Evaluate $\lim_{n\rightarrow \infty} \frac{n^{n+1}}{(1^{1\over n}+2^{1\over n}+\dots+n^{1\over n})^n}$ 
Evaluate $$\lim_{n\rightarrow \infty} \frac{n^{n+1}}{(1^{1\over n}+2^{1\over n}+\dots+n^{1\over n})^n}.
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I know it's easy to find the answer  $e$， but I wanna collect some good ways.
 A: Since $x\to x^{1/n}$  is increasing in $[0,+\infty)$, it follows that
$$\frac{n^{1+1/n}}{1+1/n}=\int_{x=0}^{n}x^{1/n}\,dx\leq 1^{1\over n}+2^{1\over n}+\dots+n^{1\over n}\leq \int_{x=1}^{n+1}x^{1/n}\,dx=
\frac{(n+1)^{1+1/n}-1}{1+1/n}.$$
Hence
$$\frac{1}{(1+1/n)^n}\leq \frac{(1^{1\over n}+2^{1\over n}+...+n^{1\over n})^n}{n^{n+1}}\leq \frac{((n+1)^{1+1/n}-1)^n}{n^{n+1}(1+1/n)^n}=
\frac{((1+1/n)^{1+1/n}-1/n^{1+1/n})^n}{(1+1/n)^n}$$
As $n$ goes to infinity, the left-hand side and the right-side tend to $1/e$. Therefore the given limit is $e$.
A: Here is my way:
$$ \lim_{n \rightarrow \infty}n^{1\over n}-1\sim\lim_{n \rightarrow \infty} \frac{\ln n}{n}\\ n!\approx\sqrt{2n\pi}({n\over \mathrm e})^n      $$So we have:\begin{align}
  \lim_{ n \rightarrow \infty} \frac{n}
{\left(1+\frac{1^{1\over n}+2^{1\over n}+...+n^{1\over n}-n}{n}\right)^{\left(\frac{n}{1^{1\over n}+2^{1\over n}+...+n^{1\over n}-n}\right)(1^{1\over n}+2^{1\over n}+...+n^{1\over n}-n)}}\;\;\;\; \\ =\lim_{n \rightarrow \infty}\frac{n}{\exp[1+1^{1\over n}+2^{1\over n}+...+n^{1\over n}-n]}\\=\lim_{n \rightarrow \infty}\frac{n}{\exp[\frac{1}{n}\left(\ln1+\ln2+...\ln n\right)]}\\=\lim_{n \rightarrow \infty}\frac{n}{\sqrt[n]{n!}} =\mathrm e
\end{align}
A: Another quick way to see it by geometric mean
$$\frac{n^{n+1}}{(1^{1\over n}+2^{1\over n}+\dots+n^{1\over n})^n}= \frac{n}{\left(\frac{1^{1\over n}+2^{1\over n}+\dots+n^{1\over n}}{n}\right)^n}\sim \frac{n}{\sqrt[n]{n!}} \to\mathrm e
$$
