Recent content by transfear

Thanks for the hint, but I think I may have found another way to prove it. 2^n may be splitted into n-1 products of n: 2 * 2 * 2 * 2 * 2 * ... * 2 <-- doing this n times sqrt(n!) may also be splitted into products like: sqrt(1) * sqrt(2) * sqrt(3) * ... * sqrt(n) Thus, I may recreate...

Stirling brought me to: lim n-->inf [2^n * e^(n/2)]/n^(n/2+1/4) I've also been able to express it as: lim n-->inf [2 * e^(1/2)]^n / n^(n/2+1/4) And I won't show you L'Hospital's results since it gets horrible... Anyways, I feel there's something obvious about it that I cannot see. L'Hospital...

So I've been asked to prove that: lim (n-->infinity) [2^n]/sqrt(n!) = 0 I've tried fiddling with Stirling and L'Hospital, but can't find my way through it. Any thoughts?