Sum of series with algebraic function The finite sum of the series given by,
$\sum\limits_{k=1}^{\infty}\frac{k}{(a^2 + k^2)^{3/2}} = ?$
The series converges. Mathematica does not give any results. I tried searching extensively in books like Gradshteyn & Ryzhik (2007), Jolley (1961), but got no answer. Closest I got to was for indefinite integral of the function, but doesn't serve the purpose. Thanks a lot for any pointers to solve this series.
 A: Assuming $a>0$, you may use Poisson summation formula to convert
$$ \frac{1}{2}\sum_{k\in\mathbb{Z}}\frac{|k|}{(a^2+k^2)^{3/2}} $$
into
$$ \frac{1}{2}\sum_{k\in\mathbb{Z}}\left[\frac{2}{a}+2\pi^2 k L_0(2\pi a k)-2\pi^2|k|I_0(2\pi a k)\right]\\=\frac{1}{a}+\frac{2\pi}{a}\sum_{k\geq 1}\underbrace{\left[2\pi ak L_0(2\pi a k)-2\pi a k I_0(2\pi a k)+\frac{2}{\pi}\right]}_{\sim -\frac{1}{2\pi^3 a^2 k^2}\text{ as } k\to +\infty}$$
where $I_0$ and $L_0$ are modified Bessel/Struve functions of the first kind,
$$ I_0(z)=\sum_{n\geq 0}\frac{(z/2)^{2n}}{\Gamma(n+1)^2},\qquad L_0(z)=\frac{z}{2}\sum_{n\geq 0}\frac{(z/2)^{2n}}{\Gamma(n+3/2)^2}. $$
However the convergence speed of the original series is not really improved.
For such purpose it is better to employ creative telescoping or just the Maclaurin series of $\frac{x}{(a^2+x^2)^{3/2}}$, which can be computed from the extended binomial theorem. A simple inequality is given by Cauchy-Schwarz:
$$ \sum_{k\geq 1}\frac{k}{(a^2+k^2)^{3/2}}\leq \sqrt{\frac{\pi}{8a^3}\left(\pi a \coth(\pi a)-1\right)\left(\coth(\pi a)-\frac{\pi a}{\sinh^2(\pi a)}\right)}$$
this is pretty accurate when $a\approx 0$.
