Lesson 15: The Divergence and Integral Tests

Motivating question

We are covering two tests for our series convergence/divergence toolbox today, so we’ll ask two motivating questions:

• What can integrals tell us about series?
• What is the “easiest” way for a series to diverge?

Helpful reminder

Recall from Lesson 10 (link here) that for functions of the form $f(x)=\frac{1}{x^p}dx$ where $p>0$,

1. if $p>1$, then ${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx$ converges, and
2. if $0<p\le 1$ then ${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx$ diverges.

In Warmup exercise 1 p = 2 and in Warmup exercise 2, p=1.

We’ll start with the integral test and see the divergence test later in the lesson

Integral test

In Lesson 10 (link here) we saw that improper integrals represent areas of unbounded regions and they either converge or diverge. In Lesson 14 (link here) we saw that series are infinite sums and they either converge or diverge. The integral test tells us exactly what these two facts have to do with each other.

Before we get to the formal statement of the integral test, here are two visuals that should help give you an intuitive understanding:

1. This Desmos graph (link here) shows ${\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dx$ as the red shaded region.  The series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ is represented by the areas of the blue rectangles plus the area of the one green rectangle. We had to separate out the first term of the series because the green rectangle with area $\frac{1}{1^2}$ doesn’t fit inside the red region, but the blue rectangles, which represent $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}$ do. Since the red region represents an integral that converges (see Warmup exercise 1), and the blue region sits inside the red region,  the blue region represents a series that also converges. So $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}$ converges and (since you can add a finite number to something that converges and get something that converges), $\frac{1}{1^2}+\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ also converges.
2. This Desmos graph (link here) shows ${\int }_1^{\mathrm{\infty }}\frac{1}{x}dx$ as the red shaded region. The series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ is represented by the areas of the blue rectangles (only the first few terms of the series are shown). Clearly, the blue region is larger than the red region. So since the red region represents an integral that diverges (see Warmup exercise 2), the blue region represents a series that also diverges.

When looking at the areas of the irregular shapes versus the sums of the areas of the rectangles, think about the blobs of different sizes we used to demonstrate how the comparison theorem for improper integrals works in Lesson 10 (link here).

Remark: As stated in the theorem above, both the improper integral and series start at the value 1, but the theorem is still true if we replace that 1 with any positive integer $N$.

Video 1 below introduces you to how the integral test works through three examples.

p-series

Two examples in that video are examples of what we call “p-series.”  These are series that look like $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$.

The corresponding improper integrals should look familiar to you: ${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx$.

You know the improper integrals converge if $p>1$ and diverge if $p\le 1$. The same is true for p-series and you can prove this using the integral test.

Videos 2, 3, and 4 below show more examples of the integral test in action.

Divergence Test

You saw the divergence test briefly in one of the videos in Lesson 14 (link here).

The divergence test is convenient when it applies. It’s always a good idea to check whether it applies to a particular series before you start trying to use another test for convergence.

Remember that when an infinite series converges, that means that a sum of infinitely many numbers is actually (in the limit) finite. The only way this can possibly happen is if the individual numbers are getting smaller and smaller and smaller. If the individual numbers (terms of the sequence) are not getting smaller, there’s no way for the series to converge.

Compare this to improper integrals of Type 1 from Lesson 10 (link here): if your function $f(x)$ does not have the $x$-axis as a horizontal asymptote as $x$ approaches $\mathrm{\infty }$, then there’s no way that ${\int }_a^{\mathrm{\infty }}f(x)dx$ (which calculates the area under the curve) can possibly converge.

Notice which direction the implication goes: if the individual terms of the sequence don’t approach zero, then the infinite series diverges. The test does not say that that if the individual terms do approach zero, then the infinite series converges. Compare this to the improper integral again: if $f(x)=\frac{1}{x}$ then $f(x)$ has the $x$-axis as a horizontal asymptote ($\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}=0$). But having the $x$-axis as a horizontal asymptote is not enough for convergence; we know that ${\int }_1^{\mathrm{\infty }}\frac{1}{x}dx$ diverges.

To reiterate: the divergence test cannot tell you if a series converges, but it might tell you if the series diverges.

Video 5 below show how the divergence test works in some examples where it applies.

Video 6 below shows how the divergence test works in some other examples where it applies. But note that these are examples of geometric series and we know from Lesson 14 (link here) exactly when geometric series converge or diverge. The divergence test can only confirm divergence when $|r|\ge 1$.

Back to today’s motivating questions

• What can integrals tell us about series?
  • If the individual terms $a_n$ of a series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ correspond to a function $f(x)$ satisfying the conditions of the integral test, then the convergence or divergence of the corresponding improper integral of $f(x)$ tells us whether the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges or diverges.
• What is the “easiest” way for a series to diverge?
  • Some people may argue about how “easy” the divergence test is, but the point is that it can tell you if a particular series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ doesn’t have a hope of converging: if the individual terms in an infinite sum don’t approach zero, there is no way for that infinite sum to converge. If they do approach zero, you’ll have to use another test to determine whether the series converges or diverges.