When should I use direct comparison versus limit comparison?

Use direct comparison when you can cleanly bound your series above (to prove convergence) or below (to prove divergence) by a known series, usually a p-series. Use limit comparison when constants or added terms make a clean inequality awkward — it only requires that the ratio of terms approach a finite positive number, so it handles rational and root expressions gracefully.

Do these tests work for series with negative terms?

No. The integral test, p-series test, direct comparison, and limit comparison all require positive (or eventually positive) terms. For series with alternating or mixed signs, use the alternating series test or the ratio test with absolute convergence covered in later lessons.

Why does the harmonic series diverge if its terms go to zero?

Terms approaching zero is necessary but not sufficient for convergence. The harmonic series is the p-series with p=1, and the integral _1^∞1/x\,dx=_b→∞ b diverges. The terms shrink too slowly for the partial sums to level off.

How much justification does the AP exam expect for these tests?

You must state the hypotheses you rely on. For the integral test, confirm the function is continuous, positive, and decreasing. For any comparison, explicitly name the comparison series and identify it as a convergent or divergent p-series or geometric series. Missing justification typically costs points even when the final answer is correct.

AP-CALCBC-10.3-10.6

U10.3 Integral Test, p-Series, and Comparison Tests

Master the integral test, p-series test, direct comparison, and limit comparison tests to determine convergence of positive-term series on AP Calculus BC.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U10.3 Integral Test, p-Series, and Comparison Tests, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Once you know a geometric series converges, the next challenge is deciding whether messier positive-term series like

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

\sum \frac{\ln n}{n}

converge. This lesson gives you four reliable tools: the integral test, the p-series test, the direct comparison test, and the limit comparison test.

These tests only work on series with positive terms, and each one shines in a different situation. The key skill the AP exam rewards is not just memorizing them but choosing the right test quickly and justifying every hypothesis. By the end you should be able to look at a series, recognize its dominant behavior, and pick a comparison or the integral test with confidence.

The Integral Test and p-Series

The integral test connects a series to an improper integral. If

f

is continuous, positive, and decreasing for

x \geq N

, and

a_n = f(n)

, then

\sum_{n=N}^{\infty} a_n

and

\int_{N}^{\infty} f(x)\, dx

either both converge or both diverge.

You must verify all three conditions — continuous, positive, decreasing — before applying it. A common mistake is skipping the decreasing check. Often you show a function is decreasing by noting

f'(x) < 0

or that the numerator grows slower than the denominator.

The most important consequence is the p-series test. The series

\sum_{n=1}^{\infty} \frac{1}{n^p}

converges if

p > 1

and diverges if

p \leq 1

. This follows directly from evaluating

\int_1^{\infty} x^{-p}\, dx

Series	$p$	Result
$\sum \frac{1}{n^2}$	$2$	converges
$\sum \frac{1}{\sqrt{n}}$	$1/2$	diverges
$\sum \frac{1}{n}$	$1$	diverges (harmonic)

Memorize that the harmonic series (

p=1

) diverges even though its terms shrink to zero — shrinking terms are necessary but not sufficient for convergence.

The Direct Comparison Test

The direct comparison test says: suppose

0 \leq a_n \leq b_n

for all large

n

. If

\sum b_n

converges, then

\sum a_n

converges (a smaller positive series can't outrun a convergent one). If

\sum a_n

diverges, then

\sum b_n

diverges.

The logic runs one direction only. Knowing

\sum b_n

diverges tells you nothing about a smaller series

\sum a_n

, and knowing

\sum a_n

converges tells you nothing about a larger

\sum b_n

.

To use it, you compare your series to a known p-series or geometric series. The trick is building the correct inequality. For example, since

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

and

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

converges, the original series converges.

The main danger is getting the inequality backwards. If you want to prove convergence, you need your series to be smaller than something that converges. If you want to prove divergence, your series must be larger than something that diverges. When the inequality points the wrong way, switch to limit comparison instead.

The Limit Comparison Test

When building an exact inequality is awkward — especially with polynomial or root expressions — the limit comparison test is cleaner. Given positive series

\sum a_n

and

\sum b_n

, compute

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

. If

0 < L < \infty

(a finite positive number), the two series both converge or both diverge.

The strategy is to choose

b_n

to capture the dominant behavior of

a_n

. Keep only the highest-power terms in numerator and denominator. For

a_n = \frac{3n^2+5}{n^4 - n + 2}

, the dominant ratio behaves like

\frac{3n^2}{n^4} = \frac{3}{n^2}

, so compare with

b_n = \frac{1}{n^2}

, a convergent p-series.

Situation	Best test
Clean inequality with p-series	direct comparison
Rational or root expression	limit comparison
$a_n = f(n)$ integrable	integral test
Pure $\frac{1}{n^p}$ form	p-series

Because you take a limit, small additive constants vanish, which is exactly why limit comparison is more forgiving than direct comparison. Just remember it still requires positive terms and a strictly positive finite limit to conclude.

Choosing and Justifying a Test

On the AP exam, most of the credit for these problems comes from justification, not just the final verdict. Whenever you use the integral test, explicitly state that

f

is continuous, positive, and decreasing. When you use a comparison, name the comparison series and identify it as a convergent or divergent p-series or geometric series.

A reliable decision process: first check whether the terms even go to zero — if not, the series diverges by the nth-term test (from an earlier lesson). Then identify the dominant term. If the series looks like

\frac{1}{n^p}

, use p-series immediately. If it is a rational/root expression, use limit comparison. If it involves

\ln n

, factorials, or nice antiderivatives, the integral test may work.

