I've known the following:


Then, I got interested in the following similar limitation: $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^n}{n^n}$$

By using computer, it seems that

$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^n}{n^n}=\frac{e}{e-1}.$$ However, I haven't been able to prove this. I need your help.


marked as duplicate by Norbert, Julian Kuelshammer, Did, Dominic Michaelis, user61527 Sep 14 '13 at 17:23

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If you denote $u_n=\frac{1}{n^n}\sum_{k=1}^n k^n$, then the idea is that you have : $$u_n=1+\left(\frac{n-1}{n}\right)^n+\left(\frac{n-2}{n}\right)^n+\dots+\left(\frac{1}{n}\right)^n$$

Because $\left(1-\frac{x}{n}\right)^n \rightarrow e^{-x}$, it "looks like" : $$u_n\rightarrow \sum_{k=0}^\infty e^{-k}=\frac{e}{e-1}$$

To show that note that $u_n=\sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^n$. With $\log(1+x)\leq x$ for $x>-1$, you have : $$0\leq u_n\leq \sum_{k=0}^{n-1}e^{-k}\leq \frac{e}{e-1}$$

Moreover let $N\geq 0$. For $n\geq N$ you have : $$u_n\geq \sum_{k=0}^N\left(1-\frac{k}{n}\right)^n=v_n$$ and $\lim_{n} v_n=l_N=\sum_{k=0}^N e^{-k}=\frac{e-e^{-N-1}}{e-1}$ and $l_N\rightarrow \frac{e}{e-1}$.

Let $\varepsilon >0$ and N such that $l_N\geq\frac{e}{e-1}-\varepsilon$. Because $v_n \rightarrow l_N$, for huge n : $$\frac{e}{e-1}-2\varepsilon\leq v_n\leq u_n\leq\frac{e}{e-1}$$



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