Conjecture $\sum_{m=1}^\infty\frac{y_{n+1,m}y_{n,k}}{[y_{n+1,m}-y_{n,k}]^3}\overset{?}=\frac{n+1}{8}$, where $y_{n,k}=(\text{BesselJZero[n,k]})^2$ While solving a quantum mechanics problem using perturbation theory I encountered the following sum
$$
S_{0,1}=\sum_{m=1}^\infty\frac{y_{1,m}y_{0,1}}{[y_{1,m}-y_{0,1}]^3},
$$
where $y_{n,k}=\left(\text{BesselJZero[n,k]}\right)^2$ is square of the $k$-th zero of Bessel function $J_n$ of the first kind. 
Numerical calculation using Mathematica showed that $S_{0,1}\approx 0.1250000$. Although I couldn't verify this with higher precision I found some other cases where analogous sums are close to rational numbers. Specifically, after some experimentation I found that the sums
$$
S_{n,k}=\sum_{m=1}^\infty\frac{y_{n+1,m}y_{n,k}}{[y_{n+1,m}-y_{n,k}]^3}
$$
are independent of $k$ and have rational values for integer $n$, and made the following conjecture

$\bf{Conjecture:}\ $ for $k=1,2,3,...$ and  arbitrary $n\geq 0$
  $$\sum_{m=1}^\infty\frac{y_{n+1,m}y_{n,k}}{[y_{n+1,m}-y_{n,k}]^3}\overset{?}=\frac{n+1}{8},\\ \text{where}\  y_{n,k}=\left(\text{BesselJZero[n,k]}\right)^2. 
$$

How one can prove it?
It seems this conjecture is correct also for negative values of $n$. For example for $n=-\frac{1}{2}$ one has $y_{\frac{1}{2},m}=\pi^2 m^2$, $y_{-\frac{1}{2},k}=\pi^2 \left(k-\frac{1}{2}\right)^2$ and the conjecture becomes (see Claude Leibovici's answer for more details)
$$
\sum_{m=1}^\infty\frac{m^2\left(k-\frac{1}{2}\right)^2}{\left(m^2-\left(k-\frac{1}{2}\right)^2\right)^3}=\frac{\pi^2}{16}.
$$
 A: There is a rather neat proof of this.
First, note that there is already an analogue for this:
DLMF §10.21 says that a Rayleigh
function $\sigma_n(\nu)$ is defined as a similar power series
$$ \sigma_n(\nu) = \sum_{m\geq1} y_{\nu, m}^{-n}. $$
It links to http://arxiv.org/abs/math/9910128v1 among others as an example of how
to evaluate such things.
In your case, call $\zeta_m = y_{\nu,m}$ and $z=y_{\nu-1,k}$ ($\nu$ is $n$ shifted by $1$), so that after
expanding in partial fractions your sum is
$$ \sum_{m\geq1} \frac{\zeta_m z}{(\zeta_m-z)^3} = \sum_{m\geq1}
\frac{z^2}{(\zeta_m-z)^3} + \frac{z}{(\zeta_m-z)^2}. $$
Introduce the function
$$ y_\nu(z) = z^{-\nu/2}J_\nu(z^{1/2}). $$
By DLMF 10.6.5 its derivative
satisfies the two relations
$$\begin{aligned}
 y'_\nu(z) &= (2z)^{-1} y_{\nu-1}(z) - \nu z^{-1} y_\nu(z) 
\\&=
-\tfrac12 y_{\nu+1}(z).
\end{aligned} $$
It also has the infinite product
expansion
$$ y_\nu(z) = \frac{1}{2^\nu\nu!}\prod_{k\geq1}(1 - z/\zeta_k). $$
Therefore, each partial sum of $(\zeta_k-z)^{-s}$, $s\geq1$ can be evaluated in
terms of derivatives of $y_\nu$:
$$ \sum_{k\geq1}(\zeta_k-z)^{-s} = \frac{-1}{(s-1)!}\frac{d^s}{dz^s}\log
y_\nu(z). $$
When evaluating this logarithmic derivative, the derivative $y'_\nu$
can be expressed in terms of $y_{\nu-1}$, going down in $\nu$, but the derivative
$y'_{\nu-1}$ can be expressed in terms of $y_\nu$ using the other
relation that goes up in the index $\nu$. So even higher-order derivatives contain only $y_\nu$ and $y_{\nu-1}$.
