TITLE: A Ramanujan sum involving $\sinh$
QUESTION [25 upvotes]: Today, in a personal communication, I was asked to prove the classical result
$$\boxed{ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)} = \frac{\pi^3}{360}}\tag{CR} $$
which I believe is due to Ramanujan. My proof can be found here and it is based on the closed form for $\sum_{n\geq 1}\frac{1}{m^2+n^2}$ and double counting. I would use this question for collecting alternative proofs. I am already aware that standard techniques for tackling series with a similar structure are

The Poisson summation formula;
The residue theorem coupled with the Laplace transform of $\text{Li}_k$ (I used this approach due to Simon Plouffe here, for instance);
Eisenstein series related to Gaussian integers;
Identities involving Dirichlet's series, since $$r_2(N)=\left|\{(a,b)\in\mathbb{Z}^2:a^2+b^2=N\}\right|=4 \sum_{d\mid N}\chi_4(d)$$
(this can be regarded as an analytic-combinatorial equivalent of the statement "$\mathbb{Z}[i]$ is a UFD"). 

I also know that a cornerstone is given by Zucker's The Summation of Series of Hyperbolic Functions, 1979. Let us see if we can devise a very short proof of $(\text{CR})$ through these ingredients or other ones.

REPLY [13 votes]: Suppose we seek to show that
$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{1}{\sinh(\pi n)}
= \frac{\pi^3}{360}.$$
Using $$\frac{1}{\sinh(x)} = \frac{2}{e^x-e^{-x}}
= 2\frac{e^{-x}}{1-e^{-2x}}$$
this is the same as
$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3}
\frac{e^{-n\pi}}{1-e^{-2n\pi}}
= \frac{\pi^3}{720}.$$
The sum  term may  be evaluated  using harmonic  summation techniques.
Since  this  method  has  not   been  presented  I  will  detail  this
calculation here.
Let $p$ be a positive integer and introduce
$$S(x;p) =
\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{2p+1}}
\frac{e^{-nx}}{1-e^{-2nx}}.$$
We will evaluate  $S(\pi;p)$ using a functional  equation for $S(x;p)$
that is obtained by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{(-1)^{k+1}}{k^{2p+1}}, \quad
\mu_k = k
\quad \text{and} \quad
g(x) = \frac{e^{-x}}{1-e^{-2x}}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{e^{-x}}{1-e^{-2x}} x^{s-1} dx
= \int_0^\infty 
\sum_{q\ge 0} e^{- (2 q+1) x} x^{s-1} dx
=  \sum_{q\ge 0}  \int_0^\infty e^{-(2q+1)x} x^{s-1} dx 
\\= \Gamma(s) \sum_{q\ge 0} \frac{1}{(2q+1)^s}
= \left(1-\frac{1}{2^s}\right) \Gamma(s) \zeta(s)$$
with fundamental strip $\langle 1, \infty\rangle.$

It follows that the Mellin transform $Q(s)$ of the harmonic sum 
$S(x;p)$ is given by
$$Q(s) = \left(1-\frac{1}{2^s}\right) 
\left(1-\frac{1}{2^{s+2p}}\right)
\Gamma(s) \zeta(s) \zeta(s+2p+1)
\\ \text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 1} \frac{(-1)^{k+1}}{k^{2p+1}} \frac{1}{k^s}
= \left(1-\frac{2}{2^{s+2p+1}}\right) \zeta(s+2p+1)$$
for $\Re(s+2p+1) > 1$ or $\Re(s) \gt -2p.$

The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate  by shifting it to the left  for an expansion about
zero.

Fortunately the  trivial zeros of  the two zeta function  terms cancel
the poles of the gamma function  term. The first term cancels those at
$-2m$ where $m\ge 1$ and the second  one the odd ones from $-2p-3$ on,
which leaves  the poles at $s=1$,  and $-2q-1$ with $0\le  q\le p.$ It
would appear there is a pole at $s=-2p$ but this is not the case since
we  have two  simple poles  among  the five  terms but  also two  zero
values,  making for  cancelation. The  pole  at $s=0$  is canceled  as
well. 
For the residue at $s=1$ we find
$$\frac{1}{2} \frac{2^{2p+1}-1}{2^{2p+1}} 
\times 1 \times \zeta(2p+2) \frac{1}{x}
= \frac{2^{2p+1}-1}{2^{2p+2}} 
\frac{(-1)^{p} B_{2p+2}  (2\pi)^{2p+2}}{2 (2p+2)!} \frac{1}{x}
\\ = (2^{2p+1}-1) 
\frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}.$$
The negative odd values at $s=-2q-1$ yield
$$\left(1-\frac{1}{2^{-2q-1}}\right) 
\left(1-\frac{1}{2^{2p-2q-1}}\right)
\frac{(-1)^{2q+1}}{(2q+1)!} \zeta(-2q-1) \zeta(2p-2q) x^{2q+1}
\\ = (1-2^{2q+1}) 
\left(1-\frac{1}{2^{2p-2q-1}}\right)
\frac{1}{(2q+1)!} \frac{B_{2q+2}}{2q+2}
\frac{(-1)^{p-q+1} B_{2p-2q} (2\pi)^{2p-2q}}{2(2p-2q)!}
x^{2q+1}
\\ = \frac{1}{2} (1-2^{2q+1}) 
(2^{2p-2q-1}-1)
\frac{(-1)^{p-q+1}}{(2q+1)!}
\frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)}
x^{2q+1}.$$
Shifting to  $\Re(s) = -2p -3/2$ we get
$$S(x;p) =
(2^{2p+1}-1) 
\frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}
\\ +
\frac{1}{2} \sum_{q=0}^p
(1-2^{2q+1}) 
(2^{2p-2q-1}-1)
\frac{(-1)^{p-q+1}}{(2q+1)!}
\frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)}
x^{2q+1}
\\ + \frac{1}{2\pi i} 
\int_{-2p-3/2-i\infty}^{-2p-3/2+i\infty} Q(s)/x^s ds.$$
We will turn this into the promised functional equation.
Substitute $s = -2p - t$ in the remainder integral to get
$$- \frac{1}{2\pi i} 
\int_{3/2+i\infty}^{3/2-i\infty}
\left(1-2^{2p+t}\right) 
\left(1-2^t\right)
\Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t)
x^{t+2p} dt$$
which is
$$\frac{x^{2p}}{2\pi i} 
\int_{3/2-i\infty}^{3/2+i\infty}
\left(1-2^{2p+t}\right) 
\left(1-2^t\right)
\Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t)
x^{t} dt.$$
In view of  the desired functional equation we  now use the functional
equation  of the Riemann  zeta function  on $Q(s)$  to prove  that the
integrand of the last integral is in fact $(-1)^p Q(t)/\pi^{2p+2t}.$

