The ratio test

3 The ratio test

This test, which is one of the most useful and widely used convergence tests, applies only to series of positive terms.

Key Point 7

The Ratio Test

Let $\sum_{p = 1}^{\infty} a_{p}$ be a series of positive terms such that, as $p$ increases, the limit of $\frac{a_{p + 1}}{a_{p}}$ equals

a number $λ$ . That is $lim_{p \to \infty} \frac{a_{p + 1}}{a_{p}} = λ$ .

It can be shown that:

$∙$ if $λ > 1,$ then $\sum_{p = 1}^{\infty} a_{p}$ diverges

$∙$ if $λ < 1,$ then $\sum_{p = 1}^{\infty} a_{p}$ converges

$∙$ if $λ = 1,$ then $\sum_{p = 1}^{\infty} a_{p}$ may converge or diverge.

That is, the test is inconclusive in this case.

Example 1

Use the ratio test to examine the convergence of the series

$1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + &ctdot;$
$1 + x + x^{2} + x^{3} + &ctdot;$

Solution

The general term in this series is $\frac{1}{p!}$ i.e.
$1 + \frac{1}{2!} + \frac{1}{3!} + &ctdot; = \sum_{p = 1}^{\infty} \frac{1}{p!} a_{p} = \frac{1}{p!} ∴ a_{p + 1} = \frac{1}{(p + 1)!}$

and the ratio

$\frac{a_{p + 1}}{a_{p}} = \frac{p!}{(p + 1)!} = \frac{p (p - 1) \dots (3) (2) (1)}{(p + 1) p (p - 1) \dots (3) (2) (1)} = \frac{1}{(p + 1)}$

$∴ lim_{p \to \infty} \frac{a_{p + 1}}{a_{p}} = lim_{p \to \infty} \frac{1}{(p + 1)} = 0$

Since $0 < 1$ the series is convergent. In fact, it will be easily shown, using the techniques outlined in HELM booklet 16.5, that

$1 + \frac{1}{2!} + \frac{1}{3!} + &ctdot; = e - 1 \approx 1.718$
Here we must assume that $x > 0$ since we can only apply the ratio test to a series of positive terms.
Now

$1 + x + x^{2} + x^{3} + &ctdot; = \sum_{p = 1}^{\infty} x^{p - 1}$

so that

$a_{p} = x^{p - 1}, a_{p + 1} = x^{p}$

and

$lim_{p \to \infty} \frac{a_{p + 1}}{a_{p}} = lim_{p \to \infty} \frac{x^{p}}{x^{p - 1}} = lim_{p \to \infty} x = x$

Thus, using the ratio test we deduce that (if $x$ is a positive number) this series will only converge if $x < 1.$

We will see in Section 16.4 that

$1 + x + x^{2} + x^{3} + &ctdot; = \frac{1}{1 - x} provided 0 < x < 1.$

Task!

Use the ratio test to examine the convergence of the series:

$\frac{1}{ln 3} + \frac{8}{{(ln 3)}^{2}} + \frac{27}{{(ln 3)}^{3}} + &ctdot;$

First, find the general term of the series:

$\frac{1}{ln 3} + \frac{8}{{(ln 3)}^{2}} + &ctdot; = \sum_{p = 1}^{\infty} \frac{p^{3}}{{(ln 3)}^{p}} so a_{p} = \frac{p^{3}}{{(ln 3)}^{p}}$

Now find $a_{p + 1}$ :

$a_{p + 1} = \frac{{(p + 1)}^{3}}{{(ln 3)}^{p + 1}}$

Finally, obtain $lim_{p \to \infty} \frac{a_{p + 1}}{a_{p}}$ :

$\frac{a_{p + 1}}{a_{p}} = {(\frac{p + 1}{p})}^{3} \frac{1}{(ln 3)}$ . Now ${(\frac{p + 1}{p})}^{3} = {(1 + \frac{1}{p})}^{3} \to 1$ as $p$ increases

$∴ lim_{p \to \infty} \frac{a_{p + 1}}{a_{p}} = \frac{1}{(ln 3)} < 1$

Hence this is a convergent series.

Note that in all of these Examples and Tasks we have decided upon the convergence or divergence of various series; we have not been able to use the tests to discover what actual number the convergent series converges to.

3 The ratio test

Key Point 7

Example 1

Solution

Task!

Answer

Answer

Answer