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

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

• 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. – Gerald Edgar Dec 26 '15 at 21:11
• but how do i show that is hasn't no closed form , is by mathematica ? – zeraoulia rafik Dec 26 '15 at 21:24
• 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. – Richard Stanley Dec 27 '15 at 0:36
• @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. – Gerald Edgar Dec 27 '15 at 1:21
• Hint for convergence with complex $s$: investigate absolute convergence. – Gerald Edgar Dec 27 '15 at 14:53

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.
• @zeraouliarafik I believe there is no closed form, and the notation means hypergeometric pFq (where $p=q=k,$ in this case). – Igor Rivin Dec 26 '15 at 21:29