I would love to know how to show the following:
$n!n^{s} = o(n^n)$ fo $n \rightarrow \infty, s \geq 0$
As for a hint, I know that $\sum_{k=1}^n \log k \leq n* \log (\frac{n+1}{2})^{n}$ can be used, but haven't really come far.
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By Stirling's approximation formula: $$n!\sim_{+\infty}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n,$$
we have $$\frac{n!n^s}{n^n}\sim\sqrt{2\pi n}e^{-n}n^{s}\to_{n\to\infty}0,$$ so we have the result.
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thank you, still I was wondering if the implication of Jensen is also a possibility to prove the "equality" – user62487 Mar 12 '13 at 14:26
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For every $1\leq k<n+s$ we have $$n!=1\cdots k \cdots n \leq k^k n^{n-k}.$$ For simplicity assume $n$ is even, and let $k=n/2$. Thus $$ n!\leq \left(\frac{n}{2}\right)^{n/2}\cdot n^{n/2}= \frac{n^n}{2^{n/2}}. $$ So $$ n!n^s\leq n^n\frac{n^s}{2^{n/2}} = o(n^n), $$ since $\lim \frac{n^s}{2^{n/2}}=0$.
Lior B-S
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thanks, yet I did not understand how you came up with $\left(\frac{n}{2}\right)^{n/2}\cdot n^{n/2}$..? – user62487 Mar 12 '13 at 14:29
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