Question

Compute
\[\lim_{n\to \infty}\frac{n!}{2^{n^2}}\]

Answer 1

I believe that \(2^{n^2} \) grows faster than \(n!\).

Answer 2

How if you "show" that? :)

Answer 3

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

Answer 4

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...