# Can this infinite sum be solved analytically?

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...

• can you do the case $a=0$? – mathworker21 Jun 23 '19 at 5:57
• No I can't, and neither can Mathematica – WillG Jun 23 '19 at 5:58
• Do you want an approximation for small or large value of parameters? – Nosrati Jun 23 '19 at 7:58
• I think that that does not have finite closed-form expression in terms of very large class of special functions. Why do you want closed-form expression? – Grešnik Jun 23 '19 at 9:18
• $$\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right)=\int_0^{\infty } \left(-\frac{1}{1-e^x}+\frac{e^{-a x} J_0(b x)}{1-e^x}\right) \, dx$$ no hope... – Mariusz Iwaniuk Jun 23 '19 at 11:31

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 $$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right)= \color{blue}{-\sum_{k=1}^\infty \left(a^2+b^2\right)^{\frac{k}{2}}\zeta(k)P_k\left(\frac{-a}{\sqrt{a^2+b^2}}\right)}$$