What is the closed form of $\sum _{n=1}^{\infty }{\frac { {{\rm J_{0}}\left(2\,n\right)} ^{2}}{{n}^{2}}}$? Using Maple I am obtaining
$$\sum _{n=1}^{\infty }{\frac { {{\rm J_0}\left(2\,n\right)}
  ^{2}}{{n}^{2}}} = 0.09845497463
$$
Please check it.  Many thanks.
 A: Yes, there is a closed form.  Expand $\arcsin(t) \chi_{[-1,1]}(t)$ as a Fourier series
on $[-\pi/2, \pi/2]$ (where $\chi_{[-1,1]}(t)$ is the indicator function of the interval $[-1,1]$.
$$ \arcsin(t) \chi_{[-1,1]} = \sum_{n=1}^\infty \dfrac{J_0(2n) - \cos(2n)}{n} \sin(2nt) $$
Now it seems that
$$ \sum_{n=1}^\infty \dfrac{\cos(2n)}{n} \sin(2nt) = \cases{
-t - \pi/2 & if $-\pi/2 < t < -1$\cr
-t & if $-1 < t < 1$\cr -t + \pi/2 & if $1 < t < \pi/2$\cr
}$$
If $$g(t) = \cases{-t - \pi/2 & if $-\pi/2 < t < -1$\cr
-t + \arcsin(t) & if $-1 < t < 1$\cr -t + \pi/2 & if $1 < t < \pi/2$\cr
}$$
Parseval's theorem says
$$ \eqalign{\sum_{n=1}^\infty \dfrac{J_0(2n)^2}{n^2} &=\dfrac{2}{\pi} \int_{-\pi/2}^{\pi/2} g(t)^2\; dt\cr
&= \dfrac{\pi^2}{6} + 1 - \dfrac{8}{\pi}} $$
EDIT:
Somewhat more generally, if $0 \le s \le \pi/2$ we have
$$\arcsin(t/s) \chi_{[-s,s]}(t) = \sum_{n=1}^\infty \dfrac{J_0(2ns) - \cos(2ns)}{n} \sin(2\pi t)$$
and $$\sum_{n=1}^\infty \dfrac{\cos(2ns)}{n} \sin(2nt) = 
\cases{-t - \pi/2 & if $-\pi/2 < t < -s$\cr
       -t & if $-s < t < s$\cr
       -t + \pi/2 & if $s < t < \pi/2$\cr}$$
so that Parseval says
$$ \sum_{n=1}^\infty \dfrac{J_0(2ns)^2}{n^2} = \dfrac{\pi^2}{6} - 8 \dfrac{s}{\pi} + s^2 $$ 
