The Ratio Test — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

A geometric series converges when each term is a fixed fraction of the one before. Many series are not geometric, but far out their terms behave as if they were: each term is close to a fixed fraction

L

of the one before. The ratio test finds that

L

and compares it with 1.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. When does the geometric series $\sum_{n=0}^{\infty} a r^n$ (with $a \ne 0$ ) converge?

When $|r| < 1$
When $r > 0$
When $|r| > 1$
When $r = 1$

Show the answer

Each term is $r$ times the last; the sum settles when $|r| < 1$ .

Correct: When $|r| < 1$ :
When $r > 0$ :
When $|r| > 1$ :
When $r = 1$ :

2. Simplify $\frac{(n+1)!}{n!}$ .

$n + 1$
$n$
$1$
$(n+1)n$

Show the answer

$(n+1)! = (n+1) \cdot n!$ , so the quotient is $n + 1$ .

Correct: $n + 1$ :
$n$ :
$1$ :
$(n+1)n$ :

3. Find $\lim_{n \to \infty} \frac{n + 1}{2n}$ .

$\frac{1}{2}$
$1$
$0$
$2$

Show the answer

Divide the top and bottom by $n$ : $\frac{1 + 1/n}{2} \to \frac{1}{2}$ .

Correct: $\frac{1}{2}$ :
$1$ :
$0$ :
$2$ :

Part 4 · See it

See it first

Dots at height (n + 1) squared over 3 n squared for n = 1 to 15, the ratio of consecutive terms of the series of n squared over 3 to the n. The first dot is at 4/3, above a solid line at 1; the rest fall below it and approach a dashed line at 1/3. The limit 1/3 is below 1, so the series converges. — The ratios $\frac{a_{n+1}}{a_n}$ for $\sum \frac{n^2}{3^n}$ start above 1 and settle at $\frac{1}{3}$ ; the limit is what counts. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Compute $\left|\frac{a_{n+1}}{a_n}\right|$ and its limit $L$ far out, each term is about $L$ times the one before, as in a geometric series with ratio $L$
$L < 1$ the terms shrink at least as fast as a convergent geometric series, so the series converges
$L > 1$ (or $L$ is infinite)the terms grow, do not approach 0, and the series diverges
$L = 1$ the test decides nothing: use another test

Part 6 · Key ideas

Key ideas

Ratio test: $L$ $= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ . $L < 1$ : converges. $L > 1$ or infinite: diverges. $L = 1$ : no conclusion.
Put $a_{n+1}$ on top and simplify before taking the limit: $\frac{(n+1)!}{n!} = n + 1$ , $\frac{3^{n+1}}{3^n} = 3$ .
Use it for factorials and exponentials. For $p$ -series and ratios of polynomials the limit is 1, so compare instead.
Only the limit matters; early ratios above or below 1 change nothing.

Part 7 · Misconception

A common mistake

The wrong idea: $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ $= 1$ for $\sum \frac{1}{n^2}$ , so the series diverges.

What actually happens: A limit of 1 decides nothing. $\sum \frac{1}{n^2}$ converges as a $p$ -series with $p = 2$ , while $\sum \frac{1}{n}$ , with the same limit, diverges.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Ratios of consecutive terms of $\sum \frac{n^2}{3^n}$

The ratio $\frac{a_{n+1}}{a_n}$ for $a_n = \frac{n^2}{3^n}$ (decimals to four places)
$n$	$1$	$2$	$5$	$10$	$100$
$\frac{a_{n+1}}{a_n}$	$1.3333$	$0.7500$	$0.4800$	$0.4033$	$0.3400$

1. Find $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ for this series.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{(n+1)^2}{3n^2}$ $= \frac{1}{3}\left(1 + \frac{1}{n}\right)^2 \to \frac{1}{3}$ , which the table's values approach.

Answer: $\frac{1}{3}$ ≈ 0.333

2. What does the ratio test conclude about $\sum_{n=1}^{\infty} \frac{n^2}{3^n}$ ?

It diverges, since the first ratio, $\frac{4}{3}$ , is greater than 1
Nothing, since some of the ratios are above 1
It converges to $\frac{1}{3}$ , the limit of the ratios
It converges, since the limit $\frac{1}{3}$ is less than 1

Show the answer

The ratio test limit is $L = \frac{1}{3} < 1$ , so the series converges. The limit says whether it converges, not what it converges to.

It diverges, since the first ratio, $\frac{4}{3}$ , is greater than 1: Only the limit decides; one early ratio above 1 changes nothing.
Nothing, since some of the ratios are above 1: The test uses the limit, not every ratio; the limit $\frac{1}{3}$ is below 1.
It converges to $\frac{1}{3}$ , the limit of the ratios: The limit of the ratios decides convergence; it is not the sum.
Correct: It converges, since the limit $\frac{1}{3}$ is less than 1: Right: a ratio test limit below 1 gives convergence.

