I know that $n!/n^{n} = o(1)$ as $n \to \infty,$ but what can I say about $e^{n}$ involved here?
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Hint: Stirling's approximation. – Nigel Overmars Mar 26 '15 at 12:13
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The relevant result is Stirling's formula, which is usually proven by comparing $\ln(n!) = \sum_{k=1}^n \ln(k)$ to $\int_1^n \ln(x) dx$. It turns out that this diverges, but "critically", in the sense that if you replace $e$ by any larger number, you don't get divergence. – Ian Mar 26 '15 at 12:15
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Ah, Stirling's, yes; thanks. – Yes Mar 26 '15 at 12:16
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Refer to http://math.stackexchange.com/questions/1196710/calculation-of-limit-without-stirling-approximation/1205727#1205727 – Mar 26 '15 at 13:47
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By the Stirling's approximation
$$n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n$$ we see that the given sequence is divergent.
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From the telescopic product: $$ n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\tag{1}$$ it follows that: $$ n! = \frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}\tag{2}$$ hence: $$ \log\left(\frac{n!e^n}{n^n}\right)=\sum_{k=1}^{n-1}\left(1-k\log\left(1+\frac{1}{k}\right)\right)\geq (1-\log 2)\sum_{k=1}^{n-1}\frac{1}{k}\geq\frac{3}{10}\log n \tag{3}$$ implies that $\frac{n!e^n}{n^n}$ diverges as $n\to +\infty$.
Jack D'Aurizio
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