The Integral Test
Textbook Reference
| Primary source | Stewart, Calculus, 9th edition, Section 11.3: “The Integral Test and Estimates of Sums,” pages 789 to 793 (the test, the $p$-series, and the worked examples) |
| Open companion | OpenStax Calculus Volume 2, Section 5.3: “The Divergence and Integral Tests” |
| Companion link | https://openstax.org/books/calculus-volume-2/pages/5-3-the-divergence-and-integral-tests |
The test rests on the partial-sum definition of a series, the Monotonic Sequence Theorem from the sequence lessons, and improper integrals from earlier calculus.
Key idea
A sum of positive terms and the area under the matching curve are two views of nearly the same quantity, so they converge or diverge together.
Picture the series $\displaystyle\sum \frac{1}{n^2}$ as a row of rectangles, each of width $1$ and height equal to a term. Lay them next to the curve $y = \dfrac{1}{x^2}$. The rectangles and the area under the curve track each other so closely that if the area is finite, the stacked rectangles cannot add to more than a little extra, and if the area is infinite, the rectangles must be infinite too. The sum is trapped by the integral.
That is the whole idea of the integral test. You already know how to decide whether an improper integral $\displaystyle\int_1^{\infty} f(x)\,dx$ converges. The integral test lets you borrow that decision for the matching series: build a function $f$ with $f(n) = a_n$, check that it is positive and decreasing, integrate, and the series follows the integral’s verdict. It is the first test that decides convergence for series whose partial sums have no tidy closed form, which is almost all of them. The price is three conditions on $f$ (continuous, positive, decreasing), and the warning that the value of the integral is not the value of the sum; only their convergence is shared.
Prerequisite Check
If these are solid, you are ready.
Quick Reference
The Integral Test (formal gloss). Suppose $f$ is continuous, positive, and decreasing on $[1, \infty)$, and let $a_n = f(n)$. Then \[ \sum_{n=1}^{\infty} a_n \text{ converges} \iff \int_1^{\infty} f(x)\,dx \text{ converges.} \] (In plain words: a positive decreasing series and the area under its matching curve share the same fate.)
The three conditions on $f$. Continuous, positive, and decreasing on $[1, \infty)$. The decreasing condition need only hold eventually (for $x$ beyond some point), since a finite number of terms does not affect convergence.
The $p$-series result (memorize). \[ \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges if } p > 1 \text{ and diverges if } p \le 1. \] (The case $p = 1$ is the harmonic series, which diverges.)
Procedure.
- Let $f(x)$ be the term formula with $n$ replaced by $x$. Check it is continuous, positive, and decreasing (compute $f'$ if not obvious).
- Evaluate $\displaystyle\int_1^{\infty} f(x)\,dx$ as a limit.
- If the integral is finite, the series converges. If the integral is infinite, the series diverges.
Warning. The sum of the series is not equal to the value of the integral. The test shares only convergence, not the number.
Key Concepts
1. Why the Test Works: Rectangles Against the Curve
The test is a picture turned into a theorem. Take a positive decreasing $f$ with $a_n = f(n)$.
Convergent direction. Draw rectangles of width $1$ whose heights are the terms $a_2, a_3, a_4, \dots$, placed so each rectangle sits under the curve $y = f(x)$. Then the total rectangle area, which is $a_2 + a_3 + \cdots$, is at most the area under the curve from $1$ to infinity: \[ a_2 + a_3 + \cdots \le \int_1^{\infty} f(x)\,dx. \] So the partial sums satisfy $s_n \le a_1 + \displaystyle\int_1^{\infty} f(x)\,dx$, a fixed bound. The partial sums are also increasing (the terms are positive). An increasing sequence bounded above converges, by the Monotonic Sequence Theorem, so the series converges.
Divergent direction. Place the rectangles so their tops lie above the curve. Then the rectangle area exceeds the area under the curve. If the integral is infinite, the rectangle area is even larger, so the partial sums grow without bound and the series diverges.
