Integral Test for Convergence

AP Calculus BC· AP Calculus BC CED — Infinite Sequences and Series· 14 min read

1. The Integral Test and Required Hypotheses★★☆☆☆⏱ 4 min

The integral test for convergence connects the behavior of an infinite discrete positive-term series to the convergence of an improper integral, leveraging your existing knowledge of integration. For a series $\sum_{n = N}^{\infty} a_{n}$ where $a_{n} = f (n)$ , convergence of the series matches convergence of the improper integral $\int_{N}^{\infty} f (x) d x$ when all three hypotheses are satisfied.

$f$ is continuous on $[N, \infty)$ : no discontinuities, jumps, or vertical asymptotes in the interval
$f$ is positive on $[N, \infty)$ : all outputs of $f$ for $x \geq N$ are greater than 0
$f$ is decreasing on $[N, \infty)$ : $f^{'} (x) \leq 0$ for all $x > N$ , so $f (x)$ does not increase as $x$ increases

📐 Worked Example

Does the integral test apply to the series $\sum_{n = 2}^{\infty} \frac{l n n}{n}$ ? Verify all hypotheses to justify your answer.

1
Define $f (x) = \frac{l n x}{x}$ for $x \geq 2$ . Check continuity: $f (x)$ is a quotient of two continuous functions, and denominator $x \neq = 0$ for $x \geq 2$ , so $f$ is continuous on $[2, \infty)$ .
2
Check positivity: For $x \geq 2$ , $ln x > ln 2 > 0$ and $x > 0$ , so $f (x) > 0$ on $[2, \infty)$ .
3
Check if $f$ is decreasing by computing the derivative:
4
$\frac{(\frac{1}{x})x - \ln x (1)}{x^2} = \frac{1 - \ln x}{x^2}$
5
For $x > e \approx 2.718$ , $ln x > 1$ , so $1 - ln x < 0$ , meaning $f^{'} (x) < 0$ on $(e, \infty)$ . We can shift the starting index to $n = 3$ (convergence only depends on the series tail), so all hypotheses are satisfied starting at $N = 3$ . The integral test can be applied.

2. Applying the Integral Test for Convergence★★☆☆☆⏱ 4 min

Once all three hypotheses are verified, the integral test gives a clear conclusion: the series $\sum_{n = N}^{\infty} a_{n}$ converges if and only if the improper integral $\int_{N}^{\infty} f (x) d x$ converges. A finite integral means the series converges; a divergent integral means the series diverges.

This relationship comes from Riemann sum bounding for decreasing positive functions:

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The most famous result from the integral test is the p-series convergence rule: $\sum_{n = 1}^{\infty} \frac{1}{n ^{p}}$ converges if $p > 1$ and diverges if $p \leq 1$ , which is proven directly using the integral test.

📐 Worked Example

Determine whether the series $\sum_{n = 1}^{\infty} \frac{1}{n ^{2} + 1}$ converges or diverges using the integral test.

1
Define $f (x) = \frac{1}{x ^{2} + 1}$ for $x \geq 1$ . Verify all hypotheses: (1) Continuous: denominator is never zero for all real $x$ , so $f$ is continuous on $[1, \infty)$ . (2) Positive: $x^{2} + 1 > 0$ for all $x$ , so $f (x) > 0$ . (3) Decreasing: $f^{'} (x) = \frac{- 2 x}{( x ^{2} + 1 ) ^{2}} < 0$ for all $x > 0$ , so $f$ is decreasing. All hypotheses are satisfied.
2
Set up and evaluate the improper integral:
3
$\int_1^\infty \frac{1}{x^2 + 1} dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2 +1} dx = \lim_{b \to \infty} \left[ \arctan x \right]_1^b$
4
Compute the limit of the integral:
5
$\lim_{b \to \infty} (\arctan b - \arctan 1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$
6
The integral evaluates to a finite value, so by the integral test, the series converges.

3. Remainder Error Bounds for Partial Sums★★★☆☆⏱ 3 min

When you approximate the sum of a convergent infinite series with its $n$ -th partial sum $S_{n}$ , the remainder $R_{n} = S - S_{n}$ is the error in your approximation. For series satisfying the integral test hypotheses, we can find explicit upper and lower bounds for this error.

If $f$ is continuous, positive, and decreasing for $x \geq n$ , and $\sum_{k = 1}^{\infty} f (k) = S$ converges, the remainder satisfies:

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📐 Worked Example

The series $\sum_{n = 1}^{\infty} \frac{1}{n ^{3}}$ is known to converge. Find an upper and lower bound for the error when the sum is approximated by the 10th partial sum $S_{10}$ .

1
Confirm $f (x) = \frac{1}{x ^{3}}$ satisfies all integral test hypotheses for $x \geq 1$ : it is continuous, positive, and $f^{'} (x) = - 3 x^{- 4} < 0$ so it is decreasing. The remainder bound formula applies.
2
Calculate the lower bound for $R_{10}$ :
3
$\int_{11}^\infty \frac{1}{x^3} dx = \lim_{b \to \infty} \int_{11}^b x^{-3} dx = \lim_{b \to \infty} \left[ -\frac{1}{2x^2} \right]_{11}^b = 0 - (-\frac{1}{2(121)}) = \frac{1}{242} \approx 0.00413$
4
Calculate the upper bound for $R_{10}$ :
5
$\int_{10}^\infty \frac{1}{x^3} dx = \frac{1}{2(10)^2} = 0.005$
6
Conclusion: The error $R_{10}$ satisfies $0.00413 \leq R_{10} \leq 0.005$ .

