Proving $\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}=25$

Question

I solve a partial differential equation (Laplace equation) with specific boundary conditions and I finally found the answer: $$U(x,y)=\frac{400}{\pi}\sum_{n=0}^{\infty}\frac{\sin\left((2n+1)\pi x\right)\sinh\left((2n+1)\pi y\right)}{(2n+1)\sinh\left((2n+1)\pi\right)}$$

If I put $x=\frac{1}{2}$ and $y=\frac{1}{2}$, then the equation becomes as follow: $$U\left(\frac{1}{2},\frac{1}{2}\right)=\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$ At the end we can draw the answer by MS. Excel software and see that the $U\left(\frac{1}{2},\frac{1}{2}\right)= 25$. Honestly, my professor asked me to prove $U\left(\frac{1}{2},\frac{1}{2}\right)= 25$. However, it is not included in my course and I do not have any idea. Could you help me? Thanks.

Answer 1

We will evaluate $$S=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$

Since the series is alternating, use $f(z)=\pi\csc(\pi z)$ and we have $$\oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh\left(\frac{\pi}{2}(2z+1)\right)}\,dz$$

The poles are at $z=-\frac{1}{2}, \;\ z=n, \;\ z=\frac{(2n+1)\,i}{2}-\frac{1}{2}$, so

$$\begin{align}&\text{Res}\left(f(z)\,;\,z=-\frac{1}{2}\right)=-\frac{\pi}{2}\\ &\text{Res}(f(z)\,;\,z=n)=\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}\\ &\text{Res}\left(f(z)\,;\,z=\frac{(2n+1)i}{2}-\frac{1}{2}\right)=\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}\end{align}$$

Hence:

$$\oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh(\frac{\pi}{2}(2z+1))}dz=-\frac{\pi}{2}+4\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$

As $N\to \infty$, then the integral on the left tends to $0$. So, we have $0=-\frac{\pi}{2}+4S$. Solving for $S$ gives $\,\dfrac{\pi}{8}$. Hence $$\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}=25$$

Answer 2

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.

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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 $c=\left\{\begin{matrix} m,& s=mk-1,k\in\mathbb{Z}.\\ 0,&\text{otherwise}. \end{matrix}\right.$.

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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.

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$\zeta=e^{\pi i/3}$, 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.

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$\zeta=e^{\pi i/4}$ has period $8$. And, $$ \begin{aligned} &\sum_{n=1}^{\infty} \frac{(-1)^nn^3}{\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )} =\frac{1}{16\pi^3}\\ &\sum_{n=1}^{\infty} \frac{(-1)^n}{n^5\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )} =\frac{\pi^5}{151200}\\ &\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{13}\sinh\pi n(\cos\sqrt{2}\pi n -\cosh\sqrt{2}\pi n )} =\frac{14527\pi^{13}}{20841060240000}\\ &\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1) \right ) \left [ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) +\cos\left ( \frac{\pi}{2}(2n+1) \sqrt{2} \right )\right ] } =\frac{\pi}{32} \\ &\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^9\cosh\left ( \frac{\pi}{2}(2n+1) \right ) \left [ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) +\cos\left ( \frac{\pi}{2}(2n+1) \sqrt{2} \right )\right ] } =\frac{17\pi^9}{5160960}\\ &\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^{17}\cosh\left ( \frac{\pi}{2}(2n+1) \right ) \left [ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) +\cos\left ( \frac{\pi}{2}(2n+1) \sqrt{2} \right )\right ] } =\frac{3718853\pi^{17}}{10712468422656000} \end{aligned} $$

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$\zeta=e^{\pi i/5}$ 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}$$

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$\zeta=e^{\pi i/6}$ has period $12$. And, $$\begin{aligned} &\sum_{n = 1}^{\infty} \frac{(-1)^nn^5}{\sinh\pi n(\cosh(\sqrt{3}\pi n )-(-1)^n) (\cosh\pi n-\cos(\sqrt{3}\pi n )) } = -\frac{1}{48\pi^5} \\ &\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^7\sinh\pi n(\cosh(\sqrt{3}\pi n )-(-1)^n) (\cosh\pi n-\cos(\sqrt{3}\pi n )) } =-\frac{691\pi^7}{3064861800} \\ &\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^{19}\sinh\pi n(\cosh(\sqrt{3}\pi n )-(-1)^n) (\cosh\pi n-\cos(\sqrt{3}\pi n )) } =-\frac{322393987\pi^{19}}{13216553801264063250000}\\ &\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1)\right ) \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\right ) +\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\right ] } =\frac{\pi}{96}\\ &\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{13}\cosh\left ( \frac{\pi}{2}(2n+1)\right ) \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\right ) +\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\right ] } =\frac{691\pi^{13}}{61312204800}\\ &\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{25}\cosh\left ( \frac{\pi}{2}(2n+1)\right ) \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\left [ \cosh\left ( \frac{\pi}{2}(2n+1)\right ) +\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right )\right ] } =\frac{14240963339\pi^{25}}{1167902873850803650560000} \end{aligned}$$

