The Ratio Test

A geometric series converges when the ratio between consecutive terms is less than $1$ in size. The ratio test asks the same question of any series: in the long run, how does each term compare with the one before it? It’s the go-to test for series with factorials and exponentials, and it’s the main tool for power series later in this unit.

Key ideas

The test

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

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

Value of $L$	Conclusion
$L \lt 1$	the series converges (in fact, absolutely)
$L \gt 1$ (or $L = \infty$)	the series diverges
$L = 1$	inconclusive: use a different test

Why it works

If $L \lt 1$, then far out in the series each term is roughly $L$ times the one before, so the tail behaves like a geometric series with ratio about $L$, which converges. If $L \gt 1$, the terms eventually grow, so they can’t approach $0$.

The absolute value means the test works even when terms are negative or alternate. When $L \lt 1$, the series converges absolutely.

The inconclusive case L = 1

For every p-series, the ratio test gives $L = 1$:

$$\left\lvert \frac{1/(n+1)}{1/n} \right\rvert = \frac{n}{n+1} \to 1, \qquad \left\lvert \frac{1/(n+1)^2}{1/n^2} \right\rvert = \frac{n^2}{(n+1)^2} \to 1 .$$

But $\sum \frac{1}{n}$ diverges and $\sum \frac{1}{n^2}$ converges. So $L = 1$ really can go either way. In practice, the ratio test is useless for series built only from powers of $n$ (rational functions and roots). Use comparison or p-series for those.

Factorial algebra

Recall $n! = 1 \cdot 2 \cdot 3 \cdots n$, and $0! = 1$. The key simplifications:

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

To find $a_{n+1}$, replace every $n$ with $n + 1$. For example, if $a_n = \dfrac{(2n)!}{5^n}$, then $a_{n+1} = \dfrac{(2n+2)!}{5^{n+1}}$.

Dividing by $a_n$ is the same as multiplying by its reciprocal, so set it up as $a_{n+1} \cdot \dfrac{1}{a_n}$ and cancel.

Worked examples

Example 1: An exponential in the denominator

Does $\displaystyle\sum_{n=1}^{\infty} \frac{n}{3^n}$ converge or diverge?

Solution.

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{n + 1}{3^{n+1}} \cdot \frac{3^n}{n} = \frac{n + 1}{3n} \;\to\; \frac{1}{3}$$

$L = \tfrac{1}{3} \lt 1$, so the series converges by the ratio test.

Example 2: Factorials

Does each series converge or diverge?

• (a) $\displaystyle\sum_{n=0}^{\infty} \frac{2^n}{n!}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{5^n}$

Solution.

(a)

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2}{n + 1} \;\to\; 0$$

$L = 0 \lt 1$, so the series converges. Factorials grow faster than any exponential.

(b)

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{(n+1)!}{5^{n+1}} \cdot \frac{5^n}{n!} = \frac{n + 1}{5} \;\to\; \infty$$

$L = \infty$, so the series diverges.

Example 3: Factorials of 2n

Does $\displaystyle\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}$ converge or diverge?

Solution. Replace every $n$ with $n + 1$: $a_{n+1} = \dfrac{\big((n+1)!\big)^2}{(2n+2)!}$.

$$\begin{aligned} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert &= \frac{\big((n+1)!\big)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2} \\ &= \left(\frac{(n+1)!}{n!}\right)^2 \cdot \frac{(2n)!}{(2n+2)!} \\ &= \frac{(n+1)^2}{(2n+2)(2n+1)} \;\to\; \frac{1}{4} && \text{leading terms } \tfrac{n^2}{4n^2} \end{aligned}$$

$L = \tfrac{1}{4} \lt 1$, so the series converges by the ratio test.

Example 4: When the ratio test is inconclusive

Does $\displaystyle\sum_{n=1}^{\infty} \frac{n + 3}{n^3 + 1}$ converge or diverge?

Solution. Try the ratio test:

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{n + 4}{(n+1)^3 + 1} \cdot \frac{n^3 + 1}{n + 3} \;\to\; 1$$

(Both the top and bottom have leading term $n^4$.) $L = 1$, so the ratio test is inconclusive. Switch tests. The dominant parts give $\frac{n}{n^3} = \frac{1}{n^2}$, so use the limit comparison test with $b_n = \frac{1}{n^2}$:

$$\lim_{n \to \infty} \frac{n + 3}{n^3 + 1} \cdot n^2 = \lim_{n \to \infty} \frac{n^3 + 3n^2}{n^3 + 1} = 1$$

Since $0 \lt 1 \lt \infty$ and $\sum \frac{1}{n^2}$ converges, the series converges. Lesson: for a rational function of $n$, skip the ratio test and go straight to a comparison.

Common mistakes

Concluding something when L = 1. ”$L = 1$, so it diverges” (or converges) is wrong. $L = 1$ means the ratio test gives no information. Say “the ratio test is inconclusive” and use another test.

Messing up factorials. $(n+1)! = (n+1) \cdot n!$, not $n! + 1$. And $(2(n+1))! = (2n+2)!$, which is $(2n+2)(2n+1) \cdot (2n)!$, not $2(n+1)!$. Write out a few factors if you’re unsure.

