Integral Test

Test series convergence by checking improper integral ∫₁^∞ f(x)dx when f is positive, continuous, and decreasing; the series ∑f(n) shares convergence behavior.

∑n=1∞f(n) and ∫1∞f(x) dx share convergence\sum_{n=1}^{\infty} f(n) \text{ and } \int_1^{\infty} f(x)\,dx \text{ share convergence}

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What each symbol means

What Integral Test takes
ff
nn
xx
Integral Test →
SymbolMeaning
ffThe general function that describes the terms of your series; it must be positive, continuous, and decreasing on the interval [1,∞) for the test to apply, or the test gives no information.
nnThe index variable counting the terms of the series as positive integers starting from 1; if you misinterpret this as continuous instead of discrete, you lose the connection between series and integral.
xxThe continuous variable in the integral ∫₁^∞ f(x)dx, replacing the discrete index n; the function f(x) must have the same formula as f(n), just with x substituted for n.

When to use it

Use the Integral Test when you have a series whose terms match a positive, decreasing, continuous function, and you can compute the corresponding improper integral.

Level

Usually taught in: Calculus II

Worked examples

1. Test a simple p-series: $\sum_{n=1}^{\infty} \frac{1}{n^2}$

Problem

Use the Integral Test to determine whether ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges.
  1. ∫1∞1x2 dx\int_1^{\infty} \frac{1}{x^2}\,dx

    Set up the improper integral using the same function as the series, with xx as the continuous variable.

  2. lim⁡t→∞∫1t1x2 dx\lim_{t \to \infty} \int_1^t \frac{1}{x^2}\,dx

    Rewrite the improper integral as a limit to evaluate it properly.

  3. lim⁡t→∞[−1x]1t\lim_{t \to \infty} \left[-\frac{1}{x}\right]_1^t

    Apply the antiderivative of 1x2\frac{1}{x^2}, which is −1x-\frac{1}{x}.

  4. lim⁡t→∞(−1t+1)\lim_{t \to \infty} \left(-\frac{1}{t} + 1\right)

    Evaluate the antiderivative at the limits; this gives −1/t−(−1/1)=−1/t+1-1/t - (-1/1) = -1/t + 1.

  5. lim⁡t→∞(−1t+1)=1\lim_{t \to \infty} \left(-\frac{1}{t} + 1\right) = 1

    As t→∞t \to \infty, the term −1/t→0-1/t \to 0, leaving the finite value 1.

Answer: ∑n=1∞1n2 converges\sum_{n=1}^{\infty} \frac{1}{n^2} \text{ converges}

The function f(x)=1/x2f(x) = 1/x^2 is positive, continuous, and decreasing on [1,∞)[1, \infty). The improper integral converges to a finite value, so by the Integral Test, the series must converge as well.

2. Test a series with a less obvious antiderivative: $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$

Problem

Determine whether ∑n=1∞1n2+1\sum_{n=1}^{\infty} \frac{1}{n^2+1} converges using the Integral Test.
  1. f(x)=1x2+1f(x) = \frac{1}{x^2+1}

    Identify the function; replace nn with xx to work with the continuous version.

  2. lim⁡t→∞∫1t1x2+1 dx\lim_{t \to \infty} \int_1^t \frac{1}{x^2+1}\,dx

    Set up the improper integral as a limit.

  3. lim⁡t→∞[arctan⁡(x)]1t\lim_{t \to \infty} \left[\arctan(x)\right]_1^t

    The antiderivative of 1x2+1\frac{1}{x^2+1} is arctan⁡(x)\arctan(x).

  4. lim⁡t→∞(arctan⁡(t)−arctan⁡(1))\lim_{t \to \infty} \left(\arctan(t) - \arctan(1)\right)

    Evaluate at the limits; note that arctan⁡(1)=π/4\arctan(1) = \pi/4.

  5. lim⁡t→∞(arctan⁡(t)−π4)\lim_{t \to \infty} \left(\arctan(t) - \frac{\pi}{4}\right)

    Substitute the known value of arctan⁡(1)\arctan(1).

