# Integral Test for Convergence

> AP Calculus BC · Unit 10: Infinite Sequences and Series
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u10-integral-test-for-convergence/

This guide covers the hypotheses, application, and error bounds for the integral test for convergence, a core AP Calculus BC Unit 10 topic that appears in both multiple-choice and free-response exam sections.

**Prerequisites:** Evaluating improper integrals of all types; Properties of positive-term infinite series; Computing derivatives and antiderivatives of basic functions

## Learning objectives

- Verify the three hypotheses of the integral test for convergence
- Apply the integral test to determine convergence or divergence of positive-term series
- Calculate upper and lower remainder error bounds for partial sum approximations
- Recognize and avoid common mistakes when applying the integral test on the AP exam

## The Integral Test and Required Hypotheses

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 $\int_N^\infty f(x) dx$ when all three hypotheses are satisfied.

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

> **tip**
>
> On AP FRQs, you must explicitly verify all three hypotheses to earn full credit—never skip this step, even if conditions seem obvious.

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

1. Define $f(x) = \frac{\ln x}{x}$ 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.

## Applying the Integral Test for Convergence

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 $\int_N^\infty f(x) dx$ 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:

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.

> **tip**
>
> The integral test only tells you if the series converges—it never tells you what the sum of the series is. A common MCQ distractor is the value of the integral, which is almost never the sum of the series.

**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: $f'(x) = \frac{-2x}{(x^2 +1)^2} < 0$ 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.

## Remainder Error Bounds for Partial Sums

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:

> **tip**
>
> If an AP question only asks for *an error bound* (not both upper and lower), it will always accept the upper bound $\int_{n}^\infty f(x) dx$ as the correct answer.

**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 $f'(x) = -3x^{-4} < 0$ 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 $0.00413 \leq R_{10} \leq 0.005$.

## AP-Style Concept Check

**Check your understanding**

Test your understanding of core integral test concepts:

1. 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) $\sum_{n=1}^\infty \frac{\cos n}{n^3}$
   - D) $\sum_{n=1}^\infty \frac{1}{1 - n^2}$

   *Why:* 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 $[1, \infty)$.

## Common pitfalls

- **Wrong:** Claiming the integral test applies to $\sum_{n=1}^\infty \frac{\sin n}{n^2}$ because $\int_1^\infty \frac{\sin x}{x^2} dx$ converges.
  - Why it fails: The positivity condition is violated, since $\frac{\sin x}{x^2}$ alternates sign.
  - Correct: 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:** Concluding $\sum_{n=1}^\infty \frac{1}{n}$ converges because $\int_1^{1000} \frac{1}{x} dx = \ln 1000$ is finite.
  - Why it fails: Students confuse definite integrals with improper integrals, forgetting we need the integral to infinity to apply the test.
  - Correct: Always evaluate the improper integral by taking the limit as the upper bound goes to infinity, never stop at a finite bound.
- **Wrong:** Claiming $\sum_{n=1}^\infty \frac{1}{n^2}$ converges to 1 because $\int_1^\infty \frac{1}{x^2} dx = 1$.
  - Why it fails: Students confuse the value of the integral with the sum of the series.
  - Correct: Remember that the integral test only determines convergence or divergence, it never gives the value of the series sum.
- **Wrong:** Stating the series diverges via the integral test when $\lim_{n \to \infty} a_n \neq 0$.
  - Why it fails: Students mix up the nth term test with the integral test, incorrectly attributing the divergence conclusion to the wrong test.
  - Correct: 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:** Using the remainder bound formula for a divergent series to bound partial sum error.
  - Why it fails: Students memorize the formula without remembering it only applies to convergent series, since divergent series have no finite sum to approximate.
  - Correct: Only use the integral test remainder bound after you have confirmed the series converges via the integral test.

## Cheatsheet

| Category | Formula/Rule | Notes |
| --- | --- | --- |
| Integral Test Hypotheses | 1. $f$ continuous on $[N, \infty)$<br>2. $f$ positive on $[N, \infty)$<br>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 $\iff \int_N^\infty f(x) dx$ 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 | $\int_{n+1}^\infty f(x) dx \leq R_n$ | $R_n = S - S_n$ = error from approximating total sum $S$ by partial sum $S_n$; only for convergent series |
| Upper Bound for Remainder Error | $R_n \leq \int_{n}^\infty f(x) dx$ | Most AP questions asking for an error bound expect this upper bound as the answer |
| What Integral Test Does Not Give | $\sum_{n=1}^\infty a_n \neq \int_1^\infty f(x) dx$ | 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) |

## 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.

- [Harmonic series and p-series](https://www.owlsprep.com/study/ap-calculus-bc-u10-harmonic-series-and-p-series/)
- [Comparison tests for convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-comparison-tests-for-convergence/)
- [Alternating series test for convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-alternating-series-test-for-convergence/)

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