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?
 A: $\frac{1}{k+ia}=\frac{k-ia}{k^2+a^2}$ $(k \in \mathbb{N})$ and $k \mapsto \frac{k}{k^2+a^2}$ is eventually decreasing to $0$. Thus
$$
\sum_{k=1}^\infty (-1)^{k+1} \frac{k}{k^2+a^2}
$$
is convergent by Leibniz and
$$
\sum_{k=1}^\infty (-1)^{k+1} \frac{a}{k^2+a^2}
$$
is absolute convergent (quite clear). Thus $\sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k+ia}$ is convergent and
$$
\sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k+ia} = \sum_{k=1}^\infty (-1)^{k+1} \frac{k}{k^2+a^2} - i\sum_{k=1}^\infty (-1)^{k+1} \frac{a}{k^2+a^2}.
$$
A: Let $a_k=(-1)^{k+1}$ and $b_k = \displaystyle{\frac{1}{k+ia}}$. Let $A_n = \sum_{k=1}^n a_k$. We do a summation by parts :
\begin{align*}
\sum_{k=1}^n (-1)^{k+1}\frac{1}{k+ia}& = \sum_{k=1}^n a_kb_k \\
&= a_1b_1 + \sum_{k=2}^n (A_k-A_{k-1})b_k\\
&=a_1b_1+\sum_{k=2}^n A_kb_k - \sum_{k=1}^{n-1} A_kb_{k+1}\\
&=a_1b_1 + A_nb_n + \sum_{k=2}^{n-1} A_k(b_k-b_{k+1})
\end{align*}
Now let's notice that
$\bullet\ $ $(A_nb_n)_{n \geq 1}$ tends to $0$ because $(A_n)_{n \geq 1}$ is bounded and $(b_n)_{n \geq 1}$ tends to $0$.\
$\bullet\ $ One has
$$ \left|A_k(b_k-b_{k+1})\right| \leq 2\left| \frac{1}{k+ia} - \frac{1}{k+1+ia}\right| \leq 2 \left| \frac{1}{(k+ia)(k+1+ia)}\right| = o \left( \frac{1}{k^{3/2}}\right)$$
Hence the series $\sum A_k(b_k-b_{k+1})$ is absolutely convergent, hence convergent, and you are done.
A: 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 $a \in \mathbb{R}$.
