What Is The Integral Test?
The Integral Test is a convergence test that decides whether an infinite series adds up to a finite number by comparing it to an improper integral. Its power is that a sum you cannot add by hand becomes an area you can evaluate with ordinary integration.
Here is the precise statement. Suppose $f$ is a function that is continuous, positive, and decreasing on the interval $[1, \infty)$, and suppose the terms of the series match the function at the whole numbers, so that $a_n = f(n)$. Then:
$$\sum_{n=1}^{\infty} a_n \quad \text{and} \quad \int_1^\infty f(x),dx \quad \text{either both converge or both diverge.}$$
All three hypotheses matter, and they only need to hold eventually, from some starting index onward. The interval does not have to begin at $1$; if the conditions hold from $n = N$, test $\int_N^\infty f(x),dx$ instead. A finite number of early terms can never change whether an infinite series converges.
One warning belongs at the very top, because it is the misconception readers carry longest: the test shares only convergence with the integral, never the numerical value. If the integral equals $1$, that does not mean the series sums to $1$. More on that below.
What Are The Conditions For The Integral Test?
Before you integrate anything, check the three hypotheses on the function $f(x)$ built from the terms. Skipping this check is the fastest way to a wrong answer.
Table: The three conditions the Integral Test requires, and how to verify each.
Condition | What it means | How to check |
|---|---|---|
Continuous | No breaks or asymptotes on $[N, \infty)$ | Confirm $f$ is defined and unbroken for all $x \geq N$ |
Positive | $f(x) > 0$ for all $x \geq N$ | The terms must stay above zero |
Decreasing | $f(x+1) \leq f(x)$ eventually | Show $f'(x) < 0$ for large $x$, or that terms shrink |
The decreasing condition is the one students verify least and lose marks on most. The cleanest way to prove it is to differentiate: if $f'(x) < 0$ for all $x$ beyond some point, the function is decreasing there. The test still applies if $f$ decreases only eventually, since the early behaviour cannot change the fate of an infinite tail.
How Does The Integral Test Work? The Rectangle Picture
The proof is a picture. Draw the decreasing curve $y = f(x)$, then draw rectangles of width $1$ whose heights are the series terms $a_n = f(n)$.
Line the rectangles up so each one sits to the right of its matching point on the curve. Because $f$ is decreasing, every rectangle lies below the curve, so the total rectangle area is less than the area under the curve:
$$\sum_{n=2}^{\infty} a_n ; \leq ; \int_1^\infty f(x),dx.$$
Now shift the rectangles to the left, so each sits above its point on the curve. Now every rectangle pokes above the curve, so the total rectangle area is greater than the area under the curve:
$$\int_1^\infty f(x),dx ; \leq ; \sum_{n=1}^{\infty} a_n.$$
Read the two inequalities together. If the integral is finite, the first inequality traps the series below a finite ceiling, so the series converges. If the integral is infinite, the second inequality pushes the series above an infinite floor, so the series diverges. The sum and the area rise and fall together.
How Do You Use The Integral Test? Worked Examples
Each example checks the three conditions first, then evaluates the improper integral as a limit. Every integral below is elementary, and each antiderivative is one you can differentiate back to confirm.
Example 1: Does $\sum_{n=1}^{\infty} \dfrac{1}{n^2}$ converge?
Set $f(x) = \dfrac{1}{x^2}$. It is continuous, positive, and decreasing on $[1, \infty)$ since $f'(x) = -\dfrac{2}{x^3} < 0$. Evaluate the integral as a limit:
$\int_1^\infty \frac{1}{x^2}\,dx=\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b=\lim_{b\to\infty}\left(-\frac{1}{b}+1\right)=1.$
The integral is finite, so by the Integral Test the series converges.
Final answer: the series converges. Note carefully that the integral equals $1$, but the series actually sums to $\dfrac{\pi^2}{6} \approx 1.6449$. The two numbers are different; only the fact of convergence is shared.
Example 2: Does the harmonic series $\sum_{n=1}^{\infty} \dfrac{1}{n}$ converge?
Set $f(x) = \dfrac{1}{x}$, which is continuous, positive, and decreasing on $[1, \infty)$. Evaluate:
$\int_1^\infty \frac{1}{x}\,dx=\lim_{b\to\infty}\left[\ln x\right]_1^b=\lim_{b\to\infty}\left(\ln b-0\right)=\infty.$
The integral diverges, so the series diverges too.
Final answer: the harmonic series diverges. This is the famous surprise of calculus: the terms $\frac{1}{n}$ shrink to zero, yet their sum grows without bound. Terms going to zero is necessary for convergence, never sufficient.
Example 3: Does $\sum_{n=2}^{\infty} \dfrac{1}{n \ln n}$ converge?
The sum starts at $n = 2$ because $\ln 1 = 0$ would divide by zero. Set $f(x) = \dfrac{1}{x \ln x}$, which is continuous, positive, and decreasing on $[2, \infty)$. Integrate with the substitution $u = \ln x$, so $du = \dfrac{1}{x},dx$:
$\int_2^\infty \frac{1}{x\ln x}\,dx=\lim_{b\to\infty}\left[\ln(\ln x)\right]_2^b=\lim_{b\to\infty}\Big(\ln(\ln b)-\ln(\ln 2)\Big)=\infty.$
The integral diverges, so the series diverges.
