Show that $\sum_{n=1}^{+\infty}\frac{1}{(n\cdot\sinh(n\pi))^2} = \frac{2}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{(2n-1)^2} - \frac{11\pi^2}{180}$

Question

What I do so far \begin{align*} \text{Show that} \quad &\sum_{n=1}^{+\infty}\frac{1}{(n\cdot\sinh(n\pi))^2} = \frac{2}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{(2n-1)^2} - \frac{11\pi^2}{180} \\ \text{Lemma 1 } &\sum_{n = - \infty }^\infty \frac{1}{{z + n}} = \frac{\pi }{{\tan (\pi z)}} \\ \text{Lemma 2 } &\frac{1}{{\sinh^2(\pi z)}} = \frac{1}{{\pi^2 z^2}} + \frac{4z^2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{(z^2 + k^2)^2}} - \frac{2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{z^2 + k^2}} \\ &\text{Because:} \nonumber \\ &\frac{\pi }{{\tan (\pi z)}} = \sum_{k = - \infty }^\infty \frac{1}{{z + k}} \Rightarrow \left( \frac{\pi }{{\tan (\pi z)}} \right)' = -\sum_{k = - \infty }^\infty \frac{1}{{(z + k)^2}} \nonumber \\ &\Rightarrow \boxed{\frac{\pi^2}{\sin^2(\pi z)}} = \sum_{k = - \infty }^\infty \frac{1}{{(z + k)^2}} \Rightarrow \nonumber \\ &\Rightarrow \frac{\pi^2}{\sin^2(\pi iz)} = \sum_{k = - \infty }^\infty \frac{1}{{(iz + k)^2}} \Rightarrow \frac{\pi^2}{\sinh^2(\pi z)} = \sum_{k = - \infty }^\infty \frac{1}{{(iz + k)^2}} \Rightarrow \boxed{\frac{\pi^2}{\sinh^2(\pi z)} = \sum_{k = - \infty }^\infty \frac{1}{{(z + ik)^2}} = \sum_{k = - \infty }^\infty \frac{1}{{(z - ik)^2}}} \nonumber \\ &\text{then} \nonumber \\ &\frac{1}{{\sinh^2(\pi z)}} = \frac{1}{{2\pi^2}}\sum_{k = - \infty }^\infty \left( \frac{1}{{(z + ik)^2}} + \frac{1}{{(z - ik)^2}} \right) \nonumber \\ &= \frac{1}{{\pi^2}}\sum_{k = - \infty }^\infty \frac{z^2 - k^2}{{(z^2 + k^2)^2}} = \frac{1}{{\pi^2}}\sum_{k = - \infty }^\infty \frac{2z^2 - (z^2 + k^2)}{{(z^2 + k^2)^2}} \nonumber \\ &= \frac{1}{{\pi^2}} \left( 2z^2\sum_{k = - \infty }^\infty \frac{1}{{(z^2 + k^2)^2}} - \frac{1}{{\pi^2}}\sum_{k = - \infty }^\infty \frac{1}{{z^2 + k^2}} \right) \nonumber \\ &\Rightarrow \boxed{\frac{1}{{\sinh^2(\pi z)}} = \frac{1}{{\pi^2 z^2}} + \frac{4z^2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{(z^2 + k^2)^2}} - \frac{2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{z^2 + k^2}}} \end{align*}

We replace $z$ with $n$ and sum, so it results:

\begin{align*} &\sum_{n=1}^\infty \frac{1}{n^2\sinh^2(\pi n)} \\ &= \frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^4} + \frac{4}{\pi^2}\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} - \frac{2}{\pi^2}\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2(n^2+k^2)} \end{align*}

But it is known that

$$ \frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^4} = \frac{1}{\pi^2}\cdot \zeta(4) = \frac{\pi^2}{90} $$

for the sum

$$ \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2(n^2+k^2)} $$

we have:

\begin{align*} S &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2(n^2+k^2)} \\ &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^2}\left(\frac{1}{n^2} - \frac{1}{n^2+k^2}\right) \\ &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{n^2} - \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^2(n^2+k^2)} \\ &\Rightarrow S = \left(\sum_{n=1}^\infty \frac{1}{n^2}\right)\left(\sum_{k=1}^\infty \frac{1}{k^2}\right) - S \\ &\Rightarrow 2S = \left(\frac{\pi^2}{6}\right)^2 \\ &\Rightarrow S = \frac{\pi^4}{72} \end{align*}

So finally

$$ \boxed{\sum_{n=1}^\infty \frac{1}{n^2\sinh^2(\pi n)} = \frac{4}{\pi^2}\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} - \frac{\pi^2}{60}} $$

The final double sum is related to functions $\zeta(2)$ & $\zeta(4)$ that are missing all add-ins of natural ones that are not written as sum of two squares. From Fermat's theorem (proved by Euler) we know that if some natural number has a prime factor of form $p = 4k + 3$ raised to odd force, then it is not written as the sum of two squares.

