Actually we have
$$\begin{aligned}
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
= \frac{\pi}{24}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{7}\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
= \frac{\pi^{7}}{23040}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{13}\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
= \frac{173\pi^{13}}{3832012800}\\
&\sum_{n=1}^{\infty} \frac{(-1)^nn^3}{\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )}
=\frac{1}{16\pi^3}
\end{aligned}$$
and so on.
Proof:
We integrate the functions along a square contour without poles on the edges. We obviously see that these four edges vanish as sides go to complex infinity.
$$\begin{aligned}
&f_1(z)=z^s\cdot\frac{\pi}{\sin(\pi z)} \prod_{k=1}^{m-1}\zeta^k\csc\left ( \zeta^k\pi z \right )\\
&f_2(z)=z^s\cdot\frac{\pi}{\cos(\pi z)} \prod_{k=1}^{m-1}\zeta^k\sec\left ( \zeta^k\pi z \right )
\end{aligned}
$$
Where $\zeta=e^{\pi i/m}$.
Now by the residue theorem,we can get
$$
\begin{aligned}
&\sum_{z_k = \text{All poles of }f_1(z)}^{\infty}
\text{Res} \left ( f_1(z),z_k \right ) = 0\\
&\sum_{z_k = \text{All poles of }f_2(z)}^{\infty}
\text{Res} \left ( f_2(z),z_k \right ) = 0
\end{aligned}
$$
Notice that $f_1(z)$ has poles at $z=(0),n,n\zeta^{-1},n\zeta^{-2}...,n\zeta^{-m+1}$,
$f_2(z)$ has poles at $z=(0),(2n+1)/2,(2n+1)/2\zeta^{-1},(2n+1)/2\zeta^{-2}...,(2n+1)/2\zeta^{-m+1}$ .
We find
$$
\begin{aligned}
&\sum_{j=0}^{m-1}(-1)^j\zeta^{-(s+1)j}
{\sum_{n=-\infty}^{\infty}}^{\prime}
(-1)^nn^s
\prod_{k=1}^{m-1}\zeta^k\csc\left ( \zeta^k\pi n \right )
=-\operatorname{Res}\left(f_1(z),z=0\right)\\
&c\sum_{n=-\infty}^{\infty}
(2n+1)^s \prod_{k=1}^{m-1}\zeta^k\sec\left ( \zeta^k\pi
\frac{2n+1}{2} \right )=2^s\operatorname{Res}\left ( f_2(z),z=0 \right )\\
\end{aligned}
$$
Where $c=\left\{\begin{matrix}
m,& s=mk-1,k\in\mathbb{Z}.\\
0,&\text{otherwise}.
\end{matrix}\right.$.
Let $\zeta=e^{\pi i/2}$, it has the period $4$. And,
$$
\begin{aligned}
&\sum_{n=1}^{\infty} \frac{(-1)^nn}{\sinh\pi n} =-\frac{1}{4\pi} \\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3\sinh\pi n} =-\frac{\pi^3}{360} \\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n^7\sinh\pi n} =-\frac{13\pi^7}{453600} \\
&\color{Red}{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1) \right ) }
= \frac{\pi}{8}}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5\cosh\left ( \frac{\pi}{2}(2n+1) \right ) }
= \frac{\pi^5}{768}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^9\cosh\left ( \frac{\pi}{2}(2n+1) \right ) }
= \frac{23\pi^9}{1720320}
\end{aligned}
$$
Remark: The second one proved here.
$\zeta=e^{\pi i/3}$, it has period $6$. And,
$$
\begin{aligned}
&\sum_{n=1}^{\infty} \frac{(-1)^nn^2}{\cosh(\sqrt{3}\pi n )-(-1)^n}
= -\frac{1}{12\pi^2} \\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4(\cosh(\sqrt{3}\pi n )-(-1)^n)}
= -\frac{\pi^4}{11340}\\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{10}(\cosh(\sqrt{3}\pi n )-(-1)^n)}
= -\frac{703\pi^{10}}{7662154500}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
= \frac{\pi}{24}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{7}\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
= \frac{\pi^{7}}{23040}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{13}\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
= \frac{173\pi^{13}}{3832012800}\\
\end{aligned}
$$
Remark: The first one proved here.
Remark 2: The fifth one proved here.
