# $\lim_{n\to\infty}(1^n+2^n+…+n^n) /n^n$. [duplicate]

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:

$$\bf{Solution\quad by\quad the\quad poster:}$$ Firstly, it is trivial that the limit exists, then I try to use $$\epsilon-N$$ language for my proof.  $$\forall N ,we denote that $$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})} Then we deduce that $$\forall \epsilon >0, \exists N_0\in\{1,2,\cdots,n\}:A As for $$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}| Hence, when $$n\to\infty$$, $$\lim_{n\to\infty}(1^n+2^n+…+n^n) /n^n=\frac{e}{e-1}$$
• 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. – angryavian Dec 30 '19 at 7:32