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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.

TryingHardToBecomeAGoodPrSlvr
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1 Answers1

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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 $1.879853862\ldots$.

Gary
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  • 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? – TryingHardToBecomeAGoodPrSlvr May 19 '21 at 04:45
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    @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 May 19 '21 at 05:46
  • Will do! Thanks for pointing them out. – TryingHardToBecomeAGoodPrSlvr May 19 '21 at 07:08
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    References for these: http://dlmf.nist.gov/25.12.ii and http://dlmf.nist.gov/6.12.E2 – Gary May 19 '21 at 09:30
  • Wow! They have a digital library of mathematical functions! Thanks once again for pointing it out. BTW I wanted to ask you one small technicality which I guess is trivial for a majority of you mathematicians. You were able to bring in the summation inside the integral. With Riemann integrals, the integral of the limit of a sequence of functions (assuming they all exist) is the integral of the limit if the sequence converges uniformly. The series $\sum ne^{-ns}s^n$ converges uniformly to $\sum e^{-s}s/(1-e^{-s}s)^s$ and that is the reason why you were able to bring in the summation. – TryingHardToBecomeAGoodPrSlvr May 19 '21 at 12:17
  • Please confirm if this is correct. I am studying undergrad analysis (from Stephen Abbott's "Understanding Analysis") which deals with Riemann integration (so no Lebesgue integrals and measure theory yet). All I have at my disposal are Abel's theorem about uniform convergence of sequence of functions, and the integrable limit theorem where the sequence or series of integrable functions converge to the integral of the limit of the sequence or series if the sequence or series converges uniformly to the limit function. – TryingHardToBecomeAGoodPrSlvr May 19 '21 at 12:23
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    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 May 19 '21 at 12:51
  • Cool! Thanks once again! – TryingHardToBecomeAGoodPrSlvr May 19 '21 at 12:52