<s>[INST] Closed form of $\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$ I'm new in the area of the series involving Bessel function of the first kind. What are
the usual tools you would recommend me for computing such a series? Thanks.
$$\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$$
 [/INST] Here is a route.
Recall that
$$
J_0(2n)=\frac 1\pi \int_0^\pi \cos (2n \sin x)\:{\rm d}x \tag1
$$
and that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^2}=\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4},\quad 0\leq t\leq 2\pi. \tag2 $$
Then, due to normal convergence of the series on $[0,2\pi] $, we may write
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0(2n)}{n^2} & =\frac 1\pi \int_0^\pi \sum_{n=1}^\infty\frac{\cos (2n \sin x)}{n^2}{\rm d}x \\\\
& =\frac 1\pi \int_0^\pi\left(\frac{\pi^2}{6}-\pi \sin x+\sin^2 x\right){\rm d}x \\\\
& =\frac{\pi^2}{6}-\frac{3}{2}\\\\
& =0.144934066848226436...
\end{align}
$$
thus

$$
\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2} = \frac{\pi^2}{6}-\frac{3}{2}.
$$

Using a similar technique, one may obtain the following result.

$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0(2 n \alpha )}{n^2}  &=\frac{\pi^2}{6}-2\alpha+\frac{\alpha^2}{2},\qquad \alpha \in [0,\pi).\tag3
\end{align}
$$

</s>