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This question related to this question in SE ,I would like to know how do I

evaluate this sum for $s$ is a complex variable :$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2s}n!}$$ .

Edit01:And I think the General complex solution of $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2s}n!}=0$$ would be $s=-1+i\beta$ for some values of $\beta$ and for $\beta =0$ it's a trivial zero

Note 01 :In wolfram alpha the series is converge but i don't know if it has a nice closed form and really the convergence in complex number ensemble is not clear !!!

Note 02 I edited the question only for the zeros of this series since it's convergent after some computation in wolfram alpha!!!

Thank you for any help

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    $\begingroup$ Even the case $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{1/2}n!} \approx 0.725065152077915927$$ seems to have no known closed form. $\endgroup$ – Gerald Edgar Dec 26 '15 at 21:11
  • $\begingroup$ but how do i show that is hasn't no closed form , is by mathematica ? $\endgroup$ – zeraoulia rafik Dec 26 '15 at 21:24
  • $\begingroup$ Note for instance that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n\cdot n!} = \int_0^1\frac{e^x-1}{x}dx$, which has no known closed form expression. $\endgroup$ – Richard Stanley Dec 27 '15 at 0:36
  • $\begingroup$ @RichardStanley: we may consider the exponential integral function $\mathrm{Ei}(x)$ to be "closed form". But the square-root case doesn't even have that. $\endgroup$ – Gerald Edgar Dec 27 '15 at 1:21
  • $\begingroup$ Hint for convergence with complex $s$: investigate absolute convergence. $\endgroup$ – Gerald Edgar Dec 27 '15 at 14:53
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Mathematica says that $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k n!} = \, _kF_k(1,\dotsc, 1;2,\dotsc,2;-1),$$ where the numbers of $1$s (and $2$s) are both equal to $k.$ This suggests to me that there is no closed form.

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  • $\begingroup$ Thank you for your answer , do you meant no general form of solution existed and what it does meant F_k ? $\endgroup$ – zeraoulia rafik Dec 26 '15 at 21:11
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    $\begingroup$ @zeraouliarafik I believe there is no closed form, and the notation means hypergeometric pFq (where $p=q=k,$ in this case). $\endgroup$ – Igor Rivin Dec 26 '15 at 21:29
  • $\begingroup$ @ Igor Rivin, This Series converge for all s , is it true ? $\endgroup$ – zeraoulia rafik Dec 28 '15 at 15:17
  • $\begingroup$ @zeraouliarafik check out Gerald Edgar's comments... $\endgroup$ – Igor Rivin Dec 28 '15 at 15:22
  • $\begingroup$ , The question has been answered here in S.E math.stackexchange.com/a/1924182/230303 $\endgroup$ – zeraoulia rafik Sep 16 '16 at 20:10

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