I read that

$$ \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} = \zeta(2) \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} - \zeta(4) $$

somewhere, but I have no proof of this, nor have I been able to discover any.

Given the above:

\begin{align*} &\sum_{n=1}^\infty \frac{1}{n^2\sinh^2(\pi n)} \\ &= \frac{4}{\pi^2}\left(\zeta(2) \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} - \zeta(4)\right) - \frac{\pi^2}{60} \\ &= \frac{4}{\pi^2}\left(\frac{\pi^2}{6} \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} - \frac{\pi^4}{90}\right) - \frac{\pi^2}{60} \\ &= \frac{2}{3}\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} - \left(\frac{2\pi^2}{45} + \frac{\pi^2}{60}\right) \\ &= \frac{2}{3}\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} - \frac{11\pi^2}{180} \end{align*}

I would be very interested to see a proof of the relationship

$$ \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} = \zeta(2) \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} - \zeta(4) $$

which is true (tested with handmade program in Visual Basic).

Answer 1

First we have, for $s>1$ and $\beta(s)$ the Dirichlet beta function, $$\DeclareMathOperator{\csch}{csch} \sum_{(x,y)\in \mathbb{Z}^2, (x,y)\neq (0,0)} \frac{1}{(x^2+y^2)^s} = 4\zeta(s)\beta(s)$$ this can be proved in many ways, the most elegant way is probably to realize LHS as $4\zeta_{\mathbb{Q}(\sqrt{-1})}(s)$ as Dedekind zeta function of a number field (which is a PID and $4$ = number of units), $\zeta_{\mathbb{Q}(\sqrt{-1})}(s)$ factors into Dirichlet L-function because the field is abelian.

Now let $s=2$ and using $$\sum_{x\in \mathbb{Z}} \frac{1}{(x^2+y^2)^2} = \frac{\pi \coth (\pi y)+\pi ^2 y \text{csch}^2(\pi y)}{2 y^3}$$ we have (with $G = \beta(2)$), $$\begin{aligned} 4\zeta(2)G &= 2\sum_{y\geq 1, x\in \mathbb{Z}} \frac{1}{(x^2+y^2)^2} + 2\zeta(4) \\ &= 2\sum_{y\geq 1} \frac{\pi \coth (\pi y)+\pi ^2 y \text{csch}^2(\pi y)}{2 y^3} + 2\zeta(4) \end{aligned}$$ the desired evaluation of $\sum_{n\geq 1} \frac{\csch^2(n\pi)}{n^2}$ follows immediately from $\sum_{n=1}^\infty \frac{\coth(n\pi)}{n^3} = \frac{7\pi^3}{180}$

Answer 2

Just out of curiosity, I wanted to see what hyperbolic series we could evaluate if we used the same approach pisco used but with the lattice sum $$\sum_{(x,y) \in \mathbb{Z}^{2}, (x,y) \ne (0,0)}\frac{(-1)^{x+y}}{(x^2+y^2)^2} = - 4 \eta(2) \beta(2)= - \frac{\pi^{2}G}{3}. $$

It turns out that we can show that $$\sum_{y=1}^{\infty} \frac{(-1)^{y}\cosh(\pi y)}{y^{2} \sinh^{2}(\pi y)} = \frac{\pi^{2}}{45}- \frac{G}{3}. $$

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Since

$$\begin{align} - \frac{\pi^{2}G}{3} &= \sum_{(x,y) \in \mathbb{Z}^{2}, (x,y) \ne (0,0)}\frac{(-1)^{x+y}}{(x^2+y^2)^2} \\ &= 2 \sum_{y=1}^{\infty} \sum_{x=-\infty}^{\infty}\frac{(-1)^{x+y}}{(x^{2}+y^{2})^{2}} - 2 \eta(4) \\ &\overset{\spadesuit}{=} 2 \sum_{y=1}^{\infty} (-1)^{y} \, \frac{\pi^{2} y \coth(\pi y)\operatorname{csch}(\pi y)+ \pi \operatorname{csch}(\pi y)}{2y^{3}} - \frac{7\pi^{4}}{360} \\ &= \pi^{2}\sum_{y=1}^{\infty} \frac{(-1)^{y}\cosh(\pi y)}{y^{2} \sinh^{2}(\pi y)} + \pi \sum_{y=1}^{\infty}\frac{(-1)^{y}}{y^{3} \sinh(\pi y)} - \frac{7\pi^{4}}{360} \\ & \overset{\clubsuit}= \pi^{2}\sum_{y=1}^{\infty} \frac{(-1)^{y}\cosh(\pi y)}{y^{2} \sinh^{2}(\pi y)} - \frac{\pi^{4}}{360} - \frac{7\pi^{4}}{360}, \end{align}$$ we have

$$\sum_{y=1}^{\infty} \frac{(-1)^{y}\cosh(\pi y)}{y^{2} \sinh^{2}(\pi y)} = \frac{1}{\pi^{2}} \left(- \frac{\pi^{2}G}{3} + \frac{\pi^{4}}{360}+ \frac{7 \pi^{4}}{360} \right) = \frac{\pi^{2}}{45}- \frac{G}{3}. $$

