I came across this series when solving a time-dependent Poisson's equation using series expansion
$$ \sum_{k=1}^\infty \frac{1}{\alpha_k J_1(\alpha_k)}=\frac{1}{2} $$ where $\alpha_k$ is the $k$-th root of $J_0(x)$. I can confirm this numerically, but I want to prove this. Any help is appreciated. Thanks a lot.
Integrate $g(z)=\frac{1}{2 \pi i z J_0(z)}$ over a large rectangle with zero as the origin. You might use the fact that $J_0(z)\sim e^{\mp iz}/\sqrt{z}$ as $\Im(z)\rightarrow\pm\infty$ to show that the contributions of the contour vanish in the limit of an infinite rectangle. This means that the the sum of all residues is just $0$. $$ \oint g(z)=0=\text{res}(g(z=0))+2\sum_{k=1}\text{res}(g(z=\alpha_k)) $$ Note that the factor 2 above is due to the symmetry of Bessel zeros $J_0(x)=0\rightarrow J_0(-x)=0$ and that all poles are simple.
The calculation of the resiudes is easy if one knows the fact that $J_0'(z)=-J_1(z)$ and your result follows: $$ \frac12=\sum_{k=1}\frac1{\alpha_k J_1(\alpha_k)} $$
Approach cleary works also for higher order Bessels...