## watchmath 5 years ago Compute $\lim_{n\to \infty}\frac{n!}{2^{n^2}}$

1. anonymous

I believe that $$2^{n^2}$$ grows faster than $$n!$$.

2. watchmath

How if you "show" that? :)

3. anonymous

Hmm I would think of using the squeeze theorem somehow to show that.

4. anonymous

let $t_{n}=\frac{n!}{2^{n^2}}$$t_{n+1}=\frac{(n+1)!}{2^{(n+1)^2}}=\frac{(n+1)}{2^{2n+1}}t_{n}$$t_{n+1}=\frac{2(n+1)}{2^{2(n+1)}}t_{n}$now obviously 2^2(n+1) grows faster than 2(n+1) so the series converges...