Resummation of the Series I wonder if there is a way to get resummation of this series? By this way , i am trying to get the integral representation of this series, it could be by Gamma function.
$$\sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ]$$
thank you.
 A: One can start with the Euler asymptotic expansion
$$
f(z)=\int_0^\infty\frac{e^{-t}}{z+t}\,dt=\sum_{k=0}^\infty(-1)^k\frac{{k!}}{z^{k+1}}.
$$
Putting 
$$
g(z)=\frac{f(-iz)-f(iz)}{2i}=
\sum _{k=0}^{\infty} (-1)^k\frac{(2k)!}{z^{2k+1}}
$$
we have formally
$$
g((i+a)^{-1})+g((-i+a)^{-1})=
\sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ].
$$
A: It looks like Andrew beat me to it
There is no guarantee that the following makes sense... But...  
There is this Borel summation
$$
\sum_{j = 0}^{\infty} \frac{(-1)^{j} j!}{x^{j + 1
}} = -\operatorname{e} ^{x} \mathrm{Ei} (-x)
\tag{1}
$$
Here, $\mathrm{Ei}$ is the exponential integral.  Put $-x$ for $x$,
$$
\sum_{j = 0}^{\infty} \frac{j!}{x^{j + 1}} = \operatorname{e} ^{-x} \mathrm{Ei} (x)
\tag{2}
$$
Add (1) and (2)
$$
\sum_{k = 0}^{\infty} \frac{(2 k)!}{x^{2 k + 1}} = -\operatorname{e} ^{x} \mathrm{Ei} (-x) + \operatorname{e} ^{-x} \mathrm{Ei} (x)
$$
Put $x=iz$,
$$
\sum_{k = 0}^{\infty} \frac{(-1)^{k} (2 k)!}{z^{(2 k + 1)}} = 
-i\operatorname{e} ^{i z} \mathrm{Ei} (-iz) + i \operatorname{e} ^{-iz} \mathrm{Ei} (i z)
$$
Put $z=a+i$ and $z=a-i$ and subtract:
$$\begin{align}
&\sum_{k = 0}^{\infty} (-1)^{k} (2 k)! \Bigl((a + i)^{(-2k - 1)} - (a - i)^{(-2k - 1)}\Bigr) = \\
&\qquad -i\operatorname{e} ^{i a - 1} \mathrm{Ei} (-ia + 1) + i \operatorname{e} ^{-ia + 1} \mathrm{Ei} (i a - 1) + 
i \operatorname{e} ^{i a + 1} \mathrm{Ei} (-ia - 1) - i \operatorname{e} ^{-ia - 1} \mathrm{Ei} (i a + 1)
\end{align}$$
