1. ## Comparison series

The question is the sum from n= 1 to infinity of n!/(n^n).
can someone tell me how to do this series...I don't know what to compare it with. I know what a factorial is, I just don't know what to do.

Thanks
John

2. do you have to use the comparison test? This is ideal for the ratio test.

an = n!/(n^n)

an+1 = (n+1)!/[(n+1)^(n+1)]

an+1/an = (n+1)!n^n/[n!(n+1)^(n+1)]

=(n+1)*n^n/[(n+1)^(n+1)]

= [n/(n+1)]^n = [1/(1+1/n)]^n which converges to 1/e

3. Thanks for the reply, but, yes, we have to use the comparison test.

4. Originally Posted by johntuan
Thanks for the reply, but, yes, we have to use the comparison test.
Ever heard of Stirling's approximation?