Table

Ratios of consecutive terms of $\sum \frac{n!}{10^n}$

The ratio $\frac{a_{n+1}}{a_n}$ $= \frac{n+1}{10}$ for $a_n = \frac{n!}{10^n}$
$n$	$1$	$5$	$9$	$20$	$100$
$\frac{a_{n+1}}{a_n}$	$0.2$	$0.6$	$1.0$	$2.1$	$10.1$

3. What does the ratio test conclude about $\sum_{n=1}^{\infty} \frac{n!}{10^n}$ ?

It converges: the first ratios are below 1
It diverges: $\frac{n+1}{10}$ grows without bound
Nothing: the ratio test limit is not a number
It converges: $\frac{n+1}{10}$ is 1 at $n = 9$

Show the answer

$\lim_{n \to \infty} \frac{n + 1}{10}$ $= \infty$ . A ratio test limit greater than 1, including an infinite one, means the series diverges.

It converges: the first ratios are below 1: Only the limit decides, and the ratios grow past every bound.
Correct: It diverges: $\frac{n+1}{10}$ grows without bound: Right: a limit of infinity counts as greater than 1.
Nothing: the ratio test limit is not a number: An infinite limit is a conclusion: the series diverges.
It converges: $\frac{n+1}{10}$ is 1 at $n = 9$ : A ratio of exactly 1 at one $n$ means nothing; the limit is infinite.

4. A student notes that the ratios for $n = 1$ to $8$ are all below 1 and concludes that the series converges. What is wrong?

The ratio test uses the limit as $n \to \infty$ , which is infinite
Nothing: ratios below 1 mean the terms shrink, so the series converges
The ratios should be compared with 0, not with 1
The ratio test needs each ratio to be below $\frac{1}{2}$

Show the answer

The ratio test looks at $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ . Here the limit is infinite, so the series diverges. Early ratios below 1 only say the first terms shrink.

Correct: The ratio test uses the limit as $n \to \infty$ , which is infinite: Right: finitely many ratios never decide the test.
Nothing: ratios below 1 mean the terms shrink, so the series converges: The terms shrink only until $n = 9$ ; after that they grow without bound.
The ratios should be compared with 0, not with 1: The threshold is 1: a limit below 1 converges, above 1 diverges.
The ratio test needs each ratio to be below $\frac{1}{2}$ : No ratio needs to be below $\frac{1}{2}$ ; the limit decides.

5. For both $\sum \frac{1}{n}$ and $\sum \frac{1}{n^2}$ the ratio test limit is 1. What does that show?

Both series converge, since their ratio test limits are equal
A limit of 1 decides nothing: the two series differ
Both series diverge, since their ratio test limits equal 1
The ratio test was done wrong, since its limit is less than 1

Show the answer

$\frac{n}{n + 1} \to 1$ and $\frac{n^2}{(n + 1)^2} \to 1$ , yet $\sum \frac{1}{n}$ diverges and $\sum \frac{1}{n^2}$ converges. A ratio test limit of 1 decides nothing; use another test (here the $p$ -series test).

Both series converge, since their ratio test limits are equal: $\sum \frac{1}{n}$ diverges, so equal limits do not mean equal behavior.
Correct: A limit of 1 decides nothing: the two series differ: Right: the two series behave differently with the same limit.
Both series diverge, since their ratio test limits equal 1: $\sum \frac{1}{n^2}$ converges ( $p = 2$ ), so a limit of 1 does not mean divergence.
The ratio test was done wrong, since its limit is less than 1: $\frac{n}{n+1} \to 1$ and $\frac{n^2}{(n+1)^2} \to 1$ : both limits are 1.

6. What does the ratio test conclude about $\sum_{n=1}^{\infty} \frac{2^n}{n^5}$ ?

It converges: the limit is $\frac{1}{2}$
It diverges: the limit is 2
Nothing: the limit is 1
It converges: $n^5$ is large

Show the answer

$\frac{a_{n+1}}{a_n}$ $= \frac{2^{n+1}}{(n+1)^5} \cdot \frac{n^5}{2^n}$ $= 2\left(\frac{n}{n+1}\right)^5 \to 2 > 1$ , so the series diverges.

It converges: the limit is $\frac{1}{2}$ : That inverts the ratio; $\frac{a_{n+1}}{a_n}$ $= 2 \cdot \frac{n^5}{(n+1)^5} \to 2$ .
Correct: It diverges: the limit is 2: Right: $2 > 1$ .
Nothing: the limit is 1: $\frac{n^5}{(n+1)^5} \to 1$ , but the factor 2 stays: the limit is 2.
It converges: $n^5$ is large: $2^n$ eventually outgrows $n^5$ ; the limit 2 shows it.

Part 9 · Summary

Summary

The ratio test compares a series with a geometric one. Find

L

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

: if

L < 1

the series converges, if

L > 1

or

L

is infinite it diverges, and if

L = 1

the test gives no conclusion. It works best on factorials and powers like

3^n

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Ratios of consecutive terms of ∑n23n\sum \frac{n^2}{3^n}

Ratios of consecutive terms of ∑n!10n\sum \frac{n!}{10^n}

Summary

What comes next

Connections

Ratios of consecutive terms of $\sum \frac{n^2}{3^n}$

Ratios of consecutive terms of $\sum \frac{n!}{10^n}$