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Since $$\lim_{n\to\infty}(\frac{n-k}{n})^n=e^{-k}$$ We can predict that: $$\lim_{n\to\infty}(1^n+2^n+…+n^n) /n^n=\sum_{k=0}^\infty e^{-k}=\frac{e}{e-1}$$ However, I think the solution is NOT strict enough. Although I have a rigorous proof for this, I want a easier solution.$$$$My solution as follows:

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$\bf{Solution\quad by\quad the\quad poster:}$ $$$$Firstly, it is trivial that the limit exists, then I try to use $\epsilon-N$ language for my proof. $$$$ $\forall N<n, N\in\mathbb{N}$ ,we denote that $$A=\sum_{i=1}^{n-N-1}(\frac{i}{n})^n$$ and $$B=\sum_{i=N}^{n}(\frac{i}{n})^n$$ Let $N$ be fixed , when $n\to\infty$, $$A<\dfrac{\int_0^{n-N}x^n dx}{n^n}=\dfrac{(n-N)^{n+1}}{(n+1)n^n}<(1-\frac{N}{n})^n=e^{\ln (1-\frac{N}{n})}<e^{-N}$$ Then we deduce that $\forall \epsilon >0, \exists N_0\in\{1,2,\cdots,n\}:A<e^{-N}<\epsilon$ $$$$As for $B$, we have: $$|e^k(1-\frac{k}{n})^n-1|=|e^{k+n\ln (1-\frac{k}{n})}-1|=\frac{k^2}{2n}+o(\frac{1}{n})<\frac{c_kN^2}{n}<\epsilon\text{ ,as $n$ is big enough.}$$ We have$$|(1-\frac{k}{n})^n-e^{-k}|<\frac{c_kN^2}{n}e^{-k}<\epsilon e^{-k}\text{ ,$c_k$ is a constant}$$ Then, we have $$|B-\sum_{k=0}^N e^{-k}|<\sum_{k=0}^N|(1-\frac{k}{n})^n-e^{-k}|<\epsilon \sum_{k=0}^Ne^{-k}<\frac{e}{e-1}\epsilon$$ We deduce that $$|(1^n+2^n+…+n^n) /n^n-\sum_{k=0}^{n-1}e^{-k}|<C\epsilon \text{ ,for some $C=$ constant}$$ Hence, when $n\to\infty$, $$\lim_{n\to\infty}(1^n+2^n+…+n^n) /n^n=\frac{e}{e-1}$$

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  • $\begingroup$ The question has already been posted on Quora. Link: quora.com/…. But I think the solution there is not what I want, since we cannot add the limits when there are infinite series. $\endgroup$ – Tamshin Dion Dec 30 '19 at 6:36
  • $\begingroup$ For the record, you are right in pointing out that you cannot simply interchange the order of the limit and sum when the number of terms grows with $n$, as the Quora answers seem to do. $\endgroup$ – angryavian Dec 30 '19 at 7:32

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