$\zeta=e^{\pi i/4}$ has period $8$. And,
$$
\begin{aligned}
&\sum_{n=1}^{\infty} \frac{(-1)^nn^3}{\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )}
=\frac{1}{16\pi^3}\\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n^5\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )}
=\frac{\pi^5}{151200}\\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{13}\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )}
=\frac{14527\pi^{13}}{20841060240000}\\
&\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1) \right )
\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) +\cos\left ( \frac{\pi}{2}(2n+1) \sqrt{2} \right )\right ]
} =\frac{\pi}{32} \\
&\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^9\cosh\left ( \frac{\pi}{2}(2n+1) \right )
\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) +\cos\left ( \frac{\pi}{2}(2n+1) \sqrt{2} \right )\right ]
} =\frac{17\pi^9}{5160960}\\
&\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^{17}\cosh\left ( \frac{\pi}{2}(2n+1) \right )
\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) +\cos\left ( \frac{\pi}{2}(2n+1) \sqrt{2} \right )\right ]
} =\frac{3718853\pi^{17}}{10712468422656000}
\end{aligned}
$$
$\zeta=e^{\pi i/5}$ has period $10$. And,
$$\begin{aligned}
&\sum_{n = 1}^{\infty}
\tfrac{(-1)^nn^4}{\left[\cosh\left ( \frac{\sqrt{10-2\sqrt{5} } }{2}\pi n \right )-\cos\left ( \frac{1+\sqrt{5} }{2} \pi n\right ) \right]
\left[\cosh\left ( \frac{\sqrt{10+2\sqrt{5} } }{2}\pi n \right )-\cos\left ( \frac{1-\sqrt{5} }{2} \pi n\right ) \right]
} = -\frac{1}{40\pi^4} \\
&\sum_{n = 1}^{\infty}
\tfrac{(-1)^n}{n^6\left[\cosh\left ( \frac{\sqrt{10-2\sqrt{5} } }{2}\pi n \right )-\cos\left ( \frac{1+\sqrt{5} }{2} \pi n\right ) \right]
\left[\cosh\left ( \frac{\sqrt{10+2\sqrt{5} } }{2}\pi n \right )-\cos\left ( \frac{1-\sqrt{5} }{2} \pi n\right ) \right]
} =-\frac{\pi^6}{3742200}\\
&\sum_{n = 1}^{\infty}
\tfrac{(-1)^n}{n^{16}\left[\cosh\left ( \frac{\sqrt{10-2\sqrt{5} } }{2}\pi n \right )-\cos\left ( \frac{1+\sqrt{5} }{2} \pi n\right ) \right]
\left[\cosh\left ( \frac{\sqrt{10+2\sqrt{5} } }{2}\pi n \right )-\cos\left ( \frac{1-\sqrt{5} }{2} \pi n\right ) \right]
} =-\frac{262177\pi^{16}}{91879767917437500}\\
&\sum_{n = 0}^{\infty}
\tfrac{(-1)^n}{(2n+1)\left[\cosh\left (\frac{\pi}{2}(2n+1) \frac{\sqrt{10-2\sqrt{5} } }{2}\right )+\cos\left ( \frac{\pi}{2}(2n+1)\frac{1+\sqrt{5} }{2} \right ) \right]
\left[\cosh\left ( \frac{\pi}{2}(2n+1)\frac{\sqrt{10+2\sqrt{5} } }{2}\right )+\cos\left ( \frac{\pi}{2}(2n+1)\frac{1-\sqrt{5} }{2} \right ) \right]
} =\frac{\pi}{80}\\
&\sum_{n = 0}^{\infty}
\tfrac{(-1)^n}{(2n+1)^{11}\left[\cosh\left (\frac{\pi}{2}(2n+1) \frac{\sqrt{10-2\sqrt{5} } }{2}\right )+\cos\left ( \frac{\pi}{2}(2n+1)\frac{1+\sqrt{5} }{2} \right ) \right]
