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I was wondering if there is a closed-form expression for

$$\sum_{n=0}^{\infty} \frac{x^n}{e^{-n}n^n},$$

although I expect there is none because Mathematica cannot compute it. However, from Stirling's approximation

$$n! \approx e^{-n} n^n$$

I would expect this sum to be $\approx e^x$ and, in particular, convergent everywhere.

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    $\begingroup$ It definitely converges everywhere (this follows from $ex$ being a fixed number while $n$ grows...that is, $\frac{ex}{n}<1$ for large enough $n$, and at that point, you could approximate using a geometric series). $\endgroup$
    – Clayton
    Commented Sep 30, 2018 at 1:45
  • $\begingroup$ As $x \to \infty$ you should get $\frac{f'(x)}{f(x)}= 1+o(1/x^2),\log f(x) = x+C+o(1/x), f(x) = A e^x (1+o(1/x))$ $\endgroup$
    – reuns
    Commented Sep 30, 2018 at 3:55

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