Math Core

Factorials and exponentials make comparison tests clumsy. The ratio test handles them directly by asking one question: in the long run, how does each term compare with the one before it? It is also the main tool you will use to find where power series converge.

The idea: eventually geometric

In a geometric series, the ratio of consecutive terms is always the same number $r$ , and the series converges exactly when $|r| < 1$ . Many other series are "geometric in the long run": the ratio $\dfrac{a_{n+1}}{a_n}$ isn't constant, but it approaches a limit. If that limit is less than $1$ in absolute value, then far out the series looks like a convergent geometric series, and it converges.

The ratio test

For a series $\sum a_n$ with nonzero terms, compute

L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.

If $L < 1$ , the series converges (in fact, it converges absolutely).
If $L > 1$ (including $L = \infty$ ), the series diverges.
If $L = 1$ , the test is inconclusive.

When $L > 1$ , the terms are eventually growing in size, so they cannot approach $0$ , and the nth term test finishes the job.

Why it works

Suppose $L = \frac12$ . Then, far enough out in the series, every ratio $\left|\frac{a_{n+1}}{a_n}\right|$ is less than, say, $0.6$ . From that point on, each term is less than $0.6$ times the one before it, so the tail of the series is smaller, term by term, than a geometric series with ratio $0.6$ :

|a_N| + |a_{N+1}| + |a_{N+2}| + \cdots \le |a_N|\left(1 + 0.6 + 0.6^2 + \cdots\right) = \frac{|a_N|}{1 - 0.6}.

That geometric series converges, so by direct comparison the tail converges, and adding back the first $N$ terms doesn't change that. The same argument works for any $L < 1$ : just pick a ratio between $L$ and $1$ .

So the ratio test is really a comparison test in disguise, with a geometric series as the benchmark. That also explains why it fails when $L = 1$ : there is no convergent geometric series with ratio $1$ to compare against, and the series may be shrinking too slowly for geometric comparison to detect.

Notice that the test uses absolute values. That means it works for series whose terms have mixed signs, like $\sum \frac{(-2)^n}{n!}$ , and when $L < 1$ it actually proves the stronger statement that $\sum |a_n|$ converges. You will make this precise in the next lesson on absolute convergence.

Looking ahead to power series

At the end of this unit, you will apply the ratio test to series with a variable in them, such as $\sum \frac{x^n}{n}$ . Then the limit $L$ depends on $x$ , here $L = |x|$ , and the condition $L < 1$ becomes an inequality that tells you exactly which values of $x$ make the series converge. The algebra is the same as in the examples below, with $x$ carried along as a constant.

Simplifying the ratio

The algebra is where students lose points. Write the ratio as a product $a_{n+1} \cdot \dfrac{1}{a_n}$ so you can cancel. These simplifications come up constantly:

\frac{(n+1)!}{n!} = n + 1, \qquad \frac{n!}{(n+1)!} = \frac{1}{n+1}, \qquad \frac{3^{n+1}}{3^n} = 3, \qquad \frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1).

Common mistake

The ratio test says nothing when $L = 1$ . This happens for every p-series and for most rational expressions in $n$ . For example, for $\sum \frac{1}{n}$ (diverges) and $\sum \frac{1}{n^2}$ (converges), the ratio limit is $1$ both times. If you get $L = 1$ , switch to a comparison, integral, or alternating series test.

When to reach for the ratio test

The ratio test works best when the terms contain

factorials, such as $n!$ or $(2n)!$ ;
exponentials, such as $3^n$ or $e^n$ , especially mixed with powers of $n$ ;
$n$ th powers, such as $n^n$ .

It is almost never the right tool for purely algebraic terms like $\dfrac{n^2 + 1}{n^4 - 3}$ .

Worked examples

Worked example: Powers over exponentials

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{3^n}$ converges.

Solution.

\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^2}{3^{n+1}} \cdot \frac{3^n}{n^2} = \frac{1}{3}\left(\frac{n+1}{n}\right)^2 \to \frac{1}{3}.

Since $L = \frac13 < 1$ , the series converges.

Worked example: Factorials beat exponentials

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{5^n}$ converges.

Solution.

\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{5^{n+1}} \cdot \frac{5^n}{n!} = \frac{n+1}{5} \to \infty.

Since $L = \infty > 1$ , the series diverges. (Its terms grow without bound.)

Worked example: A double factorial ratio

Determine whether $\displaystyle\sum_{n=0}^{\infty} \frac{(n!)^2}{(2n)!}$ converges.

Solution.

\left|\frac{a_{n+1}}{a_n}\right| = \frac{\big((n+1)!\big)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2} = \frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{n+1}{2(2n+1)} \to \frac{1}{4}.

Since $L = \frac14 < 1$ , the series converges.

Worked example: A limit that equals e

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n}$ converges.

Solution.

\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{(n+1)\,n^n}{(n+1)^{n+1}} = \left(\frac{n}{n+1}\right)^n.

Now $\left(\dfrac{n}{n+1}\right)^n = \dfrac{1}{\left(1 + \frac1n\right)^n} \to \dfrac{1}{e}$ . Since $L = \frac{1}{e} < 1$ , the series converges.

Tip

Before diving into the algebra, predict the answer with growth rates. Factorials beat exponentials, which beat powers. So $\frac{n^{10}}{2^n}$ should converge, and $\frac{100^n}{n!}$ should converge, while $\frac{n!}{100^n}$ diverges. If your computed $L$ disagrees with your prediction, recheck the algebra.

Practice

Practice 1

For $\displaystyle\sum_{n=1}^{\infty} \frac{n}{3^n}$ , compute $L = \displaystyle\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

What does the ratio test tell you about $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{10^n}$ ?

Practice 3

What does the ratio test tell you about $\displaystyle\sum_{n=0}^{\infty} \frac{2^n}{n!}$ ?

Practice 4

For $\displaystyle\sum_{n=0}^{\infty} \frac{(n!)^2}{(2n)!}$ , compute the ratio test limit $L$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

For which series is the ratio test inconclusive?

Practice 6

For $\displaystyle\sum_{n=1}^{\infty} \frac{n^n}{n!}$ , compute the ratio test limit $L$ . Give an exact value.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

For $\displaystyle\sum_{n=1}^{\infty} \frac{5^n}{n\cdot 4^n}$ , compute the ratio test limit $L$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.