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I have a question: What is the value of $\sum _{n=1}^{\infty \:}\frac{n!}{n^n}$? Only I've calculated the following identity: $$\sum _{n=1}^{\infty \:}\frac{n!}{n^n}=\int _0^{1}\left(1+x\cdot \ln \left(x\right)\right)^{-2}dx$$ I found this identity, when I watched The Sophomore's dream in wikipedia, but I would know the value exact of the serie.

Regards,

Ronald.

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    $\begingroup$ What do you mean by "the exact value"? Not everything can be written in terms of "familiar constants" and "familiar algebraic operations". The exact value of this thing might just be what ever it is. $\endgroup$
    – Yemon Choi
    Jul 4, 2014 at 19:38
  • $\begingroup$ The series grows like sqrt(6n)e^-n, so you need only 30 terms for at least 10 digits precision. You can use an inverter or a Google search to find more info. $\endgroup$ Jul 4, 2014 at 20:40
  • $\begingroup$ Yeah, I really think that with regard to Yemon's comment, the question needs to be clarified. I believe there are various notions of "numbers in closed form" and I would guess this number is not in closed form according to at least one of these notions, but to give substance to the question, one has to be more specific. $\endgroup$
    – Todd Trimble
    Jul 4, 2014 at 20:42
  • $\begingroup$ Given that there is probably no hope of a simple expression in terms of elementary functions, I'd suggest that the OP changes the question in the broader sense: are there other relevant identities for this number? $\endgroup$ Jul 4, 2014 at 23:17

1 Answer 1

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There is no known closed form for this expression. However, we have the following identity:

$$\int_0^\infty\frac{E(x)}{e^x}dx~=~\sum_{n=0}^\infty\frac{n!}{n^n}$$

where $~E(x)=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n^n}~$ and $~\displaystyle\lim_{n\to0^+}n^n=1$.

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  • $\begingroup$ $E(x)$ is defined in a manner similar to that of $e^x$, and $E(\pm1)$ yields the two expressions in the Sophomore's dream article. $\endgroup$
    – Lucian
    Jul 4, 2014 at 20:29

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