# Euler divergent series $(-1)^nn!$ in $\mathbb{R}$ and $\mathbb{Q}_p$

Consider the series: $$\sum\limits_{n=0}^{\infty}(-1)^nn!$$ Well know that if p is prime, this series convergence in $\mathbb{Q}_p$. Let $s_p\in\mathbb{Q}_p$ is sum of this series.

Also let $$s_{\infty}=\int\limits_{0}^{\infty}\dfrac{e^{-x}}{x+1}dx\in\mathbb{R}$$

In $\mathbb{R}$ the series is not convergence; but we are using the theory of divergent series.

My question: What is the connection between numbers $s_p$?

I want the relationship type $\prod_{p}\left(|s_p|_p\right)^{(1-1/p^3)}=1$.

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$\vert s_p\vert_p$ is zero for all $p<10000$ except $p=2$, $5$, $13$, $37$, and $463$ where it is $1/p$ making the product over all primes smaller than $5\cdot 10^{-7}$. Meanwhile $s_{\infty} = 0.596...$. – Chris Wuthrich Aug 10 '12 at 12:30
@Chris: is it a general fact ( that the sum is either 0 or $1/p$? – Igor Rivin Aug 10 '12 at 15:27
was the "thy" series intentional! MO is getting older by the day! – Suvrit Aug 10 '12 at 15:32
More elementary: The terms in the series are all integers, so $|s_p|_p \le 1$ for finite $p$. So if $|s_\infty|_\infty < 1$ as Chris says, then your product is certainly not $1$. – Gerald Edgar Aug 10 '12 at 16:27
If I read the comments in oeis.org/A124779 correctly, $|s_p|_p$ is one for all $p< 150,000,000$ except the five given by Chris Wuthrich. – Emil Jeřábek Aug 10 '12 at 16:46