The two pictures together are the proof. Notice the Monotonic Sequence Theorem from the sequence lessons doing the real work in the convergent case.
2. Applying the Test
Example 1 (a convergent series). Test $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ for convergence.
Goal. Confirm the three conditions, then integrate.
Work. Let $f(x) = \dfrac{1}{x^2 + 1}$. It is continuous, positive, and decreasing on $[1, \infty)$ (the denominator grows). Integrate: \[ \int_1^{\infty} \frac{1}{x^2 + 1}\,dx = \lim_{t\to\infty}\big[\tan^{-1} x\big]_1^t = \lim_{t\to\infty}\left(\tan^{-1} t - \frac{\pi}{4}\right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \] The integral is finite, so by the integral test the series converges.
Answer: the series converges.
Recap. The integral came out to $\dfrac{\pi}{4}$, a finite number, so the series converges. Note well: the sum of the series is not $\dfrac{\pi}{4}$; the test only shares the fact of convergence.
(Common error: reporting $\dfrac{\pi}{4}$ as the sum of the series. The integral and the sum are different numbers. The test transfers convergence, not value.)
Example 2 (the $p$-series). For what values of $p$ does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}$ converge?
Goal. Split into cases and use the known behavior of $\displaystyle\int_1^{\infty} x^{-p}\,dx$.
Work. If $p \le 0$, the terms do not go to $0$ (they stay at $1$ or grow), so the series diverges by the Test for Divergence. If $p > 0$, then $f(x) = \dfrac{1}{x^p}$ is continuous, positive, and decreasing on $[1, \infty)$, so the integral test applies. From earlier calculus, \[ \int_1^{\infty} \frac{1}{x^p}\,dx \text{ converges if } p > 1 \text{ and diverges if } p \le 1. \] Therefore the series converges if $p > 1$ and diverges if $0 < p \le 1$. Combined with the $p \le 0$ case, the series diverges for all $p \le 1$.
Answer: the $p$-series converges if and only if $p > 1$.
Recap. This result is used constantly for the rest of the chapter, so memorize it: $\sum 1/n^p$ converges exactly when $p > 1$. The borderline $p = 1$ is the harmonic series, which diverges.
3. When You Must Check the Decreasing Condition
For many series the term is obviously decreasing, but not always. When it is not obvious, compute the derivative.
Example 3 (check decreasing with a derivative). Determine whether $\displaystyle\sum_{n=2}^{\infty} \frac{\ln n}{n}$ converges or diverges.
Goal. The term is not clearly decreasing, so verify with $f'$, then integrate.
Work. Let $f(x) = \dfrac{\ln x}{x}$, positive and continuous for $x > 1$. Its derivative is \[ f'(x) = \frac{x \cdot \frac{1}{x} - \ln x}{x^2} = \frac{1 - \ln x}{x^2}, \] which is negative when $\ln x > 1$, that is for $x > e$. So $f$ is decreasing for $x > e$, which is enough (eventual decrease suffices). Now integrate, using the substitution that turns $\dfrac{\ln x}{x}$ into $\dfrac{(\ln x)^2}{2}$: \[ \int_{e}^{\infty} \frac{\ln x}{x}\,dx = \lim_{t\to\infty}\left[\frac{(\ln x)^2}{2}\right]_{e}^{t} = \lim_{t\to\infty}\frac{(\ln t)^2}{2} = \infty. \] The integral diverges, so by the integral test the series diverges.
Answer: the series diverges.
Recap. When the term is not visibly decreasing, the derivative settles it. Here $f$ decreases only after $x = e$, but eventual decrease is all the test needs, because the first few terms cannot change convergence.
(Common error: skipping the decreasing check. The integral test requires a positive decreasing function. If the term increases for a while, integrate from a point past where it starts decreasing, and say so.)
Inline Self-Check
Question. Without integrating, decide whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}$ and $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1/3}}$ converge.
