00:01
In this problem, our job is to determine whether the series, the summation from 2 to infinity of 1 over n times the square root of the natural log of n converges or diverges.
00:13
So we notice that this function has a natural log under a square root, and it also has an n in the denominator.
00:19
So we're going to let the corresponding function f of x be 1 over x times the square root of the natural log of x.
00:29
And what we'd like to do is to apply the interval test.
00:33
Well, for the integral test, f of x has to satisfy three conditions.
00:37
It has to be positive, and this function is positive.
00:41
The interval we're looking at is from 2 to infinity.
00:45
It's positive, it's continuous, and it's decreasing, because those terms in the denominator are increasing.
00:57
So f of x is positive, continuous, and decreasing on the interval from 2 to infinity.
01:03
That means since it checks all three of those hypotheses, we were able to use the integral test for the original series.
01:15
So the integral test says that we should calculate the following integral.
01:21
If that integral converges, then the series converges.
01:24
If that integral diverges, then the series diverges.
01:28
The integral we want to calculate is the integral from 2 to infinity of f with respect to x.
01:36
That is our function 1 over x times the square root of the natural log of x with respect to x.
01:45
So to calculate that integral, we will do a u substitution.
01:49
We will let u be the natural log of x.
01:54
So the du is 1 over x dx.
02:00
So let's take this substitution and apply it.
02:05
We have an improper integral, so we need to do the limit as b approaches infinity of the integral...