Find the value of
$$\lim_{n\to\infty} \sum_{k=1}^{n}\frac{1}{\sinh 2^k}$$
Numerical approximations gives me a value of $\frac{2}{e^2-1}$.
I tried to write the sum as
$$\sum_{k=1}^{\infty}\frac{1}{\sinh 2^k}=2\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}2e^{-2^k(1+2n)}$$
I don't see how to calculate this expression.
It is a telescopic series. The identity: $$\coth(x)-\coth(2x)=\frac{1}{\sinh(2x)}\tag{1}$$ gives: $$\sum_{k=1}^{+\infty}\frac{1}{\sinh(2^k)}=\sum_{k=1}^{+\infty}\left(\coth 2^{k-1}-\coth 2^{k}\right)=\coth 1-\coth(\infty)=\color{red}{\coth 1-1}.\tag{2}$$