Ratio Test for Convergence

AP Calculus BC · Infinite Sequences and Series

The Ratio Test for Convergence

The default whenever factorials or $n$-th powers appear — and useless the moment $L=1$.

key idea$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|;\quad L\lt 1\Rightarrow\text{converges}$

The test

Applies when

The series $\sum a_n$ has terms whose ratio simplifies — in practice, whenever factorials $n!$, exponentials $a^{n}$, or powers $n^{n}$ appear. Terms may be positive or of mixed sign; the test uses absolute values.

You compute

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

Concludes

$L\lt 1$ ⇒ $\sum a_n$ converges absolutely. $L\gt 1$ (or $L=\infty$) ⇒ $\sum a_n$ diverges.

CANNOT conclude

$L=1$ — the test is inconclusive and tells you nothing at all. You must switch to a different test. Both $\sum\tfrac1n$ (diverges) and $\sum\tfrac1{n^{2}}$ (converges) give $L=1$.

When to reach for it

The ratio test is the default for anything with a factorial or a constant raised to the $n$-th power, because those collapse beautifully when you divide consecutive terms. It is the wrong first choice for a plain rational function of $n$ — there the ratio tends to $1$ and you have learnt nothing.

Series	Ratio test?	Why
$\sum\dfrac{2^{n}}{n!}$	yes	both $a^{n}$ and $n!$ — ideal
$\sum\dfrac{n!}{100^{n}}$	yes	factorial dominates; expect $L=\infty$
$\sum\dfrac{n^{2}}{3^{n}}$	yes	the $3^{n}$ drives the ratio to $\tfrac13$
$\sum\dfrac{1}{n^{2}+1}$	no	rational in $n$ ⇒ $L=1$, so the test tells you nothing
$\sum\dfrac{(-1)^{n}}{n}$	no	$L=1$ — inconclusive here
$\sum\dfrac{n^{n}}{n!}$	yes	$n^{n}$ present; ratio → $e\gt 1$, so it diverges

If the terms are a ratio of polynomials, the ratio test will return $L=1$ every time. Recognise that before spending the algebra.

Term shape	The ratio $\left\|\dfrac{a_{n+1}}{a_n}\right\|$ simplifies to	$L$
$\dfrac{a^{n}}{n!}$	$\dfrac{a}{n+1}$	$0$ — converges
$\dfrac{n!}{a^{n}}$	$\dfrac{n+1}{a}$	$\infty$ — diverges
$\dfrac{n^{k}}{a^{n}}$	$\dfrac{1}{a}\left(1+\tfrac1n\right)^{k}$	$\dfrac1a$
$\dfrac{n^{n}}{n!}$	$\left(1+\tfrac1n\right)^{n}$	$e\gt 1$ — diverges
$\dfrac{1}{n^{p}}$	$\left(\dfrac{n}{n+1}\right)^{p}$	$1$ — inconclusive

Learn these five collapses and most ratio-test questions become one line of cancellation. The last row is the warning: any rational function of $n$ lands on $L=1$.

Worked examples

EX 1

Factorial in the denominator

$\sum\dfrac{2^{n}}{n!}$

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

Write the ratio

$\dfrac{a_{n+1}}{a_n}=\dfrac{2^{\,n+1}}{(n+1)!}\cdot\dfrac{n!}{2^{n}}$.

Cancel

$\dfrac{2^{\,n+1}}{2^{n}}=2$ and $\dfrac{n!}{(n+1)!}=\dfrac{1}{n+1}$, so the ratio is $\dfrac{2}{n+1}$.

Take the limit

$L=\lim_{n\to\infty}\dfrac{2}{n+1}=0$.

Conclude

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

Sanity check

This is the series for $e^{2}-1$, which is finite. Factorials beat exponentials, so convergence is expected.

EX 2

Factorial in the numerator

$\sum\dfrac{n!}{100^{n}}$

Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{100^{n}}$ converges.

Write the ratio

$\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!}{100^{\,n+1}}\cdot\dfrac{100^{n}}{n!}$.

Cancel

$\dfrac{(n+1)!}{n!}=n+1$ and $\dfrac{100^{n}}{100^{\,n+1}}=\dfrac{1}{100}$, giving $\dfrac{n+1}{100}$.

Take the limit

$L=\lim_{n\to\infty}\dfrac{n+1}{100}=\infty$.

Conclude

$L\gt 1$, so the series diverges.

Note the lesson

A large constant base does not save you. $n!$ eventually outgrows $a^{n}$ for every fixed $a$.

EX 3

Powers of $n$ against an exponential

$\sum\dfrac{n^{2}}{3^{n}}$

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

Write the ratio

$\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)^{2}}{3^{\,n+1}}\cdot\dfrac{3^{n}}{n^{2}}=\dfrac{1}{3}\cdot\dfrac{(n+1)^{2}}{n^{2}}$.

Simplify the polynomial part

$\dfrac{(n+1)^{2}}{n^{2}}=\left(1+\dfrac1n\right)^{2}\to 1$.

Take the limit

$L=\dfrac13\cdot 1=\dfrac13$.

Conclude

$L=\tfrac13\lt 1$, so the series converges absolutely.

Generalise

For $\sum\dfrac{n^{k}}{a^{n}}$ with $a\gt 1$, the polynomial part always tends to $1$ and $L=\tfrac1a$. Exponentials beat polynomials.

EX 4

The inconclusive case

$L=1$

Apply the ratio test to $\sum\dfrac1n$ and to $\sum\dfrac1{n^{2}}$. What do you learn?

First series

$\dfrac{a_{n+1}}{a_n}=\dfrac{n}{n+1}\to 1$, so $L=1$.

Second series

$\dfrac{a_{n+1}}{a_n}=\dfrac{n^{2}}{(n+1)^{2}}\to 1$, so $L=1$ again.

Compare the truth

The harmonic series $\sum\tfrac1n$ diverges; $\sum\tfrac1{n^{2}}$ converges ($p$-series with $p=2$).

Draw the conclusion

Identical ratio-test results, opposite behaviour. So $L=1$ carries no information whatsoever — it is not weak evidence for either outcome.

What to do instead

Recognise the form before computing. If the terms are a ratio of polynomials, the ratio test cannot help and the algebra is wasted.

What it cannot tell you

$L=1$ means start over

It is not ‘probably convergent’ or ‘borderline’. Example 4 shows both outcomes occur at $L=1$. State plainly that the ratio test is inconclusive.

Absolute values are part of the test

The limit is of $\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert$. That is why a $L\lt 1$ result gives absolute convergence, and why the test works on alternating series too.

Convergence of the series, not of the terms

$L$ is a limit of ratios, not of terms. Do not report $L$ as the sum — the ratio test never produces the value a series converges to.

It does not replace the $n$-th term test

If $a_n\not\to 0$ the series diverges immediately and no ratio work is needed. Check that cheap condition first.

Course map. Topic 10.8 · LO LIM-7.A — determine whether a series converges or diverges (EK LIM-7.A.7). BC-only.