Show answer
Both are $p$-series. The first has $p = 3 > 1$, so it converges. The second has $p = \tfrac{1}{3} \le 1$, so it diverges.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Equating the sum with the integral | “$\sum \frac{1}{n^2+1} = \frac{\pi}{4}$” | The test shares convergence, not value |
| Skipping the decreasing check | Applying the test to a non-decreasing term | Verify positive and decreasing (compute $f'$ if unsure) |
| Misreading the $p$-series rule | “$\sum 1/n$ converges” | Converges only for $p > 1$; $p = 1$ diverges |
| Forgetting the divergence test first | Integrating when $a_n \not\to 0$ | If terms do not go to $0$, the series diverges immediately |
| Demanding decrease from $n = 1$ | Rejecting $\frac{\ln n}{n}$ outright | Eventual decrease suffices; integrate from past that point |
Common Misconceptions
the sum of a convergent series equals the value of the corresponding integral.
This is the limit-equals-function-value error applied to the integral test. The test shares the verdict of convergence or divergence, not the numerical value. The series $\sum 1/n^2$ converges to $\pi^2/6 \approx 1.645$ while $\int_1^\infty x^{-2}\,dx = 1$. These are different numbers. The integral test certifies that a finite sum exists; it says nothing about what that sum equals.
the $p$-series $\sum 1/n^p$ converges for all $p > 0$.
This is the concept-image-conflicts-definition error about the cutoff. The series converges if and only if $p > 1$. The harmonic series $\sum 1/n$ has $p = 1$ and diverges, as do all $p$-series with $p \le 1$. A value such as $p = 0.99$, which is positive and close to $1$, still gives a divergent series. The boundary is strict: $p > 1$ is required.
Leveled Practice
Attempt each problem before opening the answer.
Level 1: Direct Application
Problem 1. Use the integral test on $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^6}$.
Show answer
This is a $p$-series with $p = 6 > 1$, so it converges. (Equivalently, $\displaystyle\int_1^{\infty} x^{-6}\,dx = \frac{1}{5}$ is finite.)
Problem 2. Use the integral test on $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$.
Show answer
Let $f(x) = \dfrac{1}{x \ln x}$, positive and decreasing for $x \ge 2$. With $u = \ln x$, \[ \int_2^{\infty}\frac{1}{x \ln x}\,dx = \lim_{t\to\infty}\big[\ln(\ln x)\big]_2^t = \infty. \] The integral diverges, so the series diverges.
Problem 3. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n^2}$ converges.
Show answer
Let $f(x) = \dfrac{\ln x}{x^2}$, positive and eventually decreasing. Integration by parts gives the antiderivative $\displaystyle\int \frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} - \frac{1}{x}$. Evaluate the improper integral: \[ \int_1^{\infty}\frac{\ln x}{x^2}\,dx = \lim_{t\to\infty}\left[-\frac{\ln x}{x} - \frac{1}{x}\right]_1^t. \] At the upper limit, both $\dfrac{\ln t}{t}$ and $\dfrac{1}{t}$ go to $0$. At $x = 1$, the bracket equals $-\dfrac{\ln 1}{1} - \dfrac{1}{1} = -1$. So the integral is $0 - (-1) = 1$, which is finite. The series converges.
Level 2: Check the Conditions
Problem 4. Determine whether $\displaystyle\sum_{n=1}^{\infty} n\,e^{-n}$ converges using the integral test.
Show answer
Let $f(x) = x e^{-x}$. For $x > 1$, $f'(x) = e^{-x}(1 - x) < 0$, so $f$ is decreasing and positive. Integration by parts gives $\displaystyle\int x e^{-x}\,dx = -(x+1)e^{-x}$, so \[ \int_1^{\infty} x e^{-x}\,dx = \lim_{t\to\infty}\big[-(x+1)e^{-x}\big]_1^t = 0 - \big(-2e^{-1}\big) = \frac{2}{e}, \] which is finite. The series converges.
Problem 5. Explain why the integral test cannot be used on $\displaystyle\sum_{n=1}^{\infty} \frac{\cos^2 n}{1 + n^2}$.