Not replacing every n. If $a_n = \frac{n^2}{2^n}$, then $a_{n+1} = \frac{(n+1)^2}{2^{n+1}}$. Changing only the exponent and leaving $n^2$ alone is a very common slip.

Flipping the ratio. The test uses $\frac{a_{n+1}}{a_n}$ (next over current). Using $\frac{a_n}{a_{n+1}}$ gives $\frac{1}{L}$ and the opposite conclusion.

Stopping before the limit. In the harmonic series, every ratio $\frac{n}{n+1}$ is less than $1$, yet the series diverges. The test is about the limit $L$, not whether each ratio is below $1$.

Forgetting the absolute value. For a series like $\sum \frac{(-3)^n}{n!}$, take the absolute value so that the signs drop out before taking the limit.

Practice

1. (Warm-up) Does $\displaystyle\sum_{n=0}^{\infty} \frac{5^n}{n!}$ converge or diverge?

Solution

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{5^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n} = \frac{5}{n + 1} \to 0$$

$L = 0 \lt 1$, so it converges by the ratio test.

2. (Warm-up) Does $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{10^n}$ converge or diverge?

Solution

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{(n+1)!}{10^{n+1}} \cdot \frac{10^n}{n!} = \frac{n + 1}{10} \to \infty$$

$L = \infty$, so it diverges by the ratio test.

3. (Warm-up) What does the ratio test say about $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$?

Solution

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{\sqrt{n}}{\sqrt{n + 1}} = \sqrt{\frac{n}{n + 1}} \to 1$$

$L = 1$, so the ratio test is inconclusive. (It’s a p-series with $p = \tfrac{1}{2}$, so it diverges.)

4. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}$ converge or diverge?

Solution

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{(n+1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2} = \frac{1}{2}\left(\frac{n + 1}{n}\right)^2 \to \frac{1}{2}$$

$L = \tfrac{1}{2} \lt 1$, so it converges by the ratio test.

5. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n \cdot 2^n}$ converge or diverge?

Solution

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{3^{n+1}}{(n+1)2^{n+1}} \cdot \frac{n \cdot 2^n}{3^n} = \frac{3}{2} \cdot \frac{n}{n + 1} \to \frac{3}{2}$$

$L = \tfrac{3}{2} \gt 1$, so it diverges by the ratio test.

6. (Core) Does $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n)!}$ converge or diverge?

Solution

The absolute value removes the sign:

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{4^{n+1}}{(2n+2)!} \cdot \frac{(2n)!}{4^n} = \frac{4}{(2n+2)(2n+1)} \to 0$$

$L = 0 \lt 1$, so the series converges (absolutely) by the ratio test.

7. (Core) A classmate uses the ratio test on $\displaystyle\sum_{n=1}^{\infty} \frac{2n + 1}{n^2 + n}$, gets $L = 1$, and writes “diverges.” Explain what’s wrong, then decide correctly.

Solution

$L = 1$ is inconclusive, so it can’t show divergence. Use the limit comparison test instead. The dominant parts give $\frac{2n}{n^2} = \frac{2}{n}$, so compare with $b_n = \frac{1}{n}$:

$$\lim_{n \to \infty} \frac{2n + 1}{n^2 + n} \cdot n = \lim_{n \to \infty} \frac{2n^2 + n}{n^2 + n} = 2$$

Since $0 \lt 2 \lt \infty$ and the harmonic series diverges, the series does diverge, but only the limit comparison test justifies it.

8. (Challenge) Does $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n}$ converge or diverge?

Solution

Replace $n$ with $n + 1$ everywhere, including the base and exponent of $n^n$:

$$\begin{aligned} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert &= \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \\ &= \frac{(n+1)\, n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n + 1}\right)^n \end{aligned}$$

Now $\left(\dfrac{n}{n+1}\right)^n = \dfrac{1}{\left(1 + \frac{1}{n}\right)^n} \to \dfrac{1}{e}$. Since $L = \tfrac{1}{e} \lt 1$, the series converges by the ratio test.

9. (Challenge) For which positive constants $k$ does $\displaystyle\sum_{n=1}^{\infty} \frac{(n!)^2 k^n}{(2n)!}$ converge? (Handle $k = 4$ separately.)

Solution

Using the work from Example 3:

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = k \cdot \frac{(n+1)^2}{(2n+2)(2n+1)} \to \frac{k}{4}$$

So the series converges when $\frac{k}{4} \lt 1$ (that is, $0 \lt k \lt 4$) and diverges when $k \gt 4$.

When $k = 4$, $L = 1$ and the ratio test is inconclusive. But look at the ratio itself:

$$\frac{a_{n+1}}{a_n} = \frac{4(n+1)^2}{(2n+2)(2n+1)} = \frac{2n + 2}{2n + 1} \gt 1$$

so the terms keep increasing. Since $a_1 = \frac{1 \cdot 4}{2} = 2$, every term is at least $2$, and the terms don’t approach $0$. The series diverges by the $n$th term test. So it converges exactly when $0 \lt k \lt 4$.