Analysis of a_n Series: Convergence or Divergence?

[azatkgz]

Suppose that a_n\geq 0 and there is

\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=c
If c>1,series diverges.
if c<1 series converges.

For a_n=\frac{n!}{n^n}

\lim_{n\rightarrow\infty}\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}

\lim_{n\rightarrow\infty}\frac{n^n}{(n+1)^n}

Then I used I'Hopital Rule and got answer 1.

 

[Dick]

I think you'd better check your l'Hopital. Your final limit is closely related to the limit of (1+1/n)^n, which is a famous limit and is not one.

 

[azatkgz]

A,yes I get it.


\lim_{n\rightarrow\infty}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}