Comparison Tests — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

You cannot add infinitely many terms, but you can often tell whether their sum settles by setting the series beside one you already know. If every term of a positive series is smaller than the matching term of a convergent geometric series or

p

-series, its partial sums are trapped under a ceiling, and the series converges.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Does $\sum_{n=1}^{\infty} \frac{1}{n^3}$ converge?

Yes: it is a $p$ -series with $p = 3 > 1$
No: its terms approach 0 too slowly
No: it is the harmonic series
Yes: it is geometric with ratio $\frac{1}{3}$

Show the answer

A $p$ -series $\sum \frac{1}{n^p}$ converges exactly when $p > 1$ .

Correct: Yes: it is a $p$ -series with $p = 3 > 1$ :
No: its terms approach 0 too slowly:
No: it is the harmonic series:
Yes: it is geometric with ratio $\frac{1}{3}$ :

2. Does $\sum_{n=0}^{\infty} \left(\frac{4}{5}\right)^n$ converge?

Yes: it is geometric with $|r| = \frac{4}{5} < 1$
No: its ratio is positive
No: its terms are all positive
Yes: it is a $p$ -series with $p = \frac{4}{5}$

Show the answer

A geometric series converges when $|r| < 1$ ; here its sum is $\frac{1}{1 - 4/5} = 5$ .

Correct: Yes: it is geometric with $|r| = \frac{4}{5} < 1$ :
No: its ratio is positive:
No: its terms are all positive:
Yes: it is a $p$ -series with $p = \frac{4}{5}$ :

3. Find $\lim_{n \to \infty} \frac{5n^2 - 1}{2n^2 + n}$ .

$\frac{5}{2}$
$0$
$5$
The limit is infinite

Show the answer

Divide the top and bottom by $n^2$ : $\frac{5 - 1/n^2}{2 + 1/n} \to \frac{5}{2}$ .

Correct: $\frac{5}{2}$ :
$0$ :
$5$ :
The limit is infinite:

Part 4 · See it

See it first

Partial sums S sub N for N = 1 to 12 of two series, plotted as dots. The upper dots, for the series of 1 over n squared, rise from 1 and level off just under a dashed line at pi squared over 6, about 1.645. The lower dots, for the series of 1 over (n squared + 1), rise from 0.5 and level off near 1.08, always below the upper dots. Smaller positive terms give smaller partial sums, trapped under the same ceiling. — The partial sums of $\sum \frac{1}{n^2 + 1}$ stay under those of $\sum \frac{1}{n^2}$ , which level off below $\frac{\pi^2}{6} \approx 1.645$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Every term of $\sum a_n$ is positiveits partial sums only increase, so they converge exactly when they stay below some ceiling
$0 \le a_n \le b_n$ for every $n$ and $\sum b_n$ convergeseach partial sum of $\sum a_n$ is below the sum of $\sum b_n$ , so $\sum a_n$ converges (direct comparison)
$a_n \ge b_n \ge 0$ and $\sum b_n$ divergesthe partial sums of $\sum a_n$ grow past every bound, so $\sum a_n$ diverges
$\lim_{n \to \infty} \frac{a_n}{b_n}$ $= L$ with $0 < L < \infty$ for large $n$ , $a_n$ is about $L\,b_n$ , so the two series converge together or diverge together (limit comparison)

Part 6 · Key ideas

Key ideas

Compare positive terms with a series whose behavior is known and named: geometric with its ratio, or a $p$ -series with its $p$ .
Direct comparison needs the inequality in the useful direction: smaller than convergent, or larger than divergent. Larger than convergent, or smaller than divergent, proves nothing.
Choose $b_n$ by keeping the highest powers on top and bottom; use limit comparison when the inequality points the wrong way.
Write the limit in $n$ : $\lim_{n \to \infty} \frac{a_n}{b_n}$ , and say it is finite and positive.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{1}{n^2 + n} \le \frac{1}{n}$ and the harmonic series diverges, so $\sum \frac{1}{n^2 + n}$ diverges.

What actually happens: Being smaller than a divergent series proves nothing. Compare with $\frac{1}{n^2}$ instead: $\frac{1}{n^2 + n} < \frac{1}{n^2}$ and $p = 2 > 1$ , so the series converges.

