The Ratio Test

Main idea

Ask how much each term shrinks compared with the one before. If, in the long run, each term is less than a fixed fraction of the last, the series is really a geometric series in disguise, and it converges.

{\htmlData{sym=1}{L}} = \lim_{n\to\infty} {\htmlData{sym=0}{\left|\frac{a_{n+1}}{a_n}\right|}}

What the symbols mean

$\left|\frac{a_{n+1}}{a_n}\right|$
Term-to-term ratio
How much the terms shrink or grow from one to the next.
$L$
Limit of the ratio
L < 1: converges absolutely. L > 1: diverges. L = 1: no conclusion.

Explanation

The ratio test is simply a check on how fast the terms shrink: divide each term by the one before it and see what that fraction settles on. The name just means that fraction, the ratio of one term to the next. It's good for series with factorials and powers, which are hard to compare with anything but whose ratios simplify beautifully. The formula $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$ is only three pieces: $a_{n+1}$ (the next term), $a_n$ (the current term) and the absolute value bars (ignore the signs, we only care about size). If $L$ comes out below 1, the terms are shrinking by a steady fraction, which is enough. That's all it is, so don't let the fraction-inside-a-limit scare you.

Think back to geometric series. There, every term is exactly $r$ times the one before, and the series converges when $r < 1$ . Most series aren't that tidy, but plenty of them are nearly geometric far out, with each term about $L$ times the last. The ratio test asks: what's that $L$ ?

The verdicts.

$L < 1$ : absolutely convergent.
$L > 1$ (including $L = \infty$ ): divergent.
$L = 1$ : no information. Try another test.

Why. If the ratios approach $L < 1$ , pick $r$ between $L$ and 1. Eventually each term is at most $r$ times the one before, so the tail is smaller than a geometric series with ratio $r$ , and a comparison finishes the job. If $L > 1$ , the terms eventually grow, so they can't go to 0, and the divergence test takes over.

When to use it. Factorials and exponentials. Their ratios simplify beautifully: $\frac{(n+1)!}{n!} = n + 1$ , $\frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$ , $\frac{5^{n+1}}{5^n} = 5$ .

A common mistake. It's tempting to reach for the ratio test on everything, since it's so mechanical. But for p-series and rational functions of $n$ it always gives $L = 1$ : the ratio $\frac{n^2}{(n+1)^2}$ creeps toward 1 whether the series converges or not. Those terms shrink too slowly to look geometric. Use comparison tests there instead.

See it

The terms 5ⁿ/n!, with the ratio from each term to the next shown as n moves. At first they grow, peaking around 26 at n = 4 and 5, because the ratio 5/(n + 1) is still above 1. Once it drops below 1 for good, the terms collapse, faster than any geometric series. The ratio test ignores the early bump and looks only at the limit: L = 0 < 1, so Σ 5ⁿ/n! converges (to e⁵).

Worked examples

Example 1

Does $\displaystyle\sum \frac{(-10)^n}{4^{2n+1}(n+1)}$ converge?

1
Powers everywhere, so try the ratio test. Flip the bottom fraction and multiply, and most of the powers cancel: $\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{10^{n+1}}{4^{2n+3}(n+2)}\cdot\dfrac{4^{2n+1}(n+1)}{10^n} = \dfrac{10}{16}\cdot\dfrac{n+1}{n+2}$ .

Example 2

Does $\displaystyle\sum \frac{n!}{5^n}$ converge?

1
$\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{5^{n+1}}\cdot\dfrac{5^n}{n!} = \dfrac{n+1}{5} \to \infty$ . Each term is eventually many times bigger than the last.

It diverges. Factorials beat exponentials.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the ratio-test limit $L$ for $\displaystyle\sum \frac{n}{2^n}$ .

Problem 2+5 XP

Find $L$ for $\displaystyle\sum \frac{n^n}{n!}$ .

Problem 3+5 XP

Find $L$ for $\displaystyle\sum \frac{(2n)!}{(n!)^2\,5^n}$ .

Problem 4+5 XP

What does the ratio test say about $\displaystyle\sum \frac{1}{n^2}$ ?

Problem 5+5 XP

Find $L$ for $\displaystyle\sum \frac{3^n}{n!}$ .

Problem 6+5 XP

A data packet is resent until it gets through, and each try succeeds with probability $0.6$ . The expected number of tries is $\sum_{n=1}^\infty n(0.4)^{n - 1}(0.6)$ . Check that it converges with the ratio test, then find it.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

The ratio test shows $\displaystyle\sum_{n=1}^\infty\frac{n^{100}}{1.01^n}$ converges, since $L = \frac{1}{1.01} < 1$ . But its terms grow enormously first. For which $n$ is the term $\frac{n^{100}}{1.01^n}$ largest?

Where it's used

If each stage of a process costs at most a fixed fraction of the stage before, the total cost is bounded by a geometric series. That ratio argument is how computer scientists bound the work in divide-and-conquer algorithms.

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