Final answer: the series diverges. This one is a knife-edge: the extra $\ln n$ in the denominator makes the terms shrink faster than $\frac{1}{n}$, yet not fast enough to converge.
Example 4: Does $\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 1}$ converge?
Set $f(x) = \dfrac{1}{x^2 + 1}$, continuous, positive, and decreasing on $[1, \infty)$ since $f'(x) = -\dfrac{2x}{(x^2+1)^2} < 0$. Its antiderivative is $\arctan x$:
$$\int_1^\infty \frac{1}{x^2 + 1},dx = \lim_{b \to \infty}\big[\arctan x\big]_1^{b} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \approx 0.7854.$$
The integral is finite, so the series converges.
Final answer: the series converges. As a check, $\dfrac{d}{dx}\arctan x = \dfrac{1}{x^2 + 1}$, the original integrand, so the antiderivative is correct.
What Does The Integral Test Say About p-Series?
The Integral Test is the reason the p-series rule is true. A p-series is $\sum_{n=1}^{\infty} \dfrac{1}{n^p}$, and the matching function is $f(x) = x^{-p}$. For $p \neq 1$:
$$\int_1^\infty x^{-p},dx = \lim_{b \to \infty}\left[\frac{x^{1-p}}{1-p}\right]_1^{b}.$$
When $p > 1$, the exponent $1 - p$ is negative, so $b^{1-p} \to 0$ and the integral converges to $\dfrac{1}{p-1}$. When $p < 1$, the exponent is positive, $b^{1-p} \to \infty$, and the integral diverges. The single leftover case $p = 1$ is the harmonic series from Example 2, which diverges through $\ln b$.
Table: What the Integral Test proves about the p-series $\sum \frac{1}{n^p}$.
Value of $p$ | Integral $\int_1^\infty x^{-p},dx$ | Series $\sum \frac{1}{n^p}$ |
|---|---|---|
$p > 1$ | Converges to $\frac{1}{p-1}$ | Converges |
$p = 1$ | Diverges ($\ln b \to \infty$) | Diverges (harmonic) |
$p < 1$ | Diverges | Diverges |
The rule in one line: $\sum \dfrac{1}{n^p}$ converges if and only if $p > 1$. Once proved, you can quote it directly instead of re-integrating every time.
How Accurate Is A Partial Sum? The Remainder Estimate
When a series converges, the Integral Test also bounds how far a partial sum sits from the true total. Let $S = \sum_{n=1}^{\infty} a_n$ be the full sum and $S_N = \sum_{n=1}^{N} a_n$ the sum of the first $N$ terms. The leftover tail is the remainder $R_N = S - S_N$. The same rectangle picture traps that tail between two integrals:
$$\int_{N+1}^{\infty} f(x),dx ; \leq ; R_N ; \leq ; \int_{N}^{\infty} f(x),dx.$$
Take $\sum \dfrac{1}{n^2}$ and stop after $N = 10$ terms. Since $\int_N^\infty x^{-2},dx = \dfrac{1}{N}$, the bounds are:
$$\frac{1}{11} \leq R_{10} \leq \frac{1}{10}, \qquad \text{that is} \qquad 0.0909 \leq R_{10} \leq 0.1000.$$
The partial sum $S_{10} \approx 1.5498$, and the true total is $\dfrac{\pi^2}{6} \approx 1.6449$, so the genuine tail is $R_{10} \approx 0.0952$. It lands neatly inside the predicted window, which is exactly what the estimate promises: you know the error before you know the answer.
Why Does The Integral Test Work?
The test feels almost too convenient the first time, because a sum and an integral look like different objects. The bridge is geometric, and it rests on three ideas.
A sum of terms is a sum of rectangle areas. Each term $a_n$ is the area of a width-one rectangle of height $a_n$. Adding the terms is adding those rectangle areas, which is a crude version of the area under the curve.
Decreasing keeps the rectangles honest. Because $f$ falls steadily, every rectangle either sits fully under the curve or fully over it, with no crossing. That is what lets the two inequalities squeeze the series from both sides.
A finite area caps an endless sum. If the total area under the curve is finite, the series can never exceed it, so an infinite list of positive terms still lands on a finite total. If the area is infinite, nothing holds the sum back.
This is why the value is not shared. The rectangles only approximate the area; they overshoot or undershoot by a bounded amount. That bounded gap is enough to match convergence, but it is exactly why the integral and the series settle on different numbers.
Who Discovered The Integral Test?
The idea of comparing a sum to an area is old, but two names are attached to the test, which is why it is formally called the Maclaurin-Cauchy test.
Colin Maclaurin (1698-1746, Scotland) was a prodigy who became a professor at nineteen. Augustin-Louis Cauchy (1789-1857, France) gave calculus much of its modern rigour, including the careful definitions of limit and convergence the test depends on.
