Convergence/Divergence of series

[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!

 

[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!.

 

[Thomas_]

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!.

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.

 

[Gib Z]

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

 

[Office_Shredder]

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.

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