The Integral Test

The integral test is simply a way to decide whether a series adds up to something finite by looking at the area under a matching curve instead. It works because adding up $f(1) + f(2) + f(3) + \cdots$ and finding the area under the graph of $f$ are close cousins, so if one is finite, so is the other. It's good because integrals are often much easier to work out than sums. The statement $\sum f(n)$ converges $\iff \int_k^{\infty} f(x)\,dx$ converges looks intimidating, but it's just three pieces: $\sum f(n)$ (the series, the heights of the graph at whole numbers added up), $\int_k^{\infty} f(x)\,dx$ (the area under that same graph, all the way out) and $\iff$ (they go together, either both finite or both infinite). It's a lot of symbols for "the sum and the area share a fate," so bear with it.

Why would you want this? Last lesson you met the harmonic series, and we needed a clever grouping trick to show it diverges. Most series don't have a trick like that. But you already know how to handle improper integrals, and those are often easy. This test lets you trade a hard series question for an integral question you already know how to answer.

The picture. Draw rectangles of width 1 and height $f(n)$. Each rectangle has area $f(n)$, so the total area of the rectangles is the series. Placed to the left of each $n$, they sit under the curve, so $\sum_{n=2}^{\infty} f(n) \le \int_1^{\infty} f(x)\,dx$. Placed to the right, they cover the curve, so $\int_1^{\infty} f(x)\,dx \le \sum_{n=1}^{\infty} f(n)$. A finite area traps the series underneath it; an infinite area sits under the series and drags it to infinity too.

Requirements. $f$ must be positive, continuous and decreasing, at least from some point on. If the terms aren't decreasing, the rectangles don't line up with the curve and the test doesn't apply.

A common mistake. It's tempting to think the series and the integral are equal, since they're tied so closely together. They're not. $\int_1^{\infty} \frac{dx}{x^2} = 1$, but $\sum \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645$. The rectangles stick out above the curve, so they hold extra area. The test only says the two share a fate: both finite or both infinite.

p-series. Now we can settle a whole family at once. $\int_1^{\infty} x^{-p}\,dx$ is finite exactly when $p > 1$, so $\sum \frac{1}{n^p}$ converges exactly when $p > 1$. Notice the harmonic series, $p = 1$, sits right on the edge and falls on the divergent side. This family, together with geometric series, supplies most of the comparisons in the next lesson.

Reach for the integral test when $f(x)$ has an easy antiderivative, often through a substitution, like $\frac{1}{n\ln n}$ or $ne^{-n^2}$.

The integral test is behind running-time estimates like $1 + \frac12 + \cdots + \frac1n \approx \ln n$, which is why quicksort takes about $n\ln n$ steps on average.