Evaluate $\lim_{n\rightarrow \infty} \frac{n^{n+1}}{(1^{1\over n}+2^{1\over n}+\dots+n^{1\over n})^n}$

Question

Evaluate $$\lim_{n\rightarrow \infty} \frac{n^{n+1}}{(1^{1\over n}+2^{1\over n}+\dots+n^{1\over n})^n}. $$

I know it's easy to find the answer $e$， but I wanna collect some good ways.

Answer 1

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 $e$.

Answer 2

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}

Answer 3

Another quick way to see it by geometric mean

$$\frac{n^{n+1}}{(1^{1\over n}+2^{1\over n}+\dots+n^{1\over n})^n}= \frac{n}{\left(\frac{1^{1\over n}+2^{1\over n}+\dots+n^{1\over n}}{n}\right)^n}\sim \frac{n}{\sqrt[n]{n!}} \to\mathrm e $$