Evaluate the limit $$\large{\lim_{n\to \infty} {e^n n!\over n^n}}$$
Does the limit have a defined value? If yes then please provide me a solution to it.
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Hint
When you have problems involving factorials, Stirling approximation is very useful. A simple form is $$n! \simeq \sqrt{2\pi n}\Big(\frac{n}{e}\Big)^n$$
I am sure that you can take from here.
Claude Leibovici
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Havent heard bout the formula.. Nyways thank you...Let me try it from here. – Naive Sep 07 '14 at 08:00
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@AbhishekBhat. It is a very classical and very useful and simple approximation. There are more accurate approximations but this one is so simple. Keep it in your memory for ever since you will use it a lot with success ! – Claude Leibovici Sep 07 '14 at 08:02
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$$a_n:=\frac{n^n}{e^nn!}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)^n}{e^{n+1}(n+1)!}\frac{e^nn!}{n^n}=\frac1e\left(1+\frac1n\right)^n\frac1{(n+1)}\xrightarrow[n\to\infty]{}\frac1ee\cdot0=0$$
Thus, the positive series $\;\sum_{n=1}^\infty a_n\;$ converges by D'Alembert's test, so
$$a_n\xrightarrow[n\to\infty]{}0\implies\frac1{a_n}=\frac{e^nn!}{n^n}\xrightarrow[n\to\infty]{}\infty$$
Timbuc
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2This is wrong, we have $$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{e^{n+1}(n+1)!} \frac{e^n n!}{n^n} = \frac{(n+1)^n}{e^{n+1} n!} \frac{e^n n!}{n^n} = \frac{1}{e}\biggl(1 + \frac{1}{n}\biggr)^n \to 1.$$ The ratio test is inconclusive. In fact, the series is divergent, since $a_n \sim \frac{1}{\sqrt{2\pi n}}$. (cc @AbhishekBhat) – Daniel Fischer Nov 05 '16 at 12:46
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yes it does. The easiest way to see it is to use Stirling's formula for $n!$. Another way is to write $\frac{1 \cdot 2 \ldots n}{n \cdot n \ldots n}$ and take the upper bound. Can you handle from here?
Alex
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Hint
Stirling's formula states that $$n!\sim n^ne^{-n}\sqrt{2n\pi}.$$ I think from here you can do