Where Is The Integral Test Used In The Real World?
The test is a decision tool wherever an endless sum has to be judged finite or infinite before anyone trusts it.
Algorithm analysis: the running cost of many computer algorithms is a sum like $\sum \frac{1}{n}$ or $\sum \frac{1}{n^2}$, and knowing whether it converges tells engineers whether a total stays bounded as the input grows.
Physics and engineering: energy stored across infinitely many modes, or the tail of a field summed over distance, is accepted only if the corresponding integral converges, exactly the check the Integral Test performs.
Probability and statistics: deciding whether an infinite collection of probabilities or an expected value is finite often reduces to a p-series comparison settled by this test.
Numerical computing: the remainder estimate tells a programmer how many terms to add for a target accuracy, turning "sum forever" into "sum until the bound is small enough."
One geometric idea, comparing a sum to an area, lets fields as different as computing and physics certify that an infinite process produces a finite answer.
What Are The Most Common Mistakes With The Integral Test?
These four errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and university common-error keys.
Skipping the positive-and-decreasing check.
Where it slips in:
A student sees a $\frac{1}{n}$-shaped term, jumps straight to integrating, and never confirms the function actually decreases and stays positive.
Don't do this:
Do not apply the test to a function that is not decreasing on the interval, such as one that oscillates. The whole rectangle argument collapses.
The correct way:
Verify all three conditions first. The reliable move for "decreasing" is to compute $f'(x)$ and show it is negative for large $x$.
Thinking the integral equals the series sum.
Where it slips in:
A student finds $\int_1^\infty \frac{1}{x^2},dx = 1$ and writes "so the series sums to $1$."
Don't do this:
Do not report the integral as the total. The Integral Test shares convergence, not value.
The correct way:
State only that the series converges. Here the true sum is $\frac{\pi^2}{6} \approx 1.6449$, not the integral's $1$.
Forgetting to write the integral as a limit.
Where it slips in:
A student plugs $\infty$ straight into an antiderivative, for example writing $[-\frac{1}{x}]_1^\infty$ without the limit step, and mishandles the infinite endpoint.
Don't do this:
Do not evaluate an improper integral by substituting infinity directly. Infinity is not a number you can plug in.
The correct way:
Replace the upper limit with $b$, integrate, then take $\lim_{b \to \infty}$. This is what makes the improper integral rigorous.
Using the test on a series with negative or alternating terms.
Where it slips in:
A student applies the Integral Test to a series whose terms change sign, such as $\sum \frac{(-1)^n}{n}$.
Don't do this:
Do not use this test when terms are not positive. The positivity hypothesis is not decoration.
The correct way:
Choose a test built for the situation. For sign-changing series reach for the alternating series test; for other shapes the comparison test, limit comparison test, or ratio test may fit better.
Practice Problems On The Integral Test
Work each one by checking the three conditions and evaluating the integral as a limit, then check against the answer. Answers are verified.
Does $\sum_{n=1}^{\infty} \dfrac{1}{n^3}$ converge?
(Answer: yes; $\int_1^\infty x^{-3},dx = \frac{1}{2}$, finite, so it converges. It is a p-series with $p = 3 > 1$.)Does $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}$ converge?
(Answer: no; $p = \frac{1}{2} \leq 1$, and $\int_1^\infty x^{-1/2},dx = \infty$, so it diverges.)Does $\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 4}$ converge? (
Answer: yes; $\int_1^\infty \frac{1}{x^2+4},dx = \frac{1}{2}\left(\frac{\pi}{2} - \arctan\frac{1}{2}\right)$, finite, so it converges.)Does $\sum_{n=1}^{\infty} n,e^{-n^2}$ converge?
(Answer: yes; with $u = x^2$, $\int_1^\infty x,e^{-x^2},dx = \frac{1}{2}e^{-1} \approx 0.1839$, finite, so it converges.)Does $\sum_{n=2}^{\infty} \dfrac{1}{n(\ln n)^2}$ converge?
(Answer: yes; with $u = \ln x$, $\int_2^\infty \frac{dx}{x(\ln x)^2} = \frac{1}{\ln 2} \approx 1.4427$, finite, so it converges.)For $\sum_{n=1}^{\infty} \dfrac{1}{n^4}$, bound the remainder after $N = 5$ terms.
(Answer: $R_5 \leq \int_5^\infty x^{-4},dx = \frac{1}{3 \cdot 5^3} = \frac{1}{375} \approx 0.0027$.)
Where Should You Go Next After The Integral Test?
The Integral Test is one entry in a whole toolkit of convergence tests, and several natural doors open from here.
Convergence and divergence of series. See where the Integral Test sits among all the tests and when to pick which.
The comparison test. The natural next tool when an integral is awkward but the terms resemble a known series.
The divergence test. The quick first check that rules out convergence when the terms fail to approach zero.
If your child is meeting series-convergence tests for the first time, a live Bhanzu trainer teaches the Integral Test from the rectangle picture up, so the conditions feel like common sense rather than a checklist, in the Bhanzu math program.
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