Evaluating an infinite sum involving possibly hypergeometric terms I was considering the following infinite sum
$$ A(n) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+n-1}}{k^n}(0k -1)(k-1)(2k-1)...((n-1)k-1) \right] $$ 
Some cases:
$$ A(1) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+1}}{k} \right]= -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} ... = \ln(2) - 1 $$
A trickier sum to evaluate is: $A(2)$. This resolves to: 
$$ A(2) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+2}}{k^2} (k-1)\right] = \frac{1}{4} - \frac{2}{9} + \frac{3}{16} ... $$
Some jugglery is required but if we consider 
$$ 1 - x + x^2 -x^3 ... = \frac{1}{1+x}$$
And integrate
$$ \ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + ... $$ 
Divide by x and integrate
$$ \int_{0}^{1} \frac{\ln(1+x)}{x} dx   = x - \frac{1}{4}x^2 + \frac{1}{9} x^3 ... $$ 
Now multiply by (-1), divide by x and differentiate
$$ - \left( \frac{1}{x} \int_{0}^{1} \frac{\ln(1+x)}{x} dx \right)' = \frac{1}{4} - \frac{2}{9} x + \frac{3}{16} x^2 ... $$ 
Meaning 
$$ \left( \frac{1}{x^2} \int_{0}^{1} \frac{\ln(1+x)}{x} dx  - \frac{1}{x}  \frac{\ln(1+x)}{x} \right)_{@x = 1} = \frac{1}{4} - \frac{2}{9} + \frac{3}{16} ...  $$ 
Through very mysterious techniques of the omniscient Wolfram Alpha this evaluates to 
$$ \frac{\pi^2}{12} - \ln(2) = \frac{1}{4} - \frac{2}{9} + \frac{3}{16} ...  $$ 
So now this is totally in the deep end, but I have a suspicion the sum I have given can somehow be related to the riemann zeta function, in some good quality closed form. It's a wild guess but the $\ln(2)$ and now the $\pi^2$ hint me that's a good idea. 
How to show this? 

In case it's helpful
I tried to bash this with Sister Celine's method. By writing as 
$$ A(n) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+n-1}}{k^n}(0k -1)(k-1)(2k-1)...((n-1)k-1) \right] = \sum_{k=2}[ M(n,k) ]  $$
And tried to find a recurrence for $M$ but unfortunately after consider $a_0(n) M(n,k) + a_1(n) M(n+1,k) + a_2(n) M(n-1,k) $ it became clear I'm going to need some more terms (possibly a lot more) before I can find a closed recurrence. Unfortunately my maple package has expired (I did send them an email to renew!) so I couldn't throw the might of Shalosh B. Ekhad in the Zeilberger package to close it up. 
I suspect though that this would be ample for those techniques.
 A: I do not know if what follows is of some help, but:
Note that $\prod_{j=0}^{n-1}(jk-1)=P_n(k)$, is a polynomial in $k$. Put $P_n(x)=a_0+a_1x+\cdots a_{n-1}x^{n-1}$, then your sum is
$$A(n)=(-1)^{n-1}\sum_{j=0}^{n-1}a_j(\sum_{k\geq 2}\frac{(-1)^k}{k^{n-j}})$$
and $\sum_{k\geq 1}\frac{(-1)^k}{k^{s}}=\eta(s)=(1-2^{1-s})\zeta(s)$
See https://en.wikipedia.org/wiki/Riemann_zeta_function#Generalizations, ( in "the functional equation" part)
The case $j=n-1$ gives you the $\log(2)$ term, up to a constant due to the $k\geq 2$.
A: The trick should be to rewrite (for $n>0$)
\begin{align}
A(n) &= \sum_{k=2}^{\infty}\frac{(-1)^{k+n-1}}{k^n}(0k -1)(k-1)(2k-1)...((n-1)k-1)\\
&= \sum_{k=2}^{\infty}(-1)^{k-1}\left(\frac 1k\right)\left(\frac 1k-1\right)...\left(\frac 1k-(n-1)\right)\\
&= \sum_{k=2}^{\infty}(-1)^{k-1}\left(\frac 1k\right)_n\\
\end{align}
where $(x)_n$ is the Pochhammer symbol ("falling factorial") and a generating function for the signed Stirling numbers of the first kind (the term $\,i=0\,$ is omitted since $\;s(n,0)=0\,$ for $n>0$) :
$$(x)_n=\sum_{i=1}^n s(n,i)\,x^i$$
this gives (for $\,\eta\,$ the Dirichlet eta function or "alternating zeta function") :
\begin{align}
A(n) &= \sum_{k=2}^{\infty}(-1)^{k-1}\sum_{i=1}^n s(n,i)\,\frac 1{k^i}\\
&= \sum_{i=1}^n s(n,i)\,\sum_{k=2}^{\infty}(-1)^{k-1} \frac 1{k^i}\\
&= \sum_{i=1}^n s(n,i)\,(\eta(i)-1)\quad\text{but}\quad  \sum_{i=1}^n s(n,i)=0\quad\text{for}\;n>1\\
&= \sum_{i=1}^n s(n,i)\,\eta(i)\quad\text{for}\;n>1\quad\text{and}\quad s(1,1)(\eta(1)-1)\quad\text{for}\;n=1\\
\\
A(n)&= \begin{cases}s(n,1)\,\log(2)+\sum_{i=2}^n s(n,i)\left(1-2^{1-i}\right)\zeta(i),&n>1\\
\log(2)-1,&n=1\\
\end{cases}
\end{align}
Since $\;s(n,1)=(-1)^{n-1}(n-1)!\;$ these expressions are in perfect agreement with the table exposed by Claude Leibovici. 
