A Ramanujan-type identity: $11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{48}$ Out of curiosity, why it is these sums yield a rational answer?
$$11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{48}$$
I found this identity during the observing ramanujan identiy via a wolfram sum calculator.
Or it is also related to the type of $ e^{\pi{\sqrt43}}=884736743.999$ it is almost an integer?
Also,if possible can anyone verify it is correctness?

$$\sum_{n\geq 1}\frac{n^3}{e^{2n\pi}-1} = \sum_{n\geq 1}\sum_{k\geq 1} n^3 e^{-2k\pi n}=\frac{1}{8}\sum_{k\geq 1}\frac{3+2\sinh^2(k\pi)}{\sinh^4(k\pi)}\tag{1} $$
$$\sum_{n\geq 1}\frac{n^3}{e^{4n\pi}-1} = \sum_{n\geq 1}\sum_{k\geq 1} n^3 e^{-4k\pi n}=\frac{1}{8}\sum_{k\geq 1}\frac{3+2\sinh^2(2k\pi)}{\sinh^4(2k\pi)}\tag{2} $$
Now we may prove that:
$$ \sum_{k\geq 1}\frac{1}{\sinh^2(k\pi)}=\frac{1}{6}-\frac{1}{2\pi}.\tag{3}$$
Let we define $f(z)$ as $\frac{z}{e^{2\pi z}-1}$. $f$ has a removable singularity at the origin: by removing it, we are free to assume $f(0)=\frac{1}{2\pi}$. $f(z)$ has a remarkable property: $f(z)+\frac{z}{2}$ is an even function, hence in computing
$$ \sum_{n\geq 0}f(n) = \frac{1}{2\pi}+\sum_{n\geq 1}\frac{n}{e^{2\pi n}-1} $$
through the Euler-Maclaurin summation formula or through the Abel-Plana formula the remainder term greatly simplifies, leading to:
$$ \sum_{n\geq 1}\frac{n}{e^{2\pi n}-1}=\frac{1}{24}-\frac{1}{8\pi} \tag{4} $$
that proves $(3)$, since:
$$ \sum_{n\geq 1}\frac{n}{e^{2\pi n}-1} = \sum_{n\geq 1}\sum_{k\geq 1}n e^{-2k\pi n}=\frac{1}{4}\sum_{k\geq 1}\frac{1}{\sinh^2(k\pi)}.$$
In a similar fashion we get:
$$ \sum_{n\geq 1}\frac{1}{\sinh^4(n\pi)}=-\frac{11}{90}+\frac{1}{3 \pi }+\frac{\Gamma\left(\frac{1}{4}\right)^8}{1920 \pi ^6}\tag{5}$$
then by $(3)$ and $(5)$ it follows that:

$$ \sum_{n\geq 1}\frac{n^3}{e^{2n\pi}-1} = \color{red}{\frac{\Gamma\left(\frac{1}{4}\right)^8}{5120 \pi ^6}-\frac{1}{240}} \tag{6}$$

and if we tackle $(2)$ in the same way, to prove the OP's claim is just a matter of trivial algebra.
Many thanks to Zucker's The Summation of Series of Hyperbolic Functions: the last term of $(5)$ comes from the computation of an elliptic integral - these objects are deeply related with series involving $\frac{1}{\sinh},\frac{1}{\cosh}$ or their powers.
