In some problem, I need to evaluate this limit: $$\lim_{n\rightarrow \infty}\sqrt[n]{\frac{1}{n!}\sum^n_{m=0}(m^m)}.$$ I know about Taylor series and that kind of stuff. I'm not sure where to start, maybe Stirling but after using it I still could not solve it. Any help will be appreciated.

Using Stirling's equivalence, I get to: $$\lim_{n\rightarrow \infty}\frac{e}{n}\sqrt[n]{\frac{\sum^n_{m=0}(m^m)}{\sqrt{2\pi n}}}$$ I don't know if this is useful anyway.

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    $\begingroup$ The sum is bounded below by $n^n$ and above by $\sum n^m$ $\endgroup$ – Empy2 Jul 8 '20 at 11:27
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    $\begingroup$ Mathematica seems to suggest the limit is $e$. $\endgroup$ – Chrystomath Jul 8 '20 at 11:28

Using $n^n\le\sum_mm^m\le nn^n$, $$1\le\frac{1}{n}\left(\sum_{m=0}^nm^m\right)^{1/n}\le n^{1/n}$$

Since $n^{1/n}\to1$, so does the sum. Hence, (using $(\sqrt{2\pi})^{1/n}\to1$) $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{\sum_mm^m}{\sqrt{2\pi n}}}\to1$$ and the original sequence converges to $e$.


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