\left[\cosh\left ( \frac{\pi}{2}(2n+1)\frac{\sqrt{10+2\sqrt{5} } }{2}\right )+\cos\left ( \frac{\pi}{2}(2n+1)\frac{1-\sqrt{5} }{2} \right ) \right]
} =\frac{31\pi^{11}}{232243200}\\
&\sum_{n = 0}^{\infty}
\tfrac{(-1)^n}{(2n+1)^{21}\left[\cosh\left (\frac{\pi}{2}(2n+1) \frac{\sqrt{10-2\sqrt{5} } }{2}\right )+\cos\left ( \frac{\pi}{2}(2n+1)\frac{1+\sqrt{5} }{2} \right ) \right]
\left[\cosh\left ( \frac{\pi}{2}(2n+1)\frac{\sqrt{10+2\sqrt{5} } }{2}\right )+\cos\left ( \frac{\pi}{2}(2n+1)\frac{1-\sqrt{5} }{2} \right ) \right]
} =\frac{10568303\pi^{21}}{7414558501109760000}
\end{aligned}$$
$\zeta=e^{\pi i/6}$ has period $12$. And,
$$\begin{aligned}
&\sum_{n = 1}^{\infty} \frac{(-1)^nn^5}{\sinh\pi n(\cosh(\sqrt{3}\pi n )-(-1)^n)
(\cosh\pi n-\cos(\sqrt{3}\pi n ))
} = -\frac{1}{48\pi^5} \\
&\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^7\sinh\pi n(\cosh(\sqrt{3}\pi n )-(-1)^n)
(\cosh\pi n-\cos(\sqrt{3}\pi n ))
} =-\frac{691\pi^7}{3064861800} \\
&\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^{19}\sinh\pi n(\cosh(\sqrt{3}\pi n )-(-1)^n)
(\cosh\pi n-\cos(\sqrt{3}\pi n ))
} =-\frac{322393987\pi^{19}}{13216553801264063250000}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1)\right )
\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\right ) +\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\right ] }
=\frac{\pi}{96}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{13}\cosh\left ( \frac{\pi}{2}(2n+1)\right )
\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\right ) +\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\right ] }
=\frac{691\pi^{13}}{61312204800}\\
&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{25}\cosh\left ( \frac{\pi}{2}(2n+1)\right )
\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\right ) +\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\right ] }
=\frac{14240963339\pi^{25}}{1167902873850803650560000}
\end{aligned}$$
21.08.01
To find more generalizations, we should choose other functions.
If we consider $f(z)=\frac{\pi\sinh\pi}{z(\cosh\pi z\pm\cosh\pi)(\cos\pi z\pm\cosh\pi)},$then
$$\begin{aligned}
&\sum_{n=1}^{\infty} \frac{1}{(n^2+(n+1)^2)[\cosh((2n+1)\pi)
-\cosh\pi
]}=\frac{1}{2\sinh\pi} \left ( \frac{1}{\pi}+\coth\pi -\frac{\pi}{2}\tanh^2\left ( \frac{\pi}{2} \right ) \right )\\
&\sum_{n=1}^{\infty} \frac{1}{(1+4n^2)(\cosh2\pi n+\cosh\pi)}
=\frac{\pi}{16} \frac{\coth\left ( \frac{\pi}{2} \right ) }{\sinh^2\left ( \frac{\pi}{2} \right ) }
- \frac{1}{4\cosh^2\left ( \frac{\pi}{2} \right ) }
\end{aligned}$$
Another example:
$$\sum_{n=2}^{\infty} \frac{1}{(1+n^2)(\cosh2\pi n-\cosh(2\pi))}
= \frac{1}{4\sinh^2(\pi)} +\frac{1+2\pi\coth(2\pi)}{8\pi\sinh(2\pi)}
-\frac{\pi\coth(\pi)}{8\sinh^2(\pi)}.$$
Oct.27.21
These are $\tan,\cot$ cases.