Part 8 · Check yourself

Check yourself

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

Table

Two series with positive terms

The first terms of $a_n = \frac{1}{3^n + n}$ and $b_n = \frac{1}{3^n}$
$n$	$a_n = \frac{1}{3^n + n}$	$b_n = \frac{1}{3^n}$
$1$	$\frac{1}{4}$	$\frac{1}{3}$
$2$	$\frac{1}{11}$	$\frac{1}{9}$
$3$	$\frac{1}{30}$	$\frac{1}{27}$
$4$	$\frac{1}{85}$	$\frac{1}{81}$

1. Which inequality do the table's values suggest, and why does it hold for every $n \ge 1$ ?

$a_n > b_n$ , because $3^n + n > 3^n$
$a_n < b_n$ , because $3^n + n > 3^n$
$a_n < b_n$ , because $a_n$ decreases faster
$a_n = b_n$ once $n$ is large

Show the answer

For every $n \ge 1$ , $3^n + n > 3^n > 0$ , so $0 < \frac{1}{3^n + n} < \frac{1}{3^n}$ : each $a_n$ is smaller than the matching $b_n$ .

$a_n > b_n$ , because $3^n + n > 3^n$ : A bigger denominator gives a smaller fraction, so the inequality points the other way.
Correct: $a_n < b_n$ , because $3^n + n > 3^n$ : Right: a bigger denominator makes a smaller fraction.
$a_n < b_n$ , because $a_n$ decreases faster: How fast the terms shrink is not the reason; compare the denominators term by term.
$a_n = b_n$ once $n$ is large: The two are close for large $n$ but differ by the $+\,n$ in the denominator for every $n$ .

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

It diverges, since $0 < a_n < b_n$ and $\sum b_n$ is geometric
Nothing, since $a_n$ is not geometric
It converges, since $a_n$ approaches 0
It converges, since $0 < a_n < b_n$ and $\sum b_n$ converges

Show the answer

The terms are positive and $a_n < b_n$ for every $n$ ; $\sum b_n$ is geometric with $r = \frac{1}{3}$ , so it converges, and by the direct comparison test so does $\sum a_n$ .

It diverges, since $0 < a_n < b_n$ and $\sum b_n$ is geometric: Being smaller than a convergent series is the case that gives convergence.
Nothing, since $a_n$ is not geometric: The test does not need $a_n$ to be geometric; only the comparison series needs a known behavior.
It converges, since $a_n$ approaches 0: Terms approaching 0 do not prove convergence (the harmonic series has them); the comparison does.
Correct: It converges, since $0 < a_n < b_n$ and $\sum b_n$ converges: Right: smaller positive terms than a convergent series.

Table

A limit comparison, step by step

The ratio $\frac{a_n}{b_n}$ for $a_n = \frac{n^2 + 1}{n^4 + 3n}$ and $b_n = \frac{1}{n^2}$ (values to four decimal places)
$n$	$\frac{a_n}{b_n}$ $= \frac{n^2(n^2 + 1)}{n^4 + 3n}$
$1$	$0.5000$
$10$	$1.0070$
$100$	$1.0001$
$1000$	$1.0000$

3. Why is $b_n = \frac{1}{n^2}$ a good series to compare with $a_n = \frac{n^2 + 1}{n^4 + 3n}$ ?

Keeping the constant terms gives $\frac{1}{n^2}$ after cancelling
Keeping the highest powers gives $\frac{n^2}{n^4} = \frac{1}{n^2}$
Both terms are equal at $n = 1$
$\frac{1}{n^2}$ is smaller than $a_n$ for large $n$ , as the test needs

Show the answer

For large $n$ the numerator behaves like $n^2$ and the denominator like $n^4$ , so $a_n$ behaves like $\frac{n^2}{n^4} = \frac{1}{n^2}$ .

Keeping the constant terms gives $\frac{1}{n^2}$ after cancelling: The constants matter least for large $n$ ; the highest powers dominate.
Correct: Keeping the highest powers gives $\frac{n^2}{n^4} = \frac{1}{n^2}$ : Right: the dominant terms decide how the terms behave for large $n$ .
Both terms are equal at $n = 1$ : At $n = 1$ , $a_1 = \frac{2}{4}$ and $b_1 = 1$ : they are not equal, and one value decides nothing.
$\frac{1}{n^2}$ is smaller than $a_n$ for large $n$ , as the test needs: A limit comparison needs no inequality; the choice comes from the dominant powers.

