# Closed form expression for the series $\sum n!/n^n$

It is quite well known that the series $$\sum n!/n^n$$ converges. For instance, the question of convergence was addressed in this thread. However, I was wondering if there is a closed form expression for this sum.

By closed form, I mean an expression where the answer is in the form of a function of a well known kind, or an expression which could even be a definite integral of some function.

I tried to use generating functions and find an expression for this sum but I did not go very far. If someone can derive an expression for this sum starting from some known power series, that would be great.

• This is related to this post Commented May 18, 2021 at 17:42

Using $$n! = \int_0^{ + \infty } {\mathrm{e}^{ - t} t^n \mathrm{d}t} = n^{n + 1} \int_0^{ + \infty } {\mathrm{e}^{ - ns} s^n \mathrm{d}s}$$ and the fact that $$0<\mathrm{e}^{-s}s<\mathrm{e}^{-1}<1$$ for $$s>0$$, we obtain $$\sum\limits_{n = 1}^\infty {\frac{{n!}}{{n^n }}} = \sum\limits_{n = 1}^\infty {n\int_0^{ + \infty } {\mathrm{e}^{ - ns} s^n \mathrm{d}s} } = \int_0^{ + \infty } {\sum\limits_{n = 1}^\infty {n(\mathrm{e}^{ - s} s)^n } \mathrm{d}s} = \int_0^{ + \infty } {\frac{{\mathrm{e}^{ - s} s}}{{(1 - \mathrm{e}^{ - s} s)^2 }}\mathrm{d}s} .$$ The numerical value is $$1.879853862\ldots$$.

• That is sweet! Pretty much what I wanted. You did not have to use any of the "generatingfunctionology" tricks :). Interesting that $e$ made its way into this expression as well! The motivation for me asking this question is that we all know $n^n > n! > x^n > n^p$ asymptotically, where we assume $x$ and $p$ are constant and only $n$ is the variable. We know that the power series of $x^n/n!$ is $e^x$. However, the above inequality opens the possibility for five more series. Why are the rest never explored or talked about? Commented May 19, 2021 at 4:45
• @TryingHardToBecomeAGoodPrSlvr You may look up the polylogarithm for the case of $x^n/n^p$ and the divergent asymptotic series of the exponential integral for $n!/x^n$.
– Gary
Commented May 19, 2021 at 5:46
• Will do! Thanks for pointing them out. Commented May 19, 2021 at 7:08
• References for these: dlmf.nist.gov/25.12.ii and dlmf.nist.gov/6.12.E2
– Gary
Commented May 19, 2021 at 9:30
• Yes, uniform convergence is enough and the series in our case is uniformly convergent. Since the terms of the series are all non-negative, one could also refer to the monotone convergence theorem (Beppo Levi's theorem). Note however that you will encounter this latter result in measure theory.
– Gary
Commented May 19, 2021 at 12:51