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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: $$\sum_{n=2}^{\infty} \frac{1}{(1+n^2)(\cosh2\pi n-\cosh(2\pi))} = \frac{1}{4\sinh^2(\pi)} +\frac{1+2\pi\coth(2\pi)}{8\pi\sinh(2\pi)} -\frac{\pi\coth(\pi)}{8\sinh^2(\pi)}.$$

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Oct.27.21
These are $\tan,\cot$ cases. $$\begin{aligned} & \sum_{n = 0}^{\infty} \frac{\tanh\left ( \frac{\pi}{2}(2n+1) \right ) }{(2n+1)^5} \frac{ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) -\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )} {\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )+ \cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )} = \frac{\pi^5}{256}\\ & \sum_{n = 0}^{\infty} \frac{\tanh\left ( \frac{\pi}{2}(2n+1) \right ) }{(2n+1)^{13}} \frac{ \cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right ) -\cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )} {\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )+ \cos\left ( \frac{\pi}{2}(2n+1)\sqrt{2} \right )} = \frac{127\pi^{13}}{309657600} \\ \end{aligned}$$ $$\begin{aligned} &\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^5} \frac{ \cosh(\sqrt{2}\pi n )+\cos(\sqrt{2}\pi n )}{ \cosh(\sqrt{2}\pi n )-\cos(\sqrt{2}\pi n )} =\frac{127\pi^5}{37800} \\ &\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^{13}} \frac{ \cosh(\sqrt{2}\pi n )+\cos(\sqrt{2}\pi n )}{ \cosh(\sqrt{2}\pi n )-\cos(\sqrt{2}\pi n )} =\frac{444721 \pi^{13}}{1302566265000} \end{aligned}$$

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Additional Part 1:
Some functions may give us incredible relationships. For example, $$\begin{aligned} &\sum_{n = 0}^{\infty} \frac{(-1)^n(2n+1)}{\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) } -2\sqrt{3} \sum_{n = 1}^{\infty} \frac{(-1)^n n}{\sinh(\sqrt{3}\pi n )} = \frac{1}{2\pi}\\ &\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) } =\frac{\pi}{24}\\ &\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)^3\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) } +\frac{\sqrt{3} }{8} \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^3\sinh(\sqrt{3}\pi n )} = \frac{\pi^3}{240} \\ &\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)^5\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) } -\frac{\sqrt{3} }{32} \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^5\sinh(\sqrt{3}\pi n )} = \frac{13\pi^5}{30240}\\ &\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)^7\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) } =\frac{\pi^7}{23040} \\ &\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)^9\cosh\left ( \frac{\pi}{2}(2n+1)\sqrt{3} \right ) } +\frac{\sqrt{3} }{512} \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^9\sinh(\sqrt{3}\pi n )} = \frac{13\pi^9}{29568000}. \end{aligned}$$

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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 $$\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^n }{\sinh(\pi n)} \left ( \tan\left ( \frac{\pi}{2}e^{\pi n}\right ) -\tan\left ( \frac{\pi}{2}e^{-\pi n}\right )\right ) -\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{\sinh(\pi n)} +4\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(2n+1)^2-1} \frac{\cosh(\pi m)\sin(\ln(2n+1))}{\cos(2\ln(2n+1))-\cosh(2\pi m)} =-\frac{\pi^2-1}{24}+\sum_{n=1}^{\infty} \frac{1}{(2n+1)^2-1} \frac{1}{\sin(\ln(2n+1))}.$$

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Mittag-Leffler expansion part: $$\sum_{n=1}^{\infty} \left ( \frac{H_n}{n^3}\frac{2}{z^2-n^2} +\frac{1}{n^5(z+n)} \right ) =\frac{\left ( \gamma+\psi^{(0)}(-z) \right ) \left ( \gamma+\psi^{(0)}(1+z) \right ) }{z^4} -\frac{\pi^2}{6z^4}+\frac{\zeta(3)}{z^3} +\frac{\pi^4}{60z^2}+\frac{\zeta(5)}{z}.$$