Bessel Series as $-\sum _{k=1}^{\infty } \frac{2 (-1)^n}{\pi ^4 k^4 J_0(k \pi )}$ Is it possible to check the equality
$$\sum _{k=1}^{\infty } \frac{2 \csc \left(j_{0,k}\right)}{\left(j_{0,k}\right){}^4 J_1\left(j_{0,k}\right)}-\sum _{k=1}^{\infty } \frac{2 (-1)^n}{\pi ^4 k^4 J_0(k \pi )}=\frac{311}{2880}$$ where $j_0(k)$ is a zero of  $J(z)$, the BesselJ function.
Furthermore, is the following equality true?
$$\sum _{k=1}^{\infty } \frac{2 \csc \left(j_{0,k}\right)}{\left(j_{0,k}\right){}^6 J_1\left(j_{0,k}\right)}-\sum _{k=1}^{\infty } \frac{2 (-1)^n}{\pi ^6 k^6 J_0(k \pi )}=\frac{5557}{241920}$$
 A: "Obviously" this comes from looking at the residues of the function
$$ \frac{\csc{z}}{z^n J_0(z)} $$
for $n=4,6$ over some large contour. It is reasonably easy to check that
$$ \int_{C} \frac{\csc{z}}{z^n J_0(z)} \, dz = \text{Res}_{z=0} + \sum_{k=1}^p  -\frac{2\csc{(j_{0,k})}}{(j_{0,k})^n J_1(j_{0,k})} + \sum_{k=1}^{m} \frac{2(-1)^{k}}{(k\pi)^nJ_0(k\pi)}  $$
for a contour $C$ enclosing the multiples of $\pi$ up to $\pm m \pi$ and the first $p$ of the $\pm j_{0,k}$. Thus it suffices to show that the residues at $0$ are as stated, and the integral over the contour can be made to tend to zero. The first is easy enough using the series expansions of $\csc$ and $J_0$, so the issue is to choose an appropriate sequence of contours so that we can show the left-hand side $\to 0$ as we include more and more points. An obvious choice is a sequence of circles (or squares) so that the vertical sides cross the real line as far from the neighbouring poles as possible. Thankfully, the zeros of $J_0$ are asymptotic to $(m-1/4)\pi$, so it's easy to find a contour does this. Also, $\lvert \csc{z}/J_0(z) \rvert \sim K\lvert z \rvert^{1/2}e^{-2\lvert \Im(z) \rvert}$ when $\lvert z \rvert \to \infty$ for $\arg{z}$ bounded away from $0$ and $\pi$, and we can bound nearer the real line using periodicity of $\csc$ and going through the real line near enough the maxima of the Bessel function where it $\sim K' \lvert z \rvert^{-1/2}$, so we can use a modification of Jordan's lemma to bound the integral over the circle (or tackle the side of the square separately, but this is difficult with the Bessel function present). The result follows.