4. AP-Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding of core integral test concepts:

For which of the following series can the integral test for convergence be correctly applied, starting at the given lower bound of the series?
- A) $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n}$
- B) $\sum_{n = 1}^{\infty} \frac{e ^{- n}}{n ^{2}}$
- C) $\sum_{n = 1}^{\infty} \frac{c o s n}{n ^{3}}$
- D) $\sum_{n = 1}^{\infty} \frac{1}{1 - n ^{2}}$
Reveal answer
B) $\sum_{n=1}^\infty \frac{e^{-n}}{n^2}$ —
Option A has alternating negative terms, violating positivity. Option C has terms that alternate sign, violating positivity. Option D has a division by zero at $n = 1$ , violating continuity. Only Option B satisfies all three hypotheses of continuous, positive, decreasing on $[1, \infty)$ .

5. Common Pitfalls

Wrong move:

Claiming the integral test applies to $\sum_{n = 1}^{\infty} \frac{s i n n}{n ^{2}}$ because $\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x$ converges.

Why:

The positivity condition is violated, since $\frac{s i n x}{x ^{2}}$ alternates sign.

Correct move:

Always check all three hypotheses (continuous, positive, decreasing) before applying the integral test, and never use the test on alternating or term-negative series.

Wrong move:

Concluding $\sum_{n = 1}^{\infty} \frac{1}{n}$ converges because $\int_{1}^{1000} \frac{1}{x} d x = ln 1000$ is finite.

Why:

Students confuse definite integrals with improper integrals, forgetting we need the integral to infinity to apply the test.

Correct move:

Always evaluate the improper integral by taking the limit as the upper bound goes to infinity, never stop at a finite bound.

Wrong move:

Claiming $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ converges to 1 because $\int_{1}^{\infty} \frac{1}{x ^{2}} d x = 1$ .

Why:

Students confuse the value of the integral with the sum of the series.

Correct move:

Remember that the integral test only determines convergence or divergence, it never gives the value of the series sum.

Wrong move:

Stating the series diverges via the integral test when $lim_{n \to \infty} a_{n} \neq = 0$ .

Why:

Students mix up the nth term test with the integral test, incorrectly attributing the divergence conclusion to the wrong test.

Correct move:

If $lim_{n \to \infty} a_{n} \neq = 0$ , use the nth term test for divergence to conclude divergence, do not invoke the integral test for this conclusion.

Wrong move:

Using the remainder bound formula for a divergent series to bound partial sum error.

Why:

Students memorize the formula without remembering it only applies to convergent series, since divergent series have no finite sum to approximate.

Correct move:

Only use the integral test remainder bound after you have confirmed the series converges via the integral test.

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Integral Test Hypotheses	$f$ continuous on $[N, \infty)$ 2. $f$ positive on $[N, \infty)$ 3. $f$ decreasing on $[N, \infty)$	Must verify all three before applying; can shift $N$ to any starting integer if needed
Integral Test Convergence Rule	$\sum_{n = N}^{\infty} a_{n}$ converges $⟺ \int_{N}^{\infty} f (x) d x$ converges	Divergence of the integral implies divergence of the series; only for positive-term series
P-Series Convergence (proven via integral test)	$\sum_{n = 1}^{\infty} \frac{1}{n ^{p}}$ converges if $p > 1$ , diverges if $p \leq 1$	Used widely for comparison tests after this result is established
Lower Bound for Remainder Error	$\int_{n + 1}^{\infty} f (x) d x \leq R_{n}$	$R_{n} = S - S_{n}$ = error from approximating total sum $S$ by partial sum $S_{n}$ ; only for convergent series
Upper Bound for Remainder Error	$R_{n} \leq \int_{n}^{\infty} f (x) d x$	Most AP questions asking for an error bound expect this upper bound as the answer
What Integral Test Does Not Give	$\sum_{n = 1}^{\infty} a_{n} \neq = \int_{1}^{\infty} f (x) d x$	The test only tells you convergence/divergence, not the value of the series sum
Invalid Application	No conclusion possible if any hypothesis is violated	If the test can't be applied, use another test (alternating series, comparison, ratio)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Check integral test applicability
2023 · FRQ
Convergence + error bound question

What's Next

The integral test is the foundation for most other convergence tests for infinite series, and it is a prerequisite for the direct and limit comparison tests that come next in the AP Calculus BC syllabus. Understanding how the tail of a series behaves relative to an improper integral also builds intuition for all error bound questions, which appear regularly in FRQs on alternating series and power series approximation. Without mastering the hypotheses of the integral test, you will often incorrectly justify convergence tests and lose points on free-response questions. The integral test also reinforces the core AP Calculus connection between discrete sums and continuous integration.