$$\begin{aligned}
&
\sum_{n = 0}^{\infty}
\frac{\tanh\left ( \frac{\pi}{2}(2n+1) \right ) }{(2n+1)^5}
\frac{
\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )
-\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )}
{\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )+
\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )} = \frac{\pi^5}{256}\\
& \sum_{n = 0}^{\infty}
\frac{\tanh\left ( \frac{\pi}{2}(2n+1) \right ) }{(2n+1)^{13}}
\frac{
\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )
-\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )}
{\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )+
\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )} =
\frac{127\pi^{13}}{309657600} \\
\end{aligned}$$
$$\begin{aligned}
&\sum_{n=1}^{\infty}
\frac{\coth(\pi n)}{n^5} \frac{
\cosh(\sqrt{2}\pi n )+\cos(\sqrt{2}\pi n )}{
\cosh(\sqrt{2}\pi n )-\cos(\sqrt{2}\pi n )}
=\frac{127\pi^5}{37800} \\
&\sum_{n=1}^{\infty}
\frac{\coth(\pi n)}{n^{13}} \frac{
\cosh(\sqrt{2}\pi n )+\cos(\sqrt{2}\pi n )}{
\cosh(\sqrt{2}\pi n )-\cos(\sqrt{2}\pi n )}
=\frac{444721 \pi^{13}}{1302566265000}
\end{aligned}$$
Additional Part 1:
Some functions may give us incredible relationships. For example,
$$\begin{aligned}
&\sum_{n = 0}^{\infty}
\frac{(-1)^n(2n+1)}{\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
-2\sqrt{3} \sum_{n = 1}^{\infty}
\frac{(-1)^n n}{\sinh(\sqrt{3}\pi n )} = \frac{1}{2\pi}\\
&\sum_{n = 0}^{\infty}
\frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
=\frac{\pi}{24}\\
&\sum_{n = 0}^{\infty}
\frac{(-1)^n}{(2n+1)^3\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
+\frac{\sqrt{3} }{8} \sum_{n = 1}^{\infty}
\frac{(-1)^n}{n^3\sinh(\sqrt{3}\pi n )} = \frac{\pi^3}{240} \\
&\sum_{n = 0}^{\infty}
\frac{(-1)^n}{(2n+1)^5\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
-\frac{\sqrt{3} }{32} \sum_{n = 1}^{\infty}
\frac{(-1)^n}{n^5\sinh(\sqrt{3}\pi n )} = \frac{13\pi^5}{30240}\\
&\sum_{n = 0}^{\infty}
\frac{(-1)^n}{(2n+1)^7\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
=\frac{\pi^7}{23040} \\
&\sum_{n = 0}^{\infty}
\frac{(-1)^n}{(2n+1)^9\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) }
+\frac{\sqrt{3} }{512} \sum_{n = 1}^{\infty}
\frac{(-1)^n}{n^9\sinh(\sqrt{3}\pi n )} = \frac{13\pi^9}{29568000}.
\end{aligned}$$
Additional Part 2:
By considering $f(z)=\frac{\pi}{z}\frac{\gamma+\psi(-z)}{\sin(\pi z)\sinh(\pi z)}$, we have proved
$$\sum_{n=1}^{\infty} \frac{(-1)^nH_{n-1}}{n\sinh(\pi n)}
-\frac{\pi}{2} \sum_{n=1}^{\infty}\frac{(-1)^n\coth(\pi n)}
{n\sinh(\pi n)}
-\sum_{n=1}^{\infty}\frac{1}{n}
\left ( \sum_{k=1}^{\infty}\frac{(-1)^kk}{\sinh(\pi k)(n^2+k^2)} \right )
=\frac{1}{2\pi}\zeta(3).
$$
By considering $f(z)=\frac{\pi}{z} \frac{J_0(\alpha z)}{\sin(\pi z)\sinh(\pi z)}$, we have proved
