# Interval of convergence problem

1. Jan 5, 2005

### sinas

The original series is the sum from n=1 to infinity of (n!*x^n)/(n^n)

I used a ratio test to find that the interval of convergence is -e < x < e

But now I need to test the endpoints, which means I need to find if the following two series converge:

sum from n=1 to infinity of (n!*e^n)/(n^n)
sum from n=1 to infinity of (n!*(-e)^n)/(n^n)

I started with the first one, because the second is just an alternating version of it (right?) so if I proved absolute convergence I wouldn't have to do the second one as well. Here is where I am having problems though, I can't figure out a test that would prove divergence or convergence for the first one. I tried a ratio test, but I get an answer of 1 (inconclusive), which makes sense since I used a ratio to find the interval of convergence. Can someone point me in the right direction?

2. Jan 5, 2005

### Hurkyl

Staff Emeritus
Did you try applying Stirling's approximation to this one too?

3. Jan 5, 2005

### sinas

Don't know what that is but I'll look it up in a sec...

4. Jan 5, 2005

### sinas

Ahh that works out nicely...

n!=~n^n*e^-n*(2*n*pi)^(1/2)

which leaves me with...

sum from n=1 to infinity of (2*n*pi)^(1/2)
and
sum from n=1 to infinity of (-1)^n*(2*n*pi)^(1/2)

and in both cases the "infiniteth" term doesn't go to zero, so both diverge... is my reasoning sound?

5. Jan 5, 2005

### sinas

Is there a way to show that the error in stirling's approximation goes to zero as n goes to infinity?

6. Jan 5, 2005

### Hurkyl

Staff Emeritus
I had assumed Stirling's approximation when applying the ratio test -- that's usually the only way I get e's out of it.

There are indeed more detailed forms of Stirling's approximation, but now that you know it diverges, maybe there's an easier way to show the terms don't converge to zero?

7. Jan 5, 2005

### Hurkyl

Staff Emeritus
Oh, BTW, the error doesn't go to zero, but the relative error does. All you really care about, though, is getting a lower bound, so it doesn't matter if there's any sort of convergence, just that you have a useful inequality.

Mathworld or Wikipedia probably have it.

8. Jan 5, 2005

kk thanks