Show answer
The matching function $f(x) = \dfrac{\cos^2 x}{1 + x^2}$ is not decreasing: the $\cos^2 x$ factor oscillates between $0$ and $1$, so $f$ rises and falls repeatedly rather than steadily decreasing. The integral test requires an eventually decreasing function, and this one never becomes monotonic, so the test does not apply. (A comparison test handles this series instead.)
Level 3: Reasoning
Problem 6. Find all values of $p$ for which $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ converges.
Show answer
Let $f(x) = \dfrac{1}{x(\ln x)^p}$, positive and decreasing for $x \ge 2$. With $u = \ln x$, $du = \dfrac{1}{x}\,dx$, the integral becomes \[ \int_2^{\infty}\frac{1}{x(\ln x)^p}\,dx = \int_{\ln 2}^{\infty}\frac{1}{u^p}\,du, \] which is a $p$-integral in $u$. It converges if and only if $p > 1$. So the series converges exactly when $p > 1$.
Problem 7. The integral test shows $\sum 1/n^2$ converges, and separately $\sum 1/n^2 = \pi^2/6$ while $\int_1^{\infty} x^{-2}\,dx = 1$. Use this to explain why the test never claims the sum equals the integral.
Show answer
The numbers are different: the sum is $\dfrac{\pi^2}{6} \approx 1.645$ while the integral is $1$. The rectangle picture shows the partial sums are squeezed between the integral and the integral plus the first term, which bounds the sum but does not equal the integral. So the test can only conclude that both are finite (or both infinite); it gives no claim that the two finite values coincide, and in general they do not.
Mastery Checklist
You have mastered this skill when you can do all of the following without notes:
Mental Model
Picture a positive decreasing curve and a staircase of unit-width rectangles drawn alongside it. The series is the total staircase area; the integral is the area under the smooth curve. Because the curve only ever falls, the staircase and the curve hug each other: each rectangle is within one step of the curve’s height. So the two areas are within a bounded distance of each other for all time.
That hug is the whole test. If the area under the curve is finite, the staircase cannot suddenly become infinite, so the series converges. If the area under the curve is infinite, the staircase is at least as large, so the series diverges. The picture also explains the warning about value: the staircase and the curve are close, but they are not identical, so the two finite areas (the sum and the integral) are different numbers even when both are finite.
Connections
Built From
- Improper integrals: the test outsources the convergence decision to an improper integral you already know how to evaluate.
- The Monotonic Sequence Theorem: the convergent direction of the proof uses that an increasing sequence bounded above converges, applied to the partial sums.
- The Test for Divergence: always the first check; the integral test is for the cases where the terms do go to $0$ and more work is needed.
Leads To
- The remainder estimate for the integral test: the same rectangle picture bounds the error when a partial sum approximates the full sum, the next lesson.
- The comparison tests: the $p$-series result this test establishes becomes the standard yardstick the direct and limit comparison tests compare other series against.
Why It Matters
The integral test is the bridge between two halves of calculus: it lets the machinery of integration decide questions about discrete sums. The $p$-series result it produces is the single most-used reference in convergence testing, the ruler against which countless other series are measured. In numerical work the same rectangle bound estimates how many terms of a slowly converging sum are needed for a target accuracy, which is the subject of the next lesson. The Riemann zeta function, central to number theory, is exactly the $p$-series viewed as a function of $p$, and the integral test is what tells you where it is defined.
Audience Notes
For students who find math intimidating: you already know how to test an improper integral. The matching series passes or fails with that integral, plus you memorize one fact about $p$-series.
For students who want depth: the rectangle proof is worth internalizing, because the same comparison-of-areas idea reappears in the remainder estimate and underlies the comparison tests. It is also the cleanest place to see the Monotonic Sequence Theorem doing real work.
For students aimed at a career in computing or engineering: the integral test and its remainder bound are how you decide whether a series-based computation will converge and how many terms it needs. The $p$-series boundary at $p = 1$ is a sharp, practical line between sums that finish and sums that crawl.
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