4. What does the limit comparison test conclude about $\sum_{n=1}^{\infty} \frac{n^2 + 1}{n^4 + 3n}$ ?

It diverges: the limit is 1, which is not less than 1 as the test needs
Nothing: the ratio is 0.5 at $n = 1$ , so the two series differ
It converges: the terms approach 0 and the ratios approach 1
It converges, like the $p$ -series $\sum \frac{1}{n^2}$

Show the answer

Both series have positive terms and $\lim_{n \to \infty} \frac{a_n}{b_n}$ $= 1$ , a finite positive number. $\sum \frac{1}{n^2}$ is a $p$ -series with $p = 2 > 1$ , so it converges, and by the limit comparison test $\sum a_n$ converges.

It diverges: the limit is 1, which is not less than 1 as the test needs: The limit comparison test does not compare the limit with 1; any finite positive limit links the two series.
Nothing: the ratio is 0.5 at $n = 1$ , so the two series differ: Only the limit as $n \to \infty$ matters, not early values.
It converges: the terms approach 0 and the ratios approach 1: Terms approaching 0 are not enough; the comparison with $p = 2$ is the reason.
Correct: It converges, like the $p$ -series $\sum \frac{1}{n^2}$ : Right: a finite positive limit means both series do the same thing.

5. A student writes: " $\frac{1}{n+1} < \frac{1}{n}$ and the harmonic series diverges, so $\sum_{n=1}^{\infty} \frac{1}{n+1}$ diverges by the direct comparison test." What is wrong?

The series converges, because its terms are smaller than those of the harmonic series
The inequality is false: $\frac{1}{n+1} > \frac{1}{n}$ , so the comparison is reversed
Nothing: the argument is correct, since both series have positive terms
Smaller than a divergent series proves nothing

Show the answer

To show divergence by direct comparison, the terms must be at least as large as those of a divergent series. Here they are smaller, so no conclusion follows. The limit comparison with $\frac{1}{n}$ works: $\lim_{n \to \infty} \frac{n}{n+1}$ $= 1$ , so the series diverges.

The series converges, because its terms are smaller than those of the harmonic series: The series does diverge (its limit comparison with $\frac{1}{n}$ gives 1); only the reason is wrong.
The inequality is false: $\frac{1}{n+1} > \frac{1}{n}$ , so the comparison is reversed: The inequality is true; the problem is which way it points.
Nothing: the argument is correct, since both series have positive terms: A divergent series that is larger shows nothing about a smaller one.
Correct: Smaller than a divergent series proves nothing: Right: the inequality points the wrong way for divergence.

6. Which argument shows that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ converges?

Limit comparison with $\frac{1}{2^n}$ : the limit is 1 and $\sum \frac{1}{2^n}$ converges
Direct comparison with $\frac{1}{2^n}$ , since $\frac{1}{2^n - 1} < \frac{1}{2^n}$ for each $n$
Direct comparison with $\frac{1}{n}$ , since $\frac{1}{2^n - 1} < \frac{1}{n}$
Limit comparison with $\frac{1}{n}$ : the limit is 0 and $\sum \frac{1}{n}$ diverges

Show the answer

The terms $\frac{1}{2^n - 1}$ are larger than $\frac{1}{2^n}$ , so a direct comparison goes the wrong way. With limit comparison, $\lim_{n \to \infty} \frac{1/(2^n - 1)}{1/2^n}$ $= \lim_{n \to \infty} \frac{2^n}{2^n - 1}$ $= 1$ , and $\sum \frac{1}{2^n}$ is geometric with $r = \frac{1}{2}$ , so the series converges.

Correct: Limit comparison with $\frac{1}{2^n}$ : the limit is 1 and $\sum \frac{1}{2^n}$ converges: Right: the limit $\lim \frac{2^n}{2^n - 1}$ is 1.
Direct comparison with $\frac{1}{2^n}$ , since $\frac{1}{2^n - 1} < \frac{1}{2^n}$ for each $n$ : The inequality is backwards: $2^n - 1 < 2^n$ makes $\frac{1}{2^n - 1}$ the larger term.
Direct comparison with $\frac{1}{n}$ , since $\frac{1}{2^n - 1} < \frac{1}{n}$ : Smaller than a divergent series proves nothing.
Limit comparison with $\frac{1}{n}$ : the limit is 0 and $\sum \frac{1}{n}$ diverges: The limit comparison test needs a finite positive limit; 0 is not one.

Part 9 · Summary

Summary

For a series with positive terms, compare it with a known geometric series or

p

-series. Direct comparison: smaller than a convergent series converges, larger than a divergent series diverges. Limit comparison: if

\lim_{n \to \infty} \frac{a_n}{b_n}

is finite and positive, both series do the same thing.

Part 10 · Up next

What comes next

Part 11 · Connections