Shifted alternating harmonic series Suppose I have the following series
$$ \sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k+ia}, \hspace{1cm}$$
with $i$ the imaginary unit and $a \in \mathbb{R}$. Does this shifted sum converge?

It seems that you have at least two solutions.
The first one is to expand the complex number to write
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k+ia}=-\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2+a^2}+i a\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2+a^2} k$$ The first sum is given in terms of the digamma function
$$-\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2+a^2}=\frac{1}{4} \left(\psi \left(-\frac{i
   a}{2}\right)+\psi\left(\frac{i a}{2}\right)-\psi \left(\frac{1-ia}{2}\right)-\psi
   \left(\frac{1+i a}{2}\right)\right)$$ and, for the second one
$$\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2+a^2} k=\frac{\pi  a\,\text{csch}(\pi  a)-1}{2 a^2}$$
There is more compact result using special functions since
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k+ia}=\Phi (-1,1,1+i a)$$ where appears  the Hurwitz-Lerch transcendent function.
The result exists for any $a \in \mathbb{R}$.
