I am doing a long physics calculation and have arrived at the following sum:
$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right).$$
Does this have an analytic closed-form solution? An added complication is that in my problem, $a$ can take complex values, and the square root should be treated as having a branch cut along $(-\infty,0]$ (and I am only interested in the real part).
Note that a simpler version of my problem is the case where $b=0$, for which the above becomes
$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+a}\right),$$
which can be written in closed form as $\gamma+\psi(a+1)$, where $\gamma$ is the Euler-Mascheroni constant and $\psi$ is the digamma function (this comes from a standard series representation of the digamma function). So my problem might be viewed as a generalized form the digamma series representation.
Mathematica can't solve this, and I haven't seen integrands with square roots like this in any standard formulas I've looked at. Maybe there's no closed-form solution in terms of standard functions? This wouldn't surprise me since the square root seems to make things pretty messy. But on the other hand there are a lot of special functions and tricks I don't know about...
I don't know this might be useful or not, anyway by generating function for Legendre polynomials $$\dfrac{1}{\sqrt{(n+a)^2+b^2}}=\dfrac{1}{n}\dfrac{1}{\sqrt{1+2\frac{a}{n}+\left(\frac{\sqrt{(a^2+b^2}}{n}\right)^2}}=\dfrac{1}{n}\sum_{k=0}^\infty \left(\frac{\sqrt{a^2+b^2}}{n}\right)^kP_k\left(\frac{-a}{\sqrt{a^2+b^2}}\right)$$ then $$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right)= \color{blue}{-\sum_{k=1}^\infty \left(a^2+b^2\right)^{\frac{k}{2}}\zeta(k)P_k\left(\frac{-a}{\sqrt{a^2+b^2}}\right)}$$