I proved this equality for $x>0$, $$ 2(e^x-1)e^{-2x} = \sum_{n=1}^\infty \frac{\mu(n)}{\sinh nx} $$ where $\mu(n)$ is the Möbius function. The problem is my proof is not so elegant and I think I could improve it using a variation of Möbius inversion formula. I leave my proof.
\begin{align*} \sum_{n=1}^{∞} \frac{μ(n)}{\sinh (nx)} &= 2 \sum_{n=1}^{∞} \frac{μ(n)}{e^{nx} - e^{-nx}} = 2 \sum_{n=1}^{∞} \frac{μ(n)}{e^{nx}} \frac{1}{1-e^{-2nx}}\\ &= 2 \sum_{n=1}^{∞} μ(n) e^{-nx}(1+e^{-2nx}+e^{-4nx}+\dots)\\ & = 2 \sum_{n=1}^{∞} μ(n) \sum_{k=1}^{∞} e^{-(2k-1)nx}= 2 \sum_{n=1}^{∞} \sum_{k=1}^{∞} μ(n) e^{-(2k-1)nx} \end{align*} We know that series are absolutly convergent, so $$ \sum_{n=1}^{∞} \frac{μ(n)}{\sinh (nx)} = 2 \sum_{N=1}^{∞} e^{-Nx} \left(\sum_{n|N \text{ y }N/n\text{ is odd}} μ(n) \right)$$ If $N$ is odd $$ \sum_{\substack{n|N \\ N/n\text{ is odd}}} μ(n) = \begin{cases} 1, &\text{ si }N = 1\\ 0, &\text{ si }N > 1 \end{cases}$$ If $N$ is even and $N=2^am$ $$ \sum_{2^ak|2^am} μ(n) = \sum_{k|m} μ(2^a k) = \sum_{k|m} μ(2^a) μ(k) = \begin{cases} 0, &\text{ si }a ≥ 2\\ -1 &\text{ si }m=1\\ 0 &\text{ c.c.} \end{cases}$$ Finally $$ \sum_{n=1}^{∞} \frac{μ(n)}{\sinh (nx)} = 2(e^{-x}-e^{-2x})$$
