5.3 The Divergence and Integral Tests

Textbook PDF pp.471-478

Textbook HW: p. 482: 138–145 odd, 152—155, 158, 159, 161, 163

Webwork: Integral Test and Divergence Test both sets due 4/21 (these two Webwork sets are relatively short, so don’t forget to practice the textbook homework as well).

Notes from lecture and office hours

Video:

https://www.dropbox.com/s/rlz1wsvktafqjzm/Lecture%20%26%20Office%20Hours-20200415%201224-1.mp4?dl=0

Divergence Test

You saw the divergence test briefly in Monday’s lesson.

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 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: if your function $f(x)$ does not have the $x$-axis as a horizontal asymptote as $x$ approaches $\infty$, then there’s no way that $\int_a^\infty f(x)dx$ (which calculates the area under the curve) can possibly converge.

The formal statement of the divergence test is Theorem 5.8 in the text so you should read that now. 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 ($\lim_{x \to \infty} \frac{1}{x} = 0$) but $\int_1^\infty \frac{1}{x}dx$ does not converge.

These ideas lead can be formalized by the integral test, which is another tool in your toolbox to determine if a series converges or diverges.

Integral Test

This video introduces you to how the integral test works through three examples (10 minutes).

Two examples in that video are examples of what we call “p-series.”  These are series that look like:

$\sum_{n=1}^\infty \frac{1}{n^p}$.

The corresponding improper integrals should look familiar to you:

$\int_1^\infty \frac{1}{x^p}dx$.

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

Here are two visuals that might help give you an intuitive understanding:

  • This graph shows $\int_1^\infty \frac{1}{x}dx$ as the red shaded region. The series $\sum_{n=1}^\infty \frac{1}{n}$ is represented by the areas of the blue rectangles (I graphed only the first few terms of the series, but I hope you get the idea). Clearly, the blue region is larger than the red region. So since the red region represents an integral that diverges, the blue region represents a series that also diverges. (Think about the blobs of different sizes we used to demonstrate how the comparison theorem for improper integrals works.)
  • This graph shows $\int_1^\infty \frac{1}{x^2}dx$ as the red shaded region.  The series $\sum_{n=1}^\infty \frac{1}{n}$ 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 $\sum_{n=2}^\infty \frac{1}{n}$ do. Since the red region represents an integral that converges, and the blue region sits inside the red region,  the blue region represents a series that also converges. So $\sum_{n=2}^\infty \frac{1}{n}$ converges and (since you can add a finite number to something that converges and get something that converges), $\frac{1}{1^2} + \sum_{n=2}^\infty \frac{1}{n} = \sum_{n=1}^\infty \frac{1}{n}$ also converges.

Here are a few more videos showing examples of the integral test (each about 5 minutes).