manipulations with Taylor expansions for log and sinh

Question

How could we derive the equality $$ \frac14 \sum_{m=1}^\infty \frac1m \frac{1}{\sinh^2 \frac{m\alpha}2} = - \sum_{n=1}^\infty n\log (1-q^n)$$ where $q=e^{-\alpha}$ ?

Answer 1

$$ \begin{align} \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{\sinh^{2} \frac{m \alpha}{2}} &= \sum_{m=1}^{\infty} \frac{1}{m} \frac{4}{(e^{m \alpha /2}-e^{- m \alpha /2})^{2}} \\ &= 4 \sum_{m=1}^{\infty} \frac{1}{m} \frac{e^{-m \alpha }}{(1-e^{-m \alpha})^{2}} \\ &=4 \sum_{m=1}^{\infty} \frac{1}{m}\sum_{n=1}^{\infty} n(e^{-m \alpha})^{n} \tag{1} \\ &=4 \sum_{n=1}^{\infty} n \sum_{m=1}^{\infty} \frac{(e^{-\alpha n })^{m}}{m} \\ &= -4 \sum_{n=1}^{\infty} n \log(1-e^{- \alpha n}) \end{align}$$

$(1)$ If $|z| <1$, $ \displaystyle \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^{2}}$.