Convergence/Divergence of series

1. Nov 25, 2007

Thomas_

Hello,

I have to prove conv/div. for the following series:

$$\sum\frac{(2n)!}{n^n}$$

I use the "ratio-test" and get the following:

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} = \lim_{n\to\infty} \frac{(2n+2)!}{(2n)!} \frac{n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{(2n+2)(2n+1)}{(n+1)} (\frac{n}{1+n})^n = \infty \frac{1}{e} = \infty$$

This means the series diverges, however, the series should converge (I could find the finite sum online).

Where is my mistake?

Thank you!

2. Nov 26, 2007

robert Ihnot

The last half terms can be written as {(n+l)/n}{(n+2)/n}{(n+3)/n}........{2n/n}, and the first n terms are just n!.

3. Nov 26, 2007

Thomas_

Sorry, I do not quite understand what you mean or how this helps me. Could you elaborate on that?

Also, I am interested in why the test I am using does not work out like it should or if I made an algebra mistake somewhere along the way.

4. Nov 26, 2007

Gib Z

Using stirlings approximation to replace the factorial, I get the series diverges. Where did you find online its sum?

5. Nov 26, 2007

Office_Shredder

Staff Emeritus
What he's saying is that if you split it up, you get 1/n*1/n*1/n...*(2n)(2n-1)(2n-2)...(n+1)*n!

So you put one n under each 2n-k and get

2n/n*(2n-1)/n*(2n-2)/n....*(n+1)/n*n!

As each (2n-k)/n>1, and n!>1, each term in the series is >1. So there's very little reason why it would converge