I calculated your sum using this procedure with a CAS as:
$$ -\tfrac12z^2(\log y)''' -z(\log y)''
= \tfrac18\nu + z^{-1} P\big(y_{\nu-1}(z)/y_\nu(z)\big), $$
where $P$ is the polynomial
$$ P(q) = -\tfrac18 q^3 + (\tfrac38\nu-\tfrac18) q^2 + (-\tfrac14\nu^2
+ \tfrac14\nu - \tfrac18)q. $$
When $z$ is chosen to be any root of $y_{\nu-1}$,
$z=\mathsf{BesselJZero}[\nu-1, k]\hat{}2$, $P(q)=0$, your sum is equal
to
$$ \frac{\nu}{8}, $$
which is $(n+1)/8$ in your notation.
It is possible to derive a number of such closed forms for sums of
this type. For example, by differentiating $\log y$ differently
(going $\nu\to\nu+1\to\nu$), one would get
$$ \sum_{m\geq1}
\frac{y_{\nu,m}y_{\nu+1,k}}{(y_{\nu,m}-y_{\nu+1,k})^3} =
-\frac{\nu}{8}. $$
Some other examples, for which the r.h.s. is independent of $z$ ($\zeta_m=y_{\nu,m}, z=y_{\nu-1,l}$, $l$ arbitrary):
$$ \begin{gathered}
\sum_{k\geq1} \frac{\zeta_k}{(\zeta_k-z)^2} = \frac14,\\
\sum_{k\geq1} \frac{z^2}{(\zeta_k-z)^4} - \frac{1}{(\zeta_k-z)^2} + \frac1{24}\frac{5-\nu}{\zeta_k-z} = \frac{1}{48}, \\
\sum_{k\geq1} \frac{\zeta_k}{(\zeta_k-z)^4} + \frac1{96}\frac{z-\zeta_k-8+4\nu}{(\zeta_k-z)^2} = 0. 
\end{gathered} $$
or with $z=y_{\nu+1,l}$, $l$ arbitrary:
$$ \begin{gathered}
\sum_{k\geq1} \frac{z^2}{(\zeta_k-z)^3} = -\tfrac18\nu-\tfrac14,
\end{gathered} $$
and they get messier with higher degrees.
A: This is a partial answer concerning the last conjecture.
If no mistake : considering the partial fraction decomposition of $$A_m=\frac{m^2 x^2}{\left(m^2-x^2\right)^3}$$ $$A_m =\frac{m}{8 (m-x)^3}+\frac{m}{8 (m+x)^3}-\frac{1}{16 (m-x)^2}-\frac{1}{16
   (m+x)^2}-$$ $$\frac{1}{16 m (m-x)}-\frac{1}{16 m (m+x)}$$ and then $$S(x)=\sum_{m=1}^\infty A_m=\frac \pi{16x}\Big(\cot (\pi  x)+\pi  x \csc ^2(\pi  x)-2 \pi ^2 x^2 \cot (\pi  x) \csc ^2(\pi  x)\Big)$$ Then, if $k$ is an integer, $$S(k-\frac 12)=\frac {\pi^2} {16}$$
You could be interested by this paper.
A: Not an answer yet.  But the references cited for that Wikipedia page may let you get the answer.
Begin with formula
$$
J_\alpha(z) = \frac{(z/2)^\alpha}{\Gamma(\alpha+1)}
\prod_{n=1}^\infty\left(1-\frac{z^2}{j_{\alpha,n}^2}\right)
$$
Source: Wikipedia  .  Here $j_{\alpha,n}$ is the $n$th positive zero of $J_\alpha$.
Take logarithmic derivative to get
$$
\frac{\alpha}{z} - \frac{J_{\alpha+1}(z)}{J_\alpha(z)} 
= \frac{\alpha}{z} + \sum_{n=1}^\infty\frac{2z}{z^2-j_{\alpha,n}^2}
$$
Replace $\alpha$ by $-\alpha$ and simplify
$$
\frac{J_{\alpha-1}(z)}{J_\alpha(z)} = 2\sum_{n=1}^\infty\frac{z}{j_{\alpha,n}^2-z^2}
$$
Plug in $z=j_{\alpha-1,k}$ a zero of $J_{\alpha-1}$:
$$
0 = 2 \sum_{n=1}^\infty \frac{j_{\alpha-1,k}}{j_{\alpha,n}^2-j_{\alpha-1,k}^2}
\\
0 = \sum_{n=1}^\infty \frac{1}{j_{\alpha,n}^2-j_{\alpha-1,k}^2}
$$
We want something like this with the third power in the denominator.
