The Ratio Test

A geometric series multiplies each term by the same ratio to get the next. Many other series multiply by a ratio that settles down as

n

grows. The ratio test treats such a series as nearly geometric.

The test

Take positive terms with

L < 1

, and pick a number

r

with

L < r < 1

. Eventually every ratio is below

r

, so from some

a_N

on each term is at most

r

times the one before. Then

a_{N+k} \le a_N r^k

, and the series converges by comparison with a geometric series. For terms of mixed sign, the same argument shows that

\sum |a_n|

converges, and a series whose sizes have a finite sum converges too. The page on absolute and conditional convergence proves that. When

L > 1

, the sizes of the terms eventually grow, so the terms can't approach 0 and the series diverges by the nth term test.

The test works best on factorials and on powers like

5^n

, where the ratio of consecutive terms simplifies. These facts do most of the work.

\begin{aligned} \frac{(n + 1)!}{n!} &= n + 1 \\[4pt] \frac{(2n + 2)!}{(2n)!} &= (2n + 2)(2n + 1) \\[4pt] \frac{(3n + 3)!}{(3n)!} &= (3n + 3)(3n + 2)(3n + 1) \end{aligned}

Example 1

Use the ratio test to decide whether each series converges.

(a)

\displaystyle\sum_{n=1}^{\infty} \frac{(2n)!}{n!\,5^n}

(b)

\displaystyle\sum_{n=1}^{\infty} \frac{(3n)!}{n!\,(2n)!\,7^n}

(a) The terms are positive, so the absolute values can be dropped. Group the factorials and the powers of 5 before simplifying.

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(2n + 2)!}{(2n)!}\cdot\frac{n!}{(n + 1)!}\cdot\frac{1}{5} \\[4pt] &= \frac{(2n + 2)(2n + 1)}{5(n + 1)} \\[4pt] &= \frac{2(2n + 1)}{5} \end{aligned}

\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{2(2n + 1)}{5} = \infty

, and the series diverges by the ratio test.

(b) The terms are positive. Each factorial ratio is one of the facts above.

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(3n + 3)!}{(3n)!}\cdot\frac{n!}{(n + 1)!} \\[4pt] &\qquad \cdot\frac{(2n)!}{(2n + 2)!}\cdot\frac{1}{7} \\[4pt] &= \frac{(3n + 3)(3n + 2)(3n + 1)}{7(n + 1)(2n + 2)(2n + 1)} \\[4pt] &= \frac{3(3n + 2)(3n + 1)}{14(n + 1)(2n + 1)} \end{aligned}

The top and bottom are both quadratics in

n

, so the limit is the ratio of their leading coefficients.

\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{3 \cdot 9}{14 \cdot 2} = \frac{27}{28}

That's less than 1, so the series converges by the ratio test, even though the ratios come close to 1.

The ratios in (b) for n = 1 to 40. They climb toward 27/28 and stay below it, short of 1.

Every p-series has

L = 1

, whether it converges or diverges, so

L = 1

can't decide anything. The same happens for most series built only from powers of

n

, and for those a comparison test is the right tool.

Example 2

Decide whether

\sum_{n=1}^{\infty} \frac{\sqrt[3]{n^2 + 5}}{n^2 + 2}

converges.

The ratio test fails here.

\begin{aligned} \frac{a_{n+1}}{a_n} &= \sqrt[3]{\frac{(n + 1)^2 + 5}{n^2 + 5}} \\[4pt] &\qquad \cdot\frac{n^2 + 2}{(n + 1)^2 + 2} \\[4pt] &\to \sqrt[3]{1}\cdot 1 = 1 \end{aligned}

For large

n

the terms behave like

\frac{n^{2/3}}{n^2} = \frac{1}{n^{4/3}}

, so use limit comparison with

b_n = \frac{1}{n^{4/3}}

. Both series have positive terms.

\begin{aligned} &\lim_{n\to\infty} \frac{\sqrt[3]{n^2 + 5}/(n^2 + 2)}{1/n^{4/3}} \\[4pt] &\qquad = \lim_{n\to\infty} \sqrt[3]{1 + \frac{5}{n^2}}\cdot\frac{n^2}{n^2 + 2} \\[4pt] &\qquad = 1 \end{aligned}

Since

0 < 1 < \infty

and

\sum \frac{1}{n^{4/3}}

is a p-series with

p = \frac{4}{3} > 1

, the series converges by limit comparison.