Summing an alternating series of type $(-1)^{k-1} \frac1{1 + ak^{-1}}$ I have the following alternating series:
$$
   \sum_{k=1}^{+\infty} (-1)^{k-1} \dfrac{1}{1 + \alpha{}k^{-1}}
$$
with $\alpha<1$, and I'm trying to evaluate the sum. This series arose from a previous question I posted yesterday. Without citing too much from that post, the original expression is bounded, so I think this sum should also be bounded. It seems likely, due to the small coefficient and the inverse power.
I have looked around for some hints on how to proceed, but I didn't find anything useful.
Anything I can use to evaluate this?
Thanks,
 A: As stated by @RobertZ, this sum does not converge. However, one can somehow assign a principal value for it.
Note that $$\frac1{1+ak^{-1}}=1-\frac{a}{k+a}$$
If one accepts that $$\sum^\infty_{k=1} (-1)^k=0$$
it is easy to see your sum equals 
$$\color{RED}{a\cdot\Phi(-1,1,a)-1}$$ where $\Phi$ is the Lerch transcendent.
$$\Phi(z,s,q)=\sum^\infty_{k=0}\frac{z^k}{(k+q)^s}$$
A: I estimated lower and upper limits of the sum:
$S=\sum\limits_{k=1}^{+\infty} (-1)^{k-1} \dfrac{1}{1 + \alpha{}k^{-1}}$
Separate the sum according to the odd and even value of $k$s then 
$S=\sum\limits_{i=0}^{+\infty}\big( (-1)^{2i} \dfrac{1}{1 + \frac{\alpha}{2i+1}}-(-1)^{2i+1} \dfrac{1}{1 + \frac{\alpha}{2i+2}}\big)$ performing the possible simplifications I got:
$S=\sum\limits_{i=0}^{+\infty}\dfrac{-\alpha}{(2i+2+\alpha)(2i+1+\alpha)}$ then -S was forced to the following limits: 
$\sum\limits_{i=0}^{+\infty}\dfrac{\alpha}{(2i+2+\alpha)^2}\lt -S\lt \sum\limits_{i=0}^{+\infty}\dfrac{\alpha}{(2i+1+\alpha)^2}$
$\frac{\alpha}{4}\sum\limits_{i=0}^{+\infty}\dfrac{1}{(i+\frac{(2+\alpha)}{2})^2}\lt -S\lt \frac{\alpha}{4}\sum\limits_{i=0}^{+\infty}\dfrac{1}{(i+\frac{(1+\alpha)}{2})^2}$
Finally $\frac{\alpha}{4}\zeta(2,\frac{(2+\alpha)}{2})\lt-S \lt \frac{\alpha}{4}\zeta(2,\frac{(1+\alpha)}{2})$ where $\zeta(s,q)$ is the Hurwitz zeta function.
