Is the series $\sum_{k=1}^\infty\zeta(1/2,k)-\zeta(1/2,k+x)$ convergent?

Question

Consider the series $$\sum_{k=1}^\infty\zeta(1/2,k)-\zeta(1/2,k+x)$$ where $\zeta(s,k)$ is the Hurwitz zeta function.

Is the series convergent point wise or uniformly on $(-1, +\infty)$?

Does the series have a special closed form?

Answer 1

We have: $$ \zeta\left(\frac{1}{2},k\right)=\zeta\left(\frac{1}{2}\right)+\sum_{n\geq 0}\left[\frac{1}{\sqrt{n+k}}-\frac{1}{\sqrt{n+1}}\right]\tag{1} $$ that through the inverse Laplace transform turns into the following integral representation: $$ \zeta\left(\frac{1}{2},k\right)=\zeta\left(\frac{1}{2}\right)+\int_{0}^{+\infty}\frac{e^{(1-k)s}-1}{\sqrt{\pi s}(e^s-1)}\,ds \tag{2}$$ In particular: $$\begin{eqnarray*} \sum_{k\geq 1}\left[\zeta\left(\frac{1}{2},k\right)-\zeta\left(\frac{1}{2},k+x\right)\right]&=&\int_{0}^{+\infty}\sum_{k\geq 1}\frac{e^{(1-k)s}-e^{(1-k-x)s}}{\sqrt{\pi s}(e^s-1)}\,ds\\&=&\int_{0}^{+\infty}\frac{1-e^{-sx}}{4\sqrt{\pi s}\sinh^2\left(\frac{s}{2}\right)}\,ds\end{eqnarray*}\tag{3}$$ is divergent due to the behaviour of the integrand function in a right neighbourhood of the origin, $\frac{x}{\sqrt{\pi s^3}}$, leading to a non-integrable singularity. That also follows from $$\zeta\left(\frac{1}{2},k\right)-\zeta\left(\frac{1}{2},k+x\right)\sim\frac{x}{\sqrt{k}}\tag{4}$$ for large values of $k$. $\left\{\frac{1}{\sqrt{k}}\right\}_{k\geq 1}$ is not a summable sequence.