anxn where all the coefficients an are positive. If some terms are non-positive, see the section later on absolute convergence for information on what to do.

The next two examples show that the ratio test really doesn't say anything when l=1. In fact both possibilities may happen.

Example 3.2

Consider ∑n=1∞11+n. Let an=11+n. Then an+1an=1n+21n+1=n+1n+2→1 as n→∞. Thus the ratio test says nothing.

In fact the series diverges; to prove this use the comparison test with ∑n=1∞1n.

Example 3.3

Consider ∑n=1∞1(1+n)². Let an=1(1+n)2. Then an+1an=1(n+2)21(n+1)2=(n+1)2(n+2)2→1 as n→∞. Thus the ratio test says nothing.

In fact the series converges; to prove this use the comparison test with ∑n=1∞1n2.

4 Summary

You have seen the ratio test and how it is used to test convergence for series of positive terms. It does not always give an answer. (When it doesn't you should try the comparison test instead.) But when it does work it tends to be easier to use. The ratio test is used particularly for power series.