A frequent misconception is thinking terms going to zero guarantees convergence — the harmonic series disproves that. Another is applying these tests to series with negative terms; all four require positive (or eventually positive) terms. For alternating or sign-changing series you need the tests from U10.7 and U10.8. Always confirm your series qualifies before reaching for these tools.

Key terms

Integral Test.: If $f$ is continuous, positive, and decreasing with $a_n=f(n)$ , then $\sum a_n$ and $\int_N^{\infty} f(x)\,dx$ share the same convergence behavior.
p-Series.: A series of the form $\sum \frac{1}{n^p}$ ; converges when $p>1$ and diverges when $p\leq 1$ .
Harmonic Series.: The p-series with $p=1$ , namely $\sum \frac{1}{n}$ , which diverges despite terms approaching zero.
Direct Comparison Test.: For $0\leq a_n\leq b_n$ : convergence of the larger $\sum b_n$ forces convergence of $\sum a_n$ ; divergence of the smaller $\sum a_n$ forces divergence of $\sum b_n$ .
Limit Comparison Test.: If $L=\lim_{n\to\infty}\frac{a_n}{b_n}$ is finite and positive, then $\sum a_n$ and $\sum b_n$ both converge or both diverge.
Dominant term.: The highest-order part of an expression that controls large- $n$ behavior, used to pick a comparison series.
nth-term test.: If $\lim_{n\to\infty} a_n \neq 0$ , the series diverges; a necessary preliminary check before other tests.

Worked example

Determine whether

\sum_{n=1}^{\infty} \frac{2n+1}{n^3+4n}

converges or diverges, and justify your choice of test.

First note all terms are positive for

n\geq 1

, so a comparison test is legitimate. The terms clearly go to zero, so the nth-term test is inconclusive and we need more.

Identify the dominant behavior. The numerator behaves like

2n

and the denominator like

n^3

, so

a_n

behaves like

\frac{2n}{n^3}=\frac{2}{n^2}

. This suggests comparing with the convergent p-series

b_n=\frac{1}{n^2}

(here

p=2>1

).

Building a clean direct inequality is awkward because of the

+1

and

+4n

, so use the limit comparison test. Compute:

L=\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{2n+1}{n^3+4n}}{\frac{1}{n^2}}=\lim_{n\to\infty}\frac{(2n+1)n^2}{n^3+4n}.

Simplify:

\frac{2n^3+n^2}{n^3+4n}

. Divide top and bottom by

n^3

\frac{2+\frac{1}{n}}{1+\frac{4}{n^2}} \to \frac{2}{1}=2

.

Since

L=2

is finite and positive, the two series share the same behavior. Because

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

converges (p-series with

p=2>1

), the original series

\sum \frac{2n+1}{n^3+4n}

converges by the limit comparison test.

Practice questions

Which test most efficiently determines the convergence of

\sum_{n=2}^{\infty} \frac{1}{n\ln n}

The nth-term test
The integral test
The direct comparison test with $\sum \frac{1}{n^2}$
The p-series test directly

Answer: The integral test

Let

f(x)=\frac{1}{x\ln x}

, which is continuous, positive, and decreasing for

x\geq 2

. Then

\int_2^{\infty}\frac{1}{x\ln x}\,dx=\lim_{b\to\infty}[\ln(\ln x)]_2^b

diverges. So the series diverges by the integral test. The nth-term test fails since terms go to zero, and the series is not a plain p-series.

Determine whether

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

converges or diverges. Justify with an appropriate test.

Answer: Diverges by limit comparison with the divergent p-series $\sum \frac{1}{\sqrt{n}}$ .

All terms are positive. The dominant term of the denominator is

\sqrt{n}=n^{1/2}

, so compare with

b_n=\frac{1}{n^{1/2}}

. Compute

L=\lim_{n\to\infty}\frac{1/(\sqrt{n}+3)}{1/\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n}+3}=1

, which is finite and positive. Since

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

is a p-series with

p=\frac12\leq 1

, it diverges, so the original series diverges too.

For which values of

p

does

\sum_{n=1}^{\infty} \frac{1}{n^p}

converge?

$p<1$
$p\leq 1$
$p>1$
all real $p$

Answer: $p>1$

By the p-series test (a direct consequence of the integral test applied to

\int_1^{\infty} x^{-p}\,dx

), the series converges exactly when

p>1

and diverges when

p\leq 1

, including the harmonic case

p=1

FAQ

When should I use direct comparison versus limit comparison?: Use direct comparison when you can cleanly bound your series above (to prove convergence) or below (to prove divergence) by a known series, usually a p-series. Use limit comparison when constants or added terms make a clean inequality awkward — it only requires that the ratio of terms approach a finite positive number, so it handles rational and root expressions gracefully.
Do these tests work for series with negative terms?: No. The integral test, p-series test, direct comparison, and limit comparison all require positive (or eventually positive) terms. For series with alternating or mixed signs, use the alternating series test or the ratio test with absolute convergence covered in later lessons.
Why does the harmonic series diverge if its terms go to zero?: Terms approaching zero is necessary but not sufficient for convergence. The harmonic series is the p-series with $p=1$ , and the integral $\int_1^{\infty}\frac{1}{x}\,dx=\lim_{b\to\infty}\ln b$ diverges. The terms shrink too slowly for the partial sums to level off.
How much justification does the AP exam expect for these tests?: You must state the hypotheses you rely on. For the integral test, confirm the function is continuous, positive, and decreasing. For any comparison, explicitly name the comparison series and identify it as a convergent or divergent p-series or geometric series. Missing justification typically costs points even when the final answer is correct.

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