TITLE: What is the closed form of $\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(n\right)} ^2}{{n}^4}}$?
QUESTION [8 upvotes]: Using Maple I am obtaining the numerical approximation
$$0.5902373619$$
Please, let me know what is the closed form.  Many thanks.

REPLY [4 votes]: Using the wonderful method of Olivier implemented by Maple it is possible to prove that
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)} ^{2}}{{n}^{4}}}=-{\frac {3
}{64}}\,{\alpha}^{4}+{\frac {1}{90}}\,{\pi }^{4}-{\frac {1}{12}}\,{\alpha}^{2}{
\pi }^{2}+{\frac {32}{27}}\,{\frac {{\alpha}^{3}}{\pi }}
$$
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)}  ^{2}}{{n}^{2}}}=\frac{1}{4}\,{
\alpha}^{2}+\frac{1}{6}\,{\pi }^{2}-{\frac {4\alpha}{\pi }}
$$
For this last sum, when $\alpha =2 $ we obtain
$$\sum _{n=1}^{\infty }{\frac {  {{\it J}_{0}\left(\,2\,n\right)}
  ^{2}}{{n}^{2}}}=1+\frac{1}{6}\,{\pi }^{2}-8\,{\pi }^{-1}
$$
that is justly the result derived  here.
Other results:
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)}  ^{2}}{{n}^{6}}}={\frac {1}
{945}}\,{\pi }^{6}+{\frac {1}{64}}\,{\pi }^{2}{\alpha}^{4}-{\frac {1}{
180}}\,{\pi }^{4}{\alpha}^{2}+{\frac {5}{1152}}\,{\alpha}^{6}-{\frac {
512}{3375}}\,{\frac {{\alpha}^{5}}{\pi }}
$$
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)}  ^{2}}{{n}^{8}}}=-{\frac {{\pi }^{6}{\alpha}^{2}
}{1890}}\,+{\frac {1}{960}}\,{\pi }^{4}{\alpha}^
{4}-{\frac {35}{147456}}\,{\alpha}^{8}-{\frac {5}{3456}}\,{\pi }^{2}{
\alpha}^{6}+{\frac {1}{9450}}\,{\pi }^{8}+{\frac {4096}{385875}}\,{
\frac {{\alpha}^{7}}{\pi }}
$$