  6. π2−π4=π4≈0.785\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \approx 0.785

    As t→∞t \to \infty, arctan⁡(t)→π/2\arctan(t) \to \pi/2, giving a finite result.

Answer: ∑n=1∞1n2+1 converges\sum_{n=1}^{\infty} \frac{1}{n^2+1} \text{ converges}

The function f(x)=1/(x2+1)f(x) = 1/(x^2+1) is positive, continuous, and decreasing on [1,∞)[1, \infty) because larger values of xx make the denominator larger. The improper integral converges to π/4\pi/4, confirming convergence of the series.

3. A word problem: fencing materials for expanding garden layers

Problem

A rectangular garden bed is being expanded with successively smaller rectangular layers. The nn-th layer requires 1n(ln⁡n)2\frac{1}{n(\ln n)^2} units of fencing material for n≥2n \geq 2. Will the total fencing needed for infinitely many layers converge to a finite amount?
  1. ∫2∞1x(ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx

    Set up the improper integral starting from x=2x = 2, matching the series starting index.

  2. u=ln⁡x,du=1x dxu = \ln x, \quad du = \frac{1}{x}\,dx

    Use substitution to simplify the integral; the factor 1x dx\frac{1}{x}\,dx becomes dudu.

  3. ∫1u2 du=−1u=−1ln⁡x\int \frac{1}{u^2}\,du = -\frac{1}{u} = -\frac{1}{\ln x}

    Integrate 1/u21/u^2 to get −1/u-1/u, then substitute back u=ln⁡xu = \ln x.

  4. lim⁡t→∞[−1ln⁡x]2t=lim⁡t→∞(−1ln⁡t+1ln⁡2)=1ln⁡2\lim_{t \to \infty} \left[-\frac{1}{\ln x}\right]_2^t = \lim_{t \to \infty} \left(-\frac{1}{\ln t} + \frac{1}{\ln 2}\right) = \frac{1}{\ln 2}

    Evaluate at the limits; as t→∞t \to \infty, ln⁡t→∞\ln t \to \infty, so −1/ln⁡t→0-1/\ln t \to 0.

  5. ∑n=2∞1n(ln⁡n)2 converges\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} \text{ converges}

    Since the integral converges to the finite value 1/ln⁡2≈1.441/\ln 2 \approx 1.44, the series converges.

Answer: Total fencing converges to approximately 1ln⁡2≈1.44 units\text{Total fencing converges to approximately } \frac{1}{\ln 2} \approx 1.44 \text{ units}

This classic example shows why the Integral Test is powerful: the series ∑1/(n(ln⁡n)2)\sum 1/(n(\ln n)^2) decreases slowly enough that simpler tests might miss it, but the integral test makes it clear. The function f(x)=1/(x(ln⁡x)2)f(x) = 1/(x(\ln x)^2) is positive, continuous, and decreasing on [2,∞)[2, \infty). Since the improper integral converges, the infinite series converges, meaning the gardener will eventually need only a finite amount of fencing material.

