Let we have the following sequence $$x_n=\left(\frac{n^n}{n!}\right)^{\frac{1}{n}}$$ Determine if the following sequence is convergent and find its limit
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It converges to $e$, to shhow it you can use stirlings approximation. – Redundant Aunt Sep 27 '15 at 13:55
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@user109899 how can I do that ? – Mhmd Diyene Sep 27 '15 at 13:58
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1This is the same limit as here: Finding the limit of $\frac {n}{\sqrt[n]{n!}}$. (You just use $(n^n)^{1/n}=n$ in the numerator.) It seems to be asked quite often. – Martin Sleziak Sep 27 '15 at 20:43
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For convergence you can use the bounds $$\left(\frac{n}{3}\right)^n\leq n!\leq\left(\frac{n}{2}\right)^n$$ They are provable by induction (for $n\geq 6$ or sth like that).
If you want the limit, you'll probably have to use a version of the deeper result: $$\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}\longrightarrow 1$$ ( = Stirling's formula)
Damian Reding
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By itself it doesn't, but it's a way to start if you're interested in convergence only. – Damian Reding Sep 27 '15 at 16:59
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Taking the $\log$ gives
$$\tag 1\frac{1}{n}(n\ln n -\sum_{k=1}^{n}\ln k) = \ln n -\frac{1}{n}\sum_{k=1}^{n}\ln k.$$
Because $\ln x$ is increasing and nonnegative on $[1,\infty),$ we have
$$\tag 2\int_1^n \ln x\ dx \le \sum_{k=1}^{n}\ln k \le \int_1^n \ln x\ dx +\ln(n+1).$$
Integrating by parts gives $\int_1^n \ln x\ dx = n\ln n - n + 1.$ It's now easy to see limit of $(1)$ equals $1,$ hence the limit of the original expression is $e.$
zhw.
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