What is the closed form of $\sum_{l=1}^\infty \left(1 -\frac{l}{\sqrt{l^2 + a^2}}\right) $?

Question

How can the following infinite series be evaluated to a closed form?:

$$ \sum_{l=1}^\infty \left(1 -\frac{l}{\sqrt{l^2 + a^2}}\right) $$

I tried to evaluate this using Mathematica, but it was not able to find a closed form.

Answer 1

$$ \frac{\ell}{\sqrt{\ell^2+a^2}}=\frac{1}{\sqrt{1+\frac{a^2}{\ell^2}}}=\sum_{n=0}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\ell^{2n}} $$ Thus $$ \sum_{\ell=1}^{+\infty}\left(1-\frac{\ell}{\sqrt{\ell^2+a^2}}\right)=\sum_{\ell=1}^{+\infty}\sum_{n=1}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\ell^{2n}}=\sum_{n=1}^{+\infty}\sum_{\ell=1}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\ell^{2n}}=\sum_{n=1}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\zeta(2n)} $$ The convergence is way faster when $a\in(-1,1)$.