$$\sum_{n=1}^{\infty}
\frac{(-1)^n\left[J_0\left(\alpha n\right)-I_0(\alpha n) \right] }{n\sinh(\pi n)}
=\frac{\alpha^2}{8\pi},\qquad(\left | \alpha \right |\le\pi).$$
By considering $f(z)=\frac{\pi}{z^4}\cot(\pi z)
\operatorname{csch}\left (\sqrt{2}\pi z \right )
\coth(2\pi z)$, we have proved
$$\sum_{n=1}^{\infty}\frac{\displaystyle{\coth\left( \frac{\pi n}{2} \right )\csc\left ( \frac{\pi n}{\sqrt{2} } \right ) } }{n^4}+\frac{1}{2\sqrt{2} }\sum_{n=1}^{\infty}
\frac{\displaystyle{(-1)^n\coth\left ( \frac{\pi n}{\sqrt{2} } \right )\cot\left (\sqrt{2}\pi n\right ) } }{n^4}-\frac{1}{8}\sum_{n=1}^{\infty}\frac{\coth(2\pi n)}{n^4\sinh(\sqrt{2}\pi n )}=\frac{527\pi^4}{30240\sqrt{2} }.$$
By considering $f(z)=\frac{\pi}{z^3} \cot(\pi z)\cot\left ( \sqrt{2}\pi z \right )
\operatorname{csch}(2\pi z)\coth\left ( 2\sqrt{2}\pi z \right )$, we have proved
$$\sum_{n=1}^{\infty} \frac{\displaystyle{\csc\left ( \frac{\pi n}{\sqrt{2} } \right )\coth\left ( \frac{\pi n}{2} \right )\coth\left ( \frac{\pi n}{2\sqrt{2} } \right )}}{n^3} +\frac{1}{2} \sum_{n=1}^{\infty}\frac{(-1)^n\displaystyle{\cot\left ( \sqrt{2}\pi n \right ) \coth\left ( \frac{\pi n}{2} \right)\coth\left ( \frac{\pi n}{\sqrt{2} } \right )}}{n^3}-\frac{1}{4} \sum_{n=1}^{\infty}\frac{\displaystyle{\cot\left (\frac{\pi n}{\sqrt{2}}\right)\coth(2\pi n)}}{n^3\sinh\left ( \sqrt{2}\pi n \right ) } -\frac{1}{8} \sum_{n=1}^{\infty} \frac{\displaystyle{\cot\left (\sqrt{2} \pi n \right )\coth(2\sqrt{2} \pi n)}}{n^3\sinh\left (2\pi n \right ) }=\frac{11\pi^3}{240}.$$
Let $f(\alpha,\beta)=\sum_{n=1}^{\infty} \frac{\cot(\alpha\pi n)\cot(\beta\pi n)}{n^4}$, hence
$$f(\alpha,\beta)+\alpha^3f\left ( \frac{1}{\alpha}
,\frac{\beta}{\alpha} \right )+\beta^3
f\left ( \frac{1}{\beta},\frac{\alpha}{\beta}\right )
=L(\alpha,\beta).$$
Where $$L(\alpha,\beta)
=\frac{\pi^4 (2 \alpha^6 - 7 \alpha^4 \beta^2 - 7 \alpha^4 - 7
\alpha^2 \beta^4 + 35 \alpha^2 \beta^2 - 7 \alpha^2 + 2 \beta^6 - 7 \beta^4 - 7 \beta^2 + 2)}{1890\alpha\beta}.$$
For $|a|\le\pi$, we have
$$2\sum_{n=1}^{\infty}\frac{(-1)^nH_n\cos\sqrt{\pi^2n^2-a^2}}{n^2}
-3\sum_{n=1}^{\infty}\frac{(-1)^n\cos\sqrt{\pi^2n^2-a^2}}{n^3}-
\pi^2\sum_{n=1}^{\infty} \frac{(-1)^n\sin\sqrt{\pi^2n^2-a^2}}{n\sqrt{\pi^2n^2-a^2}}=\zeta(3)\cosh a.$$
For $|\alpha|\le\pi/2$, we have
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}J_0(2\alpha n)H_n}{n^4}
=\left(\frac{\alpha^2}{2} -\frac{\pi^2}{12} \right)\zeta(3)
-\frac{1}{2}\zeta(5) +\alpha\sum_{n=1}^{\infty}
\frac{(-1)^{n-1}J_1(2\alpha n)}{n^4} +\frac{5}{2}
\sum_{n=1}^{\infty} \frac{(-1)^{n-1}J_0(2\alpha n)}{n^5}.$$
For $\alpha\le\pi$, we have
$$\sum_{n=1}^{\infty}
\frac{(-1)^n\left [ \cosh\left (\alpha\sqrt[4]{\beta^4+n^4} \right )
