Verify the closed form of $$\sum_{k=1}^{\infty}\frac{\coth\pi k}{ke^{\pi k}}=\frac{\pi}{12}+\ln\left(1-e^{-\pi}\right)-\frac{1}{4}\ln2\tag{1}$$
This is in spirit of many such formulas like; $$\sum_{k=1}^{\infty}\frac{\coth\pi k}{k^{3}}=\frac{7\pi^{3}}{180},\hspace{.2cm}\sum_{k=1}^{\infty}\frac{\coth\pi k}{k^{7}}=\frac{19\pi^{7}}{56700}, .. \tag{for odd powers}$$
And I believe odd numbers of the form $4n-1$. Since, $\sum \coth{\pi k}/k$ is diverging, Eq $1$, could take it's place. And also is there any generalizations for the same?
We can write! $$ \frac{\coth(tk)}{k \;\exp(tk)} = - \frac{1}{k\; \exp(tk)} + \sum_{j=0}^\infty \frac{2}{k\; \exp((2j+1)tk)} $$ converging absolutely for $t > 0$. Thus
$$\eqalign{ \sum_{k=1}^\infty \frac{\coth(tk)}{k \exp(tk)} &= - \sum_{k=1}^\infty \frac{1}{k \exp(tk)} + \sum_{j=0}^\infty \sum_{k=1}^\infty \frac{2}{k\; \exp((2j+1)tk)}\cr &= -\ln(1-\exp(-t)) - 2 \sum_{j=0}^\infty \ln(1-\exp(-(2j+1)t))}$$ Now $$ \sum_{j=0}^\infty \ln(1 - \exp(-(2j+1)t)) = \ln \left(\prod_{j=0}^\infty (1 - \exp(-(2j+1)t))\right) $$ It looks to me like for $t = \pi$ that product is $\exp(-\pi/24)\; 2^{1/8}$, but I don't have a proof. If that's the case, we would have
$$ \sum_{k=1}^\infty \frac{\coth(\pi k)}{k \exp(\pi k)} = - \ln(1 - \exp(-\pi)) + \frac{\pi}{12} - \frac{1}{4} \ln(2)$$
EDIT: Hmmm. Actually we have (see formula (60) here) $$ \prod_{j=0}^\infty \left(1 - \frac{1}{x^{2j+1}}\right) = x^{-1/24} \theta_4(0,x^{-1})^{1/2} \left(\frac{2}{\theta_1'(0,x^{-1})}\right)^{1/6} $$ and I don't know if that will simplify to $\exp(-\pi/24) 2^{1/8}$ for $x = \exp(\pi)$, though it is very close numerically.
EDIT: Double hmmm: using identities for Jacobi Theta functions, the desired equality comes down to (in Maple's notation)
$$ \text{EllipticK}(\sqrt{2}/2) = \int_0^1 \frac{dt}{\sqrt{1-t^2} \sqrt{1-t^2/2}} = \frac{\pi^{3/2}}{2 \;\Gamma(3/4)^2}$$
In Mathematica notation that is $K(1/2)$, and a formula here implies that this is true.