Start with the functional equation
$$\zeta(1-s) = \frac{2}{2^s\pi^s} 
\cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$
and substitute this into $Q(s)$ to obtain
$$Q(s) = 
\left(1-\frac{1}{2^s}\right) 
\left(1-\frac{1}{2^{s+2p}}\right)
\frac{\zeta(1-s) 2^s \pi^s}{2\cos\left(\frac{\pi s}{2}\right)}
\zeta(s+2p+1)
\\ = \frac{1}{2} 
(2^s-1) \left(1-\frac{1}{2^{s+2p}}\right)
\pi^s \frac{\zeta(s+2p+1)}{\cos\left(\frac{\pi s}{2}\right)}
\zeta(1-s).$$
Apply the functional equation again  (this time to $\zeta(s+2p+1)$) to
get
$$Q(s) =  \frac{1}{2} \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)}
(2^s-1) \left(1-\frac{1}{2^{s+2p}}\right)
\frac{2}{2^{-2p-s} \pi^{-2p-s}}
\cos\left(\frac{\pi (-2p-s)}{2}\right)
\\ \times \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)
\\ = \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)}
(2^s-1) (2^{2p+s}-1)
\pi^{2p+s}
(-1)^p \cos\left(\frac{-\pi s}{2}\right)
\\ \times \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)$$
and  we finally get
$$Q(s) =  (-1)^p \pi^{2p+2s} (1-2^s) (1-2^{2p+s})
\Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)$$
thus proving the claim.

Return to the remainder integral and re-write it as follows:
$$(-1)^p \frac{(x/\pi)^{2p}}{2\pi i} 
\int_{3/2-i\infty}^{3/2+i\infty}
(-1)^p \pi^{2p+2t}
\left(1-2^{2p+t}\right) 
\left(1-2^t\right)
\\ \times \Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t)
(x/\pi^2)^{t} dt.$$
so that  the fact of it being  a multiple of the  defining integral of
$S(\pi^2/x; p)$ becomes readily apparent.

We have established the functional equation
$$\color{green}
{S(x;p) =
(2^{2p+1}-1) 
\frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}
\\ +
\frac{1}{2} \sum_{q=0}^p
(1-2^{2q+1}) 
(2^{2p-2q-1}-1)
\frac{(-1)^{p-q+1}}{(2q+1)!}
\frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)}
x^{2q+1}
\\ + (-1)^p \left(\frac{x}{\pi}\right)^{2p} S(\pi^2/x;p).}$$
Now the value  $x=\pi$ is obviously special here (fixed  point) and we
get for  $p=2r+1$ with $r\ge  0$ ($p$  even yields a  Bernoulli number
identity)
$$S(\pi; 2r+1) = 
\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{4r+3}}
\frac{e^{-n\pi}}{1-e^{-2n\pi}}
= - \frac{\pi^{4r+3}}{4} 
(2^{4r+3}-1) 
\frac{B_{4r+4}}{(4r+4)!}
\\ + \frac{\pi^{4r+3}}{4} \sum_{q=0}^{2r+1}
(1-2^{2q+1}) 
(2^{4r+1-2q}-1)
\frac{(-1)^{q}}{(2q+1)!}
\frac{B_{2q+2} B_{4r+2-2q}}{(4r+2-2q)! (q+1)}.$$
We obtain a rational multiple of $\pi^{4r+3}.$ Scale by two to get
for
$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{1}{\sinh(\pi n)}, 
\quad \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^7} \frac{1}{\sinh(\pi n)},
\\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{11}} \frac{1}{\sinh(\pi n)}, 
\quad \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{15}} \frac{1}{\sinh(\pi n)},
\quad \ldots$$
the values
$$\bbox[5px,border:2px solid #00A000]{
{\frac {{\pi }^{3}}{360}},\quad
{\frac {13\,{\pi }^{7}}{453600}},\quad
{\frac {4009\,{\pi }^{11}}{13621608000}},\quad
{\frac {13739\,{\pi }^{15}}{4547140416000}},\quad \ldots}$$
These are dominated by the first term
$$\frac{1}{\sinh(\pi)}.$$