The Divergence Test, the Integral Test, and p-Series
Two workhorse tests: the divergence test instantly rules out convergence whenever the terms don't shrink to 0, and the integral test ties a positive decreasing series to the area under a curve — proving that \(\sum \frac{1}{n^p}\) converges exactly when \(p > 1\).
By the end you'll be able to screen any series with the divergence test, apply the integral test to a positive decreasing series, and state exactly where the p-series family flips from divergent to convergent.
Predict: at p = 1, does the area under \(1/x\) stay finite? Slide across p = 1 to check.
Each purple bar is one term \(a_n = 1/n^p\) of the series, drawn over the curve \(f(x) = 1/x^p\). The bars hug the curve, so the series total and the improper integral \(\int_1^\infty f(x)\,dx\) are finite or infinite together. Drag p and watch the verdict flip at the p = 1 boundary. Hover or tab onto a bar for its exact term value.
p = 1.50 · ∫₁^∞ dx/xᵖ = 1/(p−1) = 2.000 — finite, so the series converges (p > 1)
The divergence test rules out convergence when the terms don't go to 0; the integral test compares a positive decreasing series to \(\int f(x)\,dx\), and its signature payoff is the p-series benchmark: \(\sum 1/n^p\) converges if and only if \(p > 1\).
Picture the bars in the demo above: rectangle \(n\) has width 1 and height \(a_n = f(n)\), so the total bar area is the series. Because \(f\) is decreasing, the staircase of bars tracks the curve — the running total of bar areas and the area under the curve rise together, so one is finite exactly when the other is. When you slide p below 1, the curve's tail stops thinning fast enough: the shaded area balloons to infinity and drags the series down with it. Slide p above 1 and the tail pinches off — finite area, convergent series.
Divergence test. If \(\lim_{n\to\infty} a_n \neq 0\) (or the limit fails to exist), then \(\sum a_n\) diverges. It's a necessary-condition check: a convergent series has partial sums \(S_N \to S\), which forces \(a_n = S_N - S_{N-1} \to S - S = 0\). It can never prove convergence. Integral test. If \(f\) is continuous, positive, and decreasing on \([k, \infty)\) with \(f(n) = a_n\), then \(\sum_{n=k}^\infty a_n\) and \(\int_k^\infty f(x)\,dx\) either both converge or both diverge — though the integral's value is generally not the sum. Applied to \(f(x) = x^{-p}\), the integral converges exactly when \(p > 1\), giving the p-series test: \(\sum 1/n^p\) converges if \(p > 1\) and diverges if \(p \leq 1\). The boundary \(p = 1\) is the harmonic series: its terms go to 0, yet \(\int_1^\infty \frac{dx}{x} = \lim_{t\to\infty} \ln t = \infty\), so it diverges — the standard counterexample showing \(a_n \to 0\) does not imply convergence.
In practice, run the tests in that order. The divergence test is your cheap first screen: if the terms don't go to 0, you're done. If they do go to 0, the test is silent and you reach for the integral test (or, later, comparison). Together with the geometric series from the previous lesson — where \(|r| < 1\) gives an exact sum via the closed form — the p-series family becomes your second benchmark: geometric series are governed by their ratio r, p-series by their exponent p.
Use the integral test on \(\sum_{n=1}^{\infty} \frac{1}{n^2+1}\). Let \(f(x) = \frac{1}{x^2+1}\) — positive, continuous, and decreasing on \([1, \infty)\), so the test applies. \[ \int_1^{\infty} \frac{dx}{x^2+1} = \lim_{t\to\infty} \big[\arctan x\big]_1^t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}, \] a finite number. The integral converges, so \(\sum \frac{1}{n^2+1}\) converges too. (Note: the sum is not \(\pi/4\) — the test only shares the verdict, not the value.)
Test \(\sum_{n=1}^{\infty} \frac{1}{n^3}\). Let \(f(x) = x^{-3}\), which is positive, continuous, and decreasing on \([1, \infty)\). Then \[ \int_1^{\infty} x^{-3}\,dx = \lim_{t\to\infty} \left[-\frac{1}{2x^2}\right]_1^t = \;\text{____} \] Finish the evaluation — is it finite, and what's the verdict?
Reveal the answer
\(\lim_{t\to\infty}\left(-\frac{1}{2t^2} + \frac{1}{2}\right) = 0 + \frac{1}{2} = \) \(\frac{1}{2}\), finite — so the series converges. That matches the p-series test directly: here \(p = 3 > 1\). Set p = 2.5 (the slider's max) in the demo above and watch how thin the shaded tail already is.
More info — why the rectangles trap the integral from both sides
The demo draws each bar over \([n, n+1]\) with height \(f(n)\) — since \(f\) is decreasing, each bar sits above the curve there, so \(\sum_{n=1}^{\infty} a_n \geq \int_1^\infty f(x)\,dx\): an infinite integral forces an infinite series. Now slide every bar one unit right: bar \(n\) covers \([n, n+1]\) with height \(f(n)\), which is below the curve on \([n-1, n]\), giving \(\sum_{n=2}^{\infty} a_n \leq \int_1^\infty f(x)\,dx\) — a finite integral caps the series (the first term \(a_1\) is finite and changes nothing). The two staircases squeeze the series between the integral and the integral plus \(a_1\), which is exactly the "both converge or both diverge" statement in the Formal panel above. Paul's Online Math Notes (Dive deeper below) draws both staircases explicitly.
Check your understanding
What does the divergence test say about \(\sum_{n=1}^{\infty} \frac{n}{n+1}\)?
The terms of \(\sum_{n=1}^{\infty} \frac{1}{n}\) satisfy \(a_n = \frac{1}{n} \to 0\). What can you conclude from the divergence test alone?
For which values of \(p\) does the p-series \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) converge?
Which of these two series converge? \(A = \sum_{n=1}^{\infty} \left(\tfrac{2}{3}\right)^n\) and \(B = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\).
Recap
- Divergence test: if \(a_n \not\to 0\), then \(\sum a_n\) diverges. It only rules convergence out — \(a_n \to 0\) proves nothing (harmonic series).
- Integral test: for \(f\) continuous, positive, decreasing with \(f(n) = a_n\), the series \(\sum a_n\) and \(\int_k^\infty f(x)\,dx\) both converge or both diverge — but the integral's value is not the sum.
- p-series: \(\sum 1/n^p\) converges if \(p > 1\), diverges if \(p \leq 1\); the boundary \(p = 1\) is the divergent harmonic series.
- Workflow: divergence test first as a cheap screen; if terms \(\to 0\), escalate to the integral test or a benchmark (geometric by ratio, p-series by exponent).
Dive deeper
- Paul's Online Math Notes — Integral Test and p-series Confirm the integral test statement and the p-series boundary at p = 1.
Sources
- The Divergence Test, the Integral Test, and p-Series