The Integral Test

The most common mistake students make with the integral test is thinking it gives you the sum of a series. It does not. The integral test tells you only whether a series converges or diverges, never what it converges to. If you need the exact sum of a convergent series, say, the Basel problem value of π²/6 for Σ 1/n², the integral test is the wrong tool. For that, you need telescoping series, geometric series formulas, or Fourier series. The integral test is for one purpose: deciding if an infinite sum settles to a finite number or blows up.

The integral test works on series of the form Σ a_n where a_n is positive and eventually decreasing. You compare the series to an improper integral of the continuous function f(x) that matches the terms. If the integral converges, the series converges. If the integral diverges, the series diverges. The test itself gives you a yes-or-no answer, not a sum.

The integral test requires three conditions on the function f, where a_n = f(n). These conditions come from Stewart, Calculus, section 11.3. All three must hold on the interval [1, ∞) (or from some integer N onward). If any condition fails, the test is not valid.

Condition 1: f is continuous. No holes, jumps, or vertical asymptotes on the interval. A function with a discontinuity cannot be integrated, so the test stops here.

Condition 2: f is positive. f(x) > 0 for all x ≥ 1. If terms become zero or negative, the comparison between the integral and the series breaks down.

Condition 3: f is decreasing. f(x) eventually declines as x increases. The terms a_n must be decreasing for all n past some starting point. A series where terms increase or oscillate cannot use the integral test.

When all three are satisfied, the test makes its claim: Σ_{n=1}^∞ a_n converges if and only if ∫_1^∞ f(x) dx converges.

Before applying the integral test, run through this checklist. If any box is unchecked, the test does not apply.

• f(x) is continuous on [1, ∞)
• f(x) > 0 for x ≥ 1
• f(x) is decreasing for x ≥ 1
• a_n = f(n) for all n ≥ 1

If the function fails to be decreasing from n = 1 but becomes decreasing after some integer N (say, N = 5), the test still applies from N onward. The convergence of a finite tail does not affect the series convergence or divergence.

The easiest way to check whether f is decreasing is to compute its derivative. If f′(x) < 0 for all x ≥ 1 (or for x ≥ some N), then f is decreasing. A function whose derivative stays negative is strictly decreasing, which satisfies the condition.

Example: f(x) = 1/(x ln x). Compute f′(x) using the quotient or chain rule. f′(x) = -(ln x + 1) / (x² ln² x). For x > 1, ln x > 0, so the numerator -(ln x + 1) is always negative. The denominator is positive. Therefore f′(x) < 0 for x > 1, so f is decreasing on [2, ∞). The test is valid from n = 2 onward.

If the derivative is messy or zero at isolated points, zoom in on the tail. A function that is not decreasing on [1, 10] but monotone decreasing on [10, ∞) still works. The integral test only cares about the eventual behavior.

Example 1: Σ 1/(n ln n), Diverges

Let a_n = 1/(n ln n). Define f(x) = 1/(x ln x). f is continuous for x > 1, positive, and, as shown above, decreasing for x ≥ 2. The integral test applies.

Compute ∫_2^∞ 1/(x ln x) dx. Use substitution: u = ln x, du = (1/x) dx. When x = 2, u = ln 2. When x → ∞, u → ∞. The integral becomes ∫_{ln 2}^∞ (1/u) du = lim_{t→∞} [ln u]_{ln 2}^t = lim_{t→∞} (ln t - ln(ln 2)). The limit is infinite, so the integral diverges. By the integral test, the series Σ 1/(n ln n) diverges.

Example 2: Σ n e^{-n²}, Converges

Let a_n = n e^{-n²}. Define f(x) = x e^{-x²}. f is continuous and positive on [1, ∞). Check decreasing: f′(x) = e^{-x²} (1 - 2x²). For x ≥ 1, 2x² ≥ 2, so 1 - 2x² ≤ -1, hence f′(x) < 0. f is decreasing.

Compute ∫_1^∞ x e^{-x²} dx. Use substitution: u = x², du = 2x dx, so x dx = du/2. When x = 1, u = 1. When x → ∞, u → ∞. The integral becomes (1/2) ∫_1^∞ e^{-u} du = (1/2) lim_{t→∞} [-e^{-u}]_1^t = (1/2) (0 + e^{-1}) = 1/(2e). The integral converges to a finite number. By the integral test, the series Σ n e^{-n²} converges.

The integral test directly proves the convergence condition for p-series (Σ 1/n^p). For f(x) = 1/x^p, the three conditions hold when p > 0: f is continuous, positive, and decreasing on [1, ∞). Compute ∫_1^∞ x^{-p} dx.

If p ≠ 1, ∫_1^∞ x^{-p} dx = lim_{t→∞} [x^{1-p} / (1-p)]_1^t. If p > 1, then 1-p < 0, so x^{1-p} → 0 as x → ∞, and the integral converges to 1/(p-1). If p < 1, then 1-p > 0, so x^{1-p} → ∞, and the integral diverges. If p = 1, the harmonic series case, the integral is ∫_1^∞ (1/x) dx, which diverges as ln t.

Result: Σ 1/n^p converges if p > 1, diverges if p ≤ 1. This is the p-series rule, confirmed by Stewart, Calculus, section 11.3, Example 3. The integral test provides the proof that every calculus textbook uses.

The integral test also gives bounds on the error when you replace an infinite series with a partial sum. The remainder R_n = S - S_n, where S is the true sum and S_n is the sum of the first n terms, is trapped between two integrals.

From Stewart, section 11.3: ∫_{n+1}^∞ f(x) dx ≤ R_n ≤ ∫_n^∞ f(x) dx. This means the error in using S_n as an approximation is at most the area under f from n to ∞, and at least the area from n+1 to ∞.

Example: For Σ 1/n², p = 2, the integral test shows convergence. To estimate S_10, bound the remainder: R_10 ≤ ∫_10^∞ 1/x² dx = 1/10 = 0.1. The true sum S is between S_10 and S_10 + 0.1. S_10 ≈ 1.5498, so S is between 1.5498 and 1.6498. The actual sum is π²/6 ≈ 1.6449, which falls inside that interval.

The remainder estimate is useful when you need a numerical approximation and have no closed form for the sum. Unlike the alternating series error bound, which uses the first omitted term, the integral test remainder estimate uses integrals and works for positive-term series.

Use the integral test for Calculus II problems where terms look like a simple function and the comparison test feels awkward. AP Calculus BC students: the College Board CED (Topics 10.7-10.12) expects you to apply it under time pressure. Tutors can use the conditions checklist above to verify student work.

Skip the integral test if you need the exact sum of a convergent series. It never gives a sum; only yes or no. Skip it if terms are not positive and decreasing, or if integrating the function is harder than using another test. For series with factorials, the ratio test is faster. For series with rational functions, the comparison test or limit comparison test often requires less computation.

One failure case: applying the integral test to a non-decreasing sequence. If the terms increase, the integral test is invalid. Another failure: trying to use the remainder estimate without checking the decreasing condition, the bounds only hold when f is decreasing.