00:01
So in this question, we're going to use the integral test to determine the convergence of the series, sum from n equals 1 to infinity, of n over n squared plus 1.
00:09
So what does the integral test say to do? it says i should convert my series into the associated improper integral.
00:18
So essentially, just replace the sigma with an integral, the integral from 1 to infinity of x over x squared plus 1 dx.
00:31
Now, since this is an improper integral, i start by rewriting it as a limit.
00:38
I have a limit as b approaches infinity of the integral from 1 to b of x over x squared plus 1 dx.
00:49
Now, in order to evaluate this, i am going to do a u substitution.
00:55
That's what i need to get my antiderivative.
00:59
So what am i going to let my u be? i'm going to let u equal x squared plus 1.
01:04
If that's my u, my du is equal to 2x dx.
01:10
And dividing by 2, i have du over 2 is equal to x dx.
01:18
Now, whenever i do a u substitution within a definite integral, i have to be sure to change my limits of integration.
01:27
So when i was in the world of x, my x limits of integration, they were from 1 to b.
01:34
How about u? well, if x is 1, my u would be 1 squared plus 1, which is 2.
01:41
And if x is b, my u would be b squared plus 1.
01:47
And so when i substitute in, what do i get? i get a limit as b approaches infinity of an integral from 2 to b squared plus 1.
02:01
My denominator, x squared plus 1, that becomes u.
02:05
In my numerator, i have x dx.
02:08
That becomes du over 2...