Common mistakes

Where Integral Test usually goes wrong
Answer came out wrong
Writing ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\,dx diverges, so we conclude ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n} diverges, without computing the integral carefully.
Always compute lim⁡t→∞∫1t1x dx=lim⁡t→∞ln⁡t=∞\lim_{t \to \infty} \int_1^t \frac{1}{x}\,dx = \lim_{t \to \infty} \ln t = \infty. The integral explicitly diverges to infinity, confirming the series diverges.
Computing ∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\,dx = 1 and concluding ∑n=1∞1n2=1\sum_{n=1}^{\infty} \frac{1}{n^2} = 1.
Conclude only that the series converges. Do not claim the sum equals the integral. (In fact, the sum is π2/6≈1.645\pi^2/6 \approx 1.645, not 1.)
Applying the Integral Test to ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} but computing ∫1∞1x(ln⁡x)2 dx\int_1^{\infty} \frac{1}{x(\ln x)^2}\,dx instead of ∫2∞\int_2^{\infty}.
Ensure the lower limit of integration matches the starting index of the series: use ∫2∞1x(ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx for ∑n=2∞\sum_{n=2}^{\infty}.
The mistakeWhy it is wrongThe fix
Writing ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\,dx diverges, so we conclude ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n} diverges, without computing the integral carefully.Computing the integral carelessly—forgetting to evaluate the limit, or confusing divergence with a specific infinite value—leads to the wrong convergence conclusion even though the final answer happens to be correct here.Always compute lim⁡t→∞∫1t1x dx=lim⁡t→∞ln⁡t=∞\lim_{t \to \infty} \int_1^t \frac{1}{x}\,dx = \lim_{t \to \infty} \ln t = \infty. The integral explicitly diverges to infinity, confirming the series diverges.
Computing ∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\,dx = 1 and concluding ∑n=1∞1n2=1\sum_{n=1}^{\infty} \frac{1}{n^2} = 1.The Integral Test only tells you about convergence or divergence, not the value of the series; the numerical value of the integral is irrelevant to the test.Conclude only that the series converges. Do not claim the sum equals the integral. (In fact, the sum is π2/6≈1.645\pi^2/6 \approx 1.645, not 1.)
Applying the Integral Test to ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} but computing ∫1∞1x(ln⁡x)2 dx\int_1^{\infty} \frac{1}{x(\ln x)^2}\,dx instead of ∫2∞\int_2^{\infty}.The integral should start at the same lower limit as the series; starting at x=1x=1 causes issues because ln⁡1=0\ln 1 = 0 makes the denominator zero, and it does not match the series index.Ensure the lower limit of integration matches the starting index of the series: use ∫2∞1x(ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx for ∑n=2∞\sum_{n=2}^{\infty}.

Tips and when to use something else

  • Always check the three conditions—positive, continuous, decreasing on [1,∞)[1, \infty) (or [N,∞)[N, \infty) if the series starts at a later index)—before you set up the integral; skipping this check is the most common source of error.
  • The Integral Test tells you only whether a series converges or diverges, not what sum it converges to; resist the temptation to set the integral value equal to the series sum.
  • For series that look like 1/np1/n^p, the pp-Series Test is faster than computing an integral; reserve the Integral Test for cases where the pp-Series Test does not apply.
  • If your function oscillates, has a minimum in the middle, or fails to be monotone decreasing, use the Limit Comparison Test, Ratio Test, or Alternating Series Test instead of the Integral Test.

Related formulas

p-Series TestThe p-Series Test determines convergence of series 1/n^p: the series converges if and only if the exponent p is strictly greater than 1.Ratio TestDetermine whether an infinite series converges by checking if the ratio of consecutive terms approaches a value less than one.Harmonic SeriesThe Harmonic Series, sum of reciprocals of positive integers, diverges infinitely—a foundational benchmark for determining whether other series converge.

Frequently asked questions

Do I have to use the Integral Test, or can I use a different convergence test?
The Integral Test is just one of many tools. The pp-Series Test works directly for series like 1/np1/n^p, the Ratio Test is better for factorials and exponentials, and the Limit Comparison Test is useful when your series resembles a familiar one. Choose whichever test fits most cleanly for your particular series.
What if the function is decreasing starting from n=5n=5 instead of n=1n=1; can I still use the Integral Test?
Yes. If ff is decreasing on [N,∞)[N, \infty) for some N≥1N \geq 1, you can apply the test using ∫N∞f(x) dx\int_N^{\infty} f(x)\,dx. Convergence behavior depends only on the tail of the series, so behavior from some point onward is all that matters.
If the integral diverges, could the series still converge?
No. The Integral Test guarantees that if the integral diverges, the series diverges, and vice versa. They always share the same convergence behavior.
Why does the Integral Test require ff to be continuous and decreasing, not just eventually decreasing?
Integrals measure area under a curve, so you need a continuous function. The Integral Test compares the total area under f(x)f(x) to rectangles formed by f(n)f(n); if ff is not decreasing throughout, those rectangles do not bound the area consistently, and the comparison fails.

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Reviewed 2026-09-18