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:
$$\begin{align}
&\sum_{k = 0}^{\infty} (-1)^{k} (2 k)! \Bigl((a + i)^{(-2k - 1)} - (a - i)^{(-2k - 1)}\Bigr) = \\
&\qquad -i\operatorname{e} ^{i a - 1} \mathrm{Ei} (-ia + 1) + i \operatorname{e} ^{-ia + 1} \mathrm{Ei} (i a - 1) +
i \operatorname{e} ^{i a + 1} \mathrm{Ei} (-ia - 1) - i \operatorname{e} ^{-ia - 1} \mathrm{Ei} (i a + 1)
\end{align}$$