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?

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.
