In a (side) comment to this question it is claimed that for a certain domain the following holds:
$$-\frac{\zeta'(s)}{\zeta(s)}= \sum_{n=2}^\infty \frac{\psi(n+1/2)-\psi(n-1/2)} {n^{s}} = \sum_{n=2}^\infty \left(1-\sum_\rho \frac{(n+1/2)^\rho-(n-1/2)^\rho}{\rho\, n^{s}}\right)$$
The $\rho$'s are both the trivial and (pairs of) the non-trivial zeros of $\zeta(s)$.
When I isolate the series for the trivial zeros (at $-2k$) from the RHS, I get to:
$$\sum_{k=2}^{\infty}\frac{\frac12\ln\left(2k+3\right)-\frac12\ln\left(2k-3\right)+\frac32\ln\left(2k-1\right)-\frac32\ln\left(2k+1\right)}{k^s}$$
which is equal to: $$f(s)=\sum_{k=2}^{\infty}\frac{\coth^{-1}\left(\frac23\,k\right)-3\coth^{-1}\left(2\,k\right)}{k^s}$$
and seems to converge for all $\Re(s)>-1$.
Could there exist a closed form for this series (e.g. in terms of $\zeta'(s)\,$)?
P.S.:
Numerical evidence suggests that all complex zeros of $f(s) \pm f(1-s)$ reside on the line $\Re(s)=\frac12$ (for the domain $-1 < \Re(s) < 2$), however the distribution of the imaginary parts on the line itself seems quite irregular.