Is $ \sum_{n=1}\limits^{\infty}\frac{1}{\sinh(2^{n})} $ equal to $ \frac{2}{e^{2}-1}$?

Question

Show that $$ \sum\limits_{n=1}\limits^{\infty}\frac{1}{\sinh(2^{n})}= \frac{2}{e^{2}-1}. $$

Answer 1

$$\sum_{n\geq 1}\frac{1}{\sinh(2^n)} = 2\sum_{n\geq 1}\frac{1}{e^{2^n}-e^{-2^n}} = 2\sum_{n\geq 1}\left(e^{-2^n}+e^{-3\cdot 2^n}+e^{-5\cdot 2^n}+\ldots\right)$$ and since every even integer number $m\geq 1$ can be written in a unique way as the product of a power of two and an odd number, the last series equals: $$ 2\sum_{m\geq 1}e^{-2m} = \frac{2}{e^2-1} $$ as conjectured.