Sequences and Series, The Integral Test

The Integral Test

Let f(x) be a monotonically decreasing function for x > 0, and build the decreasing sequence s by assigning s_n = f(n). The series s converges iff f has a finite integral. The harmonic series s_n = 1/n is the textbook example. (See the previous section.) The integral of 1/x, or log(x), is unbounded, hence the series 1/n diverges.

In general, the integral of f is the area between f and the x axis. This can be overestimated by the upper sums or underestimated by the lower sums. Both estimates are partial sums of the series s. If f is integrable it dominates s, or at least a tail of s, hence s converges. And if the area under f is infinite then s, or the tail of s, overestimates this area, and is unbounded.

We've already examined the harmonic series; let's look at a series that converges. Let f(x) = 1/x², giving the sequence s_n = 1/n². Integrate f, giving -1/x, and evaluate from 1 to infinity. The result is 1, hence s converges, and it converges absolutely. By the same reasoning, 1/n^r converges for every real r > 1.