+\cos\left (\alpha\sqrt[4]{\beta^4+n^4} \right ) \right ] }{n^3\sinh(\pi n)}
= \frac{\alpha\sin(\alpha\beta)-\alpha\sinh(\alpha\beta)}{16\pi\beta^3}
-\frac{\pi^3\left (\cosh(\alpha\beta)+\cos(\alpha\beta) \right ) }{360}.$$
Another example:
$$\sum_{n=1}^{\infty}
\frac{\cot\left ( \pi\sqrt{n^2+1} \right ) }{n^2\sqrt{n^2+1} }
+\sum_{n=2}^{\infty}\frac{\cot\left ( \pi\sqrt{n^2-1} \right ) }{(n^2-1)^{3/2}}
=\frac{\pi^3}{45}-\frac{3}{16\pi}-\frac{\coth(\pi)}{2}.$$
Entry.(Ramanujan) Let $\alpha,\beta>0,\alpha\beta=4\pi^3$, we have
$$\sum_{n=1}^{\infty} \frac{1}{e^{n^2\alpha}-1}
=\frac{\pi^2}{6\alpha}+\frac{1}{4}
+\frac{\sqrt{\beta} }{4\pi}
\left ( \zeta\left ( \frac{1}{2} \right ) +\sum_{n=1}^{\infty}
\frac{1}{\sqrt{n} } \frac{\cos\left ( \sqrt{n\beta} \right )
-\sin\left ( \sqrt{n\beta} \right )-e^{-\sqrt{n\beta} } }
{\cosh\left ( \sqrt{n\beta} \right )-\cos\left ( \sqrt{n\beta} \right ) } \right ).$$
We can also verify the Entry by simple residue calculations. Let,
$$
\begin{aligned}
&f_1(z)=\left ( \frac{1}{e^{z^2\alpha}-1} -\frac{1}{\alpha z^2} \right )
\frac{e^{2\pi i z}}{e^{2\pi i z}-1}, \\
&f_2(z)=\left ( \frac{1}{e^{z^2\alpha}-1} -\frac{1}{\alpha z^2} \right )
\frac{1}{e^{2\pi i z}-1}. \\
\end{aligned}
$$
Notice that we cancel the high-order pole $z=0$ by adding $-1/(\alpha z^2)$.
Then integrate $f_1(z)$ and $f_2(z)$ in the upper-half plane and lower-half plane, respectively. It leaves
$$
\int_{-\infty}^{\infty}
\left ( \frac{1}{e^{z^2\alpha}-1} -\frac{1}{\alpha z^2} \right )\text{d}z.
$$
This is a trivial integral. Note that
$$
\int_{0}^{\infty}
x^{s-1}\left ( \frac{1}{e^x-1} -\frac{1}{x} \right ) \text{d}x
=\Gamma(s)\zeta(s).\qquad(0<s<1)
$$
Hence, the integral can be evaluated. Finally we complete the proof by calculating the residues.
By considering $$\begin{aligned}
&f_1(z): = \frac{1}{z} \left[\left ( \coth\left ( \alpha\pi z \right )
-\frac{1}{\alpha\pi z} \right )-
\left ( \coth\left ( \pi z \right )
-\frac{1}{\pi z} \right )\right]\frac{e^{2\pi iz}}{e^{2\pi iz}-1} \\
&f_2(z): = \frac{1}{z} \left[\left ( \coth\left ( \alpha\pi z \right )
-\frac{1}{\alpha\pi z} \right )-
\left ( \coth\left ( \pi z \right )
-\frac{1}{\pi z} \right )\right]\frac{1}{e^{2\pi iz}-1}
\end{aligned}$$, for $\alpha>0$, we have
$$\sum_{n=1}^{\infty}
\frac{1}{n} \left ( \frac{1}{e^{2\pi n\alpha}-1}
-\frac{1}{e^{2\pi n/\alpha}-1} \right )
=\frac{\ln \alpha}{2}-\frac{\pi \alpha}{12}+\frac{\pi}{12\alpha} .$$
We need to compute
$$\int_{-\infty}^{\infty}
\frac{1}{z} \left[\left ( \coth\left ( \alpha\pi z \right )
-\frac{1}{\alpha\pi z} \right )-
\left ( \coth\left ( \pi z \right )
-\frac{1}{\pi z} \right )\right]\text{d}z
=2\ln\alpha.$$
It easily derive through Frullani's integral.
