Shifted alternating harmonic series

Question

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?

Answer 1

$\frac{1}{k+ia}=\frac{k-ia}{k^2+a^2}$ $(k \in \mathbb{N})$ 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 $$ \sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k+ia} = \sum_{k=1}^\infty (-1)^{k+1} \frac{k}{k^2+a^2} - i\sum_{k=1}^\infty (-1)^{k+1} \frac{a}{k^2+a^2}. $$

Answer 2

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.

Answer 3

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 $a \in \mathbb{R}$.