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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe that \(2^{n^2} \) grows faster than \(n!\).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0How if you "show" that? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm I would think of using the squeeze theorem somehow to show that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let \[t_{n}=\frac{n!}{2^{n^2}}\]\[t_{n+1}=\frac{(n+1)!}{2^{(n+1)^2}}=\frac{(n+1)}{2^{2n+1}}t_{n}\]\[t_{n+1}=\frac{2(n+1)}{2^{2(n+1)}}t_{n}\]now obviously 2^2(n+1) grows faster than 2(n+1) so the series converges...
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