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$\spadesuit$ $$\sum_{x=-\infty}^{\infty} \frac{(-1)^{x}}{(x^{2}+y^{2})^{2}} = -\operatorname*{Res}_{z = i y} \frac{\pi \csc(\pi z)}{(z^{2}+y^{2})^{2}} - \operatorname*{Res}_{z=-iy}\frac{\pi \csc(\pi z)}{(z^{2}+y^{2})^{2}}$$

$\clubsuit$ A Ramanujan sum involving $\sinh$

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As Setness Ramesory pointed out in a comment, we can use this result and the fact that $$ \sum_{(x,y) \in \mathbb{Z}^{2}, (x,y) \ne (0,0)} \frac{(-1)^{x+y} \, x^{2}}{(x^{2}+y^{2})^{3}} = \sum_{(x,y) \in \mathbb{Z}^{2}, (x,y) \ne (0,0)} \frac{1}{2} \frac{(-1)^{x+y}}{(x^{2}+y^{2})^{2}} = -\frac{\pi^{2}G}{6} $$

to show that $$\sum_{y=1}^{\infty} \frac{(-1)^y}{y \sinh^{3}(\pi y)} =-\frac{17\pi}{240}+\frac{G}{6\pi}+\frac{\ln(2)}{4}. $$

We'll also need the fact that $$\sum_{y=1}^{\infty} \frac{(-1)^{y}}{y \sinh(\pi y)} = \frac{\pi}{12} - \frac{\ln (2)}{2},$$ which can be derived from the lattice sum $$\sum_{(x,y) \in \mathbb{Z}^{2}, (x,y) \ne (0,0)} \frac{(-1)^{x+y} }{x^{2}+y^{2}}= -4 \eta(1) \beta(1) = - \pi \ln (2). $$

Answer 3

Take $$\frac{\pi}{x} \coth(\pi x)= \frac{1}{x^2} + \frac{2}{x^2 + 1^2} + \frac{2}{x^2 + 2^2} + \ldots,$$ differentiate with respect to $x$ and divide across by $-x$ to obtain

$$\frac{\pi}{x^3} \coth(\pi x) + \frac{\pi^2}{x^2 \sinh^2(\pi x)}= \frac{2}{x^4} + \frac{4}{(x^2 + 1^2)^2} + \frac{4}{(x^2 + 2^2)^2} + \ldots$$

and thence $$\pi \sum_{1}^{\infty} \frac{\coth(\pi n)}{n^3} + \pi^2 \sum_{1}^{\infty} \frac{1}{(n \sinh(\pi n))^2} = 2\zeta(4) + 4 \sum_{1}^{\infty} \sum_{1}^{\infty} \frac{1}{(m^2 + n^2)^2}$$

Now

$$\sum_{1}^{\infty} \sum_{1}^{\infty} \frac{1}{(m^2 + n^2)^s} = \zeta(s) \left( \frac{1}{1^s} - \frac{1}{3^s} + \frac{1}{5^s} - \ldots \right) - \zeta(2s)$$

\begin{align*} \pi \sum_{1}^{\infty} \frac{\coth(\pi n)}{n^3} &= \sum_{1}^{\infty} \frac{1}{n^4} + 2 \sum_{1}^{\infty} \sum_{1}^{\infty} \frac{1}{n^2(n^2 + m^2)} \\ &= \sum \frac{1}{n^4} + 2 \sum \sum \frac{1}{m^2} \left( \frac{1}{n^2} - \frac{1}{m^2 + n^2} \right) \\ &= \sum \frac{1}{n^4} + 2 \sum \frac{1}{m^2} \sum \frac{1}{n^2} - 2 \sum \sum \frac{1}{m^2(m^2 + n^2)} \\ &= \sum \frac{1}{n^4} + 2 \sum \frac{1}{m^2} \sum \frac{1}{n^2} + \sum \frac{1}{m^4} - \pi \sum \frac{\coth(\pi m)}{m^3} \\ \end{align*}

And hence \begin{align*} \pi \sum_{1}^{\infty} \frac{\coth(\pi n)}{n^3} &= \frac{\pi^4}{90} + \frac{\pi^4}{36} \\ &= \frac{7\pi^4}{180} \end{align*}

The desired result follows immediately upon substitution.