Sine and cosine integral
are given by
$$
\operatorname{Si}(x)
=\int_{0}^{x} \frac{\sin(t)}{t}\text{d}t,
\operatorname{Ci}(x)
=-\int_{x}^{\infty}\frac{\cos(t)}{t}\text{d}t.
$$
Then we have following results:
$$\begin{aligned}
&\sum_{n = 1}^{\infty}\frac{\operatorname{Si}(\pi n)}{n^3} = \frac{5\pi^3}{72},\\
&\sum_{n = 1}^{\infty}\frac{\operatorname{Si}(\pi n)}{n^5} = \frac{269\pi^5}{43200},\\
&\sum_{n = 1}^{\infty}\frac{\operatorname{Si}(\pi n)}{n^7} = \frac{3919\pi^7}{6350400},\\
&\sum_{n = 1}^{\infty}\frac{\operatorname{Si}(\pi n)}{n^9} = \frac{568999\pi^9}{914457600},\\
&\sum_{n = 1}^{\infty}\frac{\operatorname{Si}(\pi n)}{n^{11}} = \frac{10451593\pi^{11}}{1659740544000},\\
&\sum_{n = 1}^{\infty}\frac{\operatorname{Si}(\pi n)}{n^{13}} =\frac{150288211487 \pi^{13}}{235616767626240000}.
\end{aligned}$$
Proof 1: Let us denote $\mathscr{S}(\alpha)=\sum_{n=1}^{\infty}\frac{\operatorname{Si}(\pi n x)}{n^{2k+1}}$, then
differentiate $\mathscr{S}$ with respect to $x$.
Solve the series after differentiating. Integrate the expression and note that $\mathscr{S}(0)=0$. The result will immediately follow.
Proof 2: Choose the function
$$f(z)=\frac{2\pi i}{z^{2k+1}} \frac{\displaystyle{\gamma+\frac{1}{2} \sum_{n=1}^{\infty}
\frac{(-1)^n(\pi z)^{2n}}{n(2n)!} +i\operatorname{Si}(\pi z)} }{e^{2\pi iz}-1}.$$
Proof 3: Choose two
functions
$$\begin{aligned}
&f_1(z)=\frac{\displaystyle{\operatorname{Si}(\pi z)-\sum_{n=1}^{c}\frac{(-1)^n(\pi z)^{2n+1}}{(2n+1)(2n+1)!} }}{z^{2k+1}
(e^{2\pi i z}-1)}e^{2\pi i z},\\
&f_2(z)=\frac{\displaystyle{\operatorname{Si}(\pi z)-\sum_{n=1}^{c}\frac{(-1)^n(\pi z)^{2n+1}}{(2n+1)(2n+1)!} }}{z^{2k+1}
(e^{2\pi i z}-1)}.\\
\end{aligned}$$
$c$ is the least value to
make sure $$\int_{0}^{\infty}\frac{\displaystyle{\operatorname{Si}(\pi z)-\sum_{n=1}^{c}\frac{(-1)^n(\pi z)^{2n+1}}{(2n+1)(2n+1)!} }}{z^{2k+1}}\text{d}z$$ converge.Thus we
only need to compute $$\int_{0}^{\infty}\frac{\displaystyle{\operatorname{Si}(\pi z)-\sum_{n=1}^{c}\frac{(-1)^n(\pi z)^{2n+1}}{(2n+1)(2n+1)!} }}{z^{2k+1}}\text{d}z.$$
This is trivial.
Consider $f(z)=\frac{\tan\left ( \frac{\pi}{2}e^z \right ) }{\sin(z)\sinh(z)}$, we obtain $$\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^n }{\sinh(\pi n)}
\left ( \tan\left ( \frac{\pi}{2}e^{\pi n}\right ) -\tan\left ( \frac{\pi}{2}e^{-\pi n}\right )\right )
-\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{\sinh(\pi n)}
+4\sum_{n=1}^{\infty} \sum_{m=1}^{\infty}
\frac{1}{(2n+1)^2-1} \frac{\cosh(\pi m)\sin(\ln(2n+1))}{\cos(2\ln(2n+1))-\cosh(2\pi m)}
=-\frac{\pi^2-1}{24}+\sum_{n=1}^{\infty}
\frac{1}{(2n+1)^2-1} \frac{1}{\sin(\ln(2n+1))}.$$
Mittag-Leffler expansion part:
$$\sum_{n=1}^{\infty}
\left ( \frac{H_n}{n^3}\frac{2}{z^2-n^2}
+\frac{1}{n^5(z+n)} \right )
=\frac{\left ( \gamma+\psi^{(0)}(-z) \right )
\left ( \gamma+\psi^{(0)}(1+z) \right ) }{z^4}
-\frac{\pi^2}{6z^4}+\frac{\zeta(3)}{z^3}
+\frac{\pi^4}{60z^2}+\frac{\zeta(5)}{z}.$$
