Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] and showing that [latex]{S}_{{2}^{k}}>1+\frac{k}{2}[/latex] for all positive integers [latex]k[/latex]. In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

To illustrate how the integral test works, use the harmonic series as an example. In Figure 1, we depict the harmonic series by sketching a sequence of rectangles with areas [latex]1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots[/latex] along with the function [latex]f\left(x\right)=\frac{1}{x}[/latex]. From the graph, we see that

Therefore, for each [latex]k[/latex], the [latex]k\text{th}[/latex] partial sum [latex]{S}_{k}[/latex] satisfies

Since [latex]\underset{k\to \infty }{\text{lim}}\text{ln}\left(k+1\right)=\infty[/latex], we see that the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] is unbounded. Therefore, [latex]\left\{{S}_{k}\right\}[/latex] diverges, and, consequently, the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] also diverges.

Figure 1. The sum of the areas of the rectangles is greater than the area between the curve [latex]f\left(x\right)=\frac{1}{x}[/latex] and the [latex]x[/latex] -axis for [latex]x\ge 1[/latex]. Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.

Now consider the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex]. We show how an integral can be used to prove that this series converges. In Figure 2, we sketch a sequence of rectangles with areas [latex]1,\frac{1}{{2}^{2}},\frac{1}{{3}^{2}},\ldots[/latex] along with the function [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex]. From the graph we see that

Therefore, for each [latex]k[/latex], the [latex]k\text{th}[/latex] partial sum [latex]{S}_{k}[/latex] satisfies

We conclude that the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] is bounded. We also see that [latex]\left\{{S}_{k}\right\}[/latex] is an increasing sequence:

Since [latex]\left\{{S}_{k}\right\}[/latex] is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] converges.

Figure 2. The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex] and the [latex]x[/latex] -axis for [latex]x\ge 1[/latex]. Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.

We can extend this idea to prove convergence or divergence for many different series. Suppose [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] is a series with positive terms [latex]{a}_{n}[/latex] such that there exists a continuous, positive, decreasing function [latex]f[/latex] where [latex]f\left(n\right)={a}_{n}[/latex] for all positive integers. Then, as in Figure 3 (a), for any integer [latex]k[/latex], the [latex]k\text{th}[/latex] partial sum [latex]{S}_{k}[/latex] satisfies

Therefore, if [latex]{\displaystyle\int }_{1}^{\infty }f\left(x\right)dx[/latex] converges, then the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] is bounded. Since [latex]\left\{{S}_{k}\right\}[/latex] is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if [latex]{\displaystyle\int }_{1}^{\infty }f\left(x\right)dx[/latex] converges, then the series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] also converges. On the other hand, from Figure 3 (b), for any integer [latex]k[/latex], the [latex]k\text{th}[/latex] partial sum [latex]{S}_{k}[/latex] satisfies

If [latex]\underset{k\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{k+1}f\left(x\right)dx=\infty[/latex], then [latex]\left\{{S}_{k}\right\}[/latex] is an unbounded sequence and therefore diverges. As a result, the series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] also diverges. Since [latex]f[/latex] is a positive function, if [latex]{\displaystyle\int }_{1}^{\infty }f\left(x\right)dx[/latex] diverges, then [latex]\underset{k\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{k+1}f\left(x\right)dx=\infty[/latex]. We conclude that if [latex]{\displaystyle\int }_{1}^{\infty }f\left(x\right)dx[/latex] diverges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] diverges.

Figure 3. (a) If we can inscribe rectangles inside a region bounded by a curve [latex]y=f\left(x\right)[/latex] and the [latex]x[/latex] -axis, and the area bounded by those curves for [latex]x\ge 1[/latex] is finite, then the sum of the areas of the rectangles is also finite. (b) If a set of rectangles circumscribes the region bounded by [latex]y=f\left(x\right)[/latex] and the [latex]x[/latex] axis for [latex]x\ge 1[/latex] and the region has infinite area, then the sum of the areas of the rectangles is also infinite.