College Board AP 2022 (BC, style)-style practice: Section I (multiple choice, no calculator). To show sum (1)/(n^2 + 1) converges by direct comparison, compare it to (A) sum (1)/(n^2) (B) sum (1)/(n) (C) sum 1 (D) sum (1)/(n^3)

The correct answer is (A), sum (1)/(n^2). Since (1)/(n^2 + 1) 1), direct comparison gives convergence of sum(1)/(n^2+1).

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How do you decide convergence by comparing a series to a known benchmark series?

Topic 10.6 Comparison Tests for Convergence: use the direct comparison test and the limit comparison test against a known p-series or geometric series (BC).

A focused answer to AP Calculus BC Topic 10.6, deciding convergence with the direct comparison test and the limit comparison test against a known p-series or geometric benchmark, with worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The two tests
Choosing the benchmark by dominant behavior
A worked direct comparison
When to prefer limit comparison
A worked limit comparison

What this topic is asking

The College Board (Topic 10.6, BC only) gives two comparison tests for series with positive terms: the direct comparison test and the limit comparison test. Both decide an unknown series by measuring it against a known benchmark, almost always a $p$ -series or a geometric series, whose behavior you already know from Topics 10.2 and 10.5.

The two tests

Choosing the benchmark by dominant behavior

The skill in both tests is choosing the comparison series, and the rule is to keep only the dominant terms of $a_n$ as $n\to\infty$ . For a rational expression, that means the highest power of $n$ in the numerator over the highest power in the denominator: $\frac{2n + 1}{n^3 + 5}$ behaves like $\frac{2n}{n^3} = \frac{2}{n^2}$ , so compare to $\frac{1}{n^2}$ (a $p$ -series with $p = 2$ ). For terms with exponentials, the exponential dominates, suggesting a geometric benchmark. Once you have the benchmark, you know its convergence from the $p$ -series or geometric rule, and the comparison transfers that verdict to $a_n$ . This dominant-behavior reasoning is exactly why $p$ -series and geometric series are the universal yardsticks.

A worked direct comparison

When to prefer limit comparison

Direct comparison needs you to prove an inequality in the right direction, which can be awkward when the benchmark is smaller where you need it larger (or vice versa). Limit comparison sidesteps this: you only compute a single limit of the ratio $\frac{a_n}{b_n}$ , and if it is a finite positive number, the two series share their fate, no inequality required. This is why limit comparison is the go-to for messy rational terms. For instance, $\sum\frac{2n+1}{n^3+5}$ is hard to bound by a clean inequality but easy by limit comparison to $\frac{1}{n^2}$ , as the worked exam question shows. The one caveat is that the limit must be finite and positive ( $0 < L < \infty$ ); a limit of $0$ or $\infty$ requires more care and usually a different choice of $b_n$ .

A worked limit comparison

Limit comparison to the harmonic series

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

step 1 Identify the dominant behavior

For large $n$ , $\sqrt{n^2 + n}\approx n$ , so the term behaves like $\frac{1}{n}$ . Compare to $b_n = \frac{1}{n}$ (harmonic, divergent).

step 2 Form the ratio

\frac{a_n}{b_n} = \frac{1/\sqrt{n^2 + n}}{1/n} = \frac{n}{\sqrt{n^2 + n}}.

step 3 Take the limit

\lim_{n\to\infty}\frac{n}{\sqrt{n^2 + n}} = \lim_{n\to\infty}\frac{1}{\sqrt{1 + 1/n}} = 1,

a finite positive number.

step 4 Conclude

Since $\sum\frac{1}{n}$ diverges and the limit is $0 < 1 < \infty$ , the given series diverges.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (BC, style)1 marksSection I (multiple choice, no calculator). To show

\sum \frac{1}{n^2 + 1}

converges by direct comparison, compare it to (A)

\sum \frac{1}{n^2}

(B)

\sum \frac{1}{n}

(C)

\sum 1

(D)

\sum \frac{1}{n^3}

Show worked answer →

The correct answer is (A), $\sum \frac{1}{n^2}$ .

Since $\frac{1}{n^2 + 1} < \frac{1}{n^2}$ for all $n$ , and $\sum\frac{1}{n^2}$ is a convergent $p$ -series ( $p = 2 > 1$ ), direct comparison gives convergence of $\sum\frac{1}{n^2+1}$ .

AP 2024 (BC, style)4 marksSection II (free response, no calculator). Use the limit comparison test to determine whether

\sum_{n=1}^{\infty} \frac{2n + 1}{n^3 + 5}

converges, naming your comparison series.

Show worked answer →

A 4-point limit-comparison problem.

(2 points) For large $n$ , $\frac{2n+1}{n^3+5}$ behaves like $\frac{2n}{n^3} = \frac{2}{n^2}$ ; compare to $b_n = \frac{1}{n^2}$ (convergent $p$ -series, $p = 2$ ).
(2 points) $\lim_{n\to\infty}\frac{(2n+1)/(n^3+5)}{1/n^2} = \lim_{n\to\infty}\frac{(2n+1)n^2}{n^3 + 5} = \lim_{n\to\infty}\frac{2n^3 + n^2}{n^3 + 5} = 2$ , a finite positive number. Since $\sum\frac{1}{n^2}$ converges, the given series converges.

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Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)