TITLE: Evaluating $\sum_{n=0}^{\infty } 2^{-n} \tanh (2^{-n})$
QUESTION [5 upvotes]: Reading in some tables pages I found 
$$\sum _{n=0}^{\infty } 2^{-n} \tanh \left(2^{-n}\right)=\tanh (1) \left(1+\coth ^2(1)-\coth (1)\right)$$ 
I try to split in two sum using the roots of the $\tanh$ but I could not to get the correct answer.

REPLY [5 votes]: Alternatively, one may notice that we have
$$
\tanh (x) = 2 \coth (2x)-\coth (x), \quad x \in (0,1], \tag1
$$ then putting $x:=\dfrac1{2^n}$, $n=0,1,2,\ldots$ in $(1)$ and multiplying by $ \dfrac1{2^n}$ implies
$$
2^{-n}\tanh(2^{-n}) =2^{-(n-1)} \coth (2^{-(n-1)}) - 2^{-n} \coth (2^{-n}) \tag2
$$ Summing $(2)$ from $n=0$ to $n=N$ gives by telescoping, 
$$
\sum_{n=0}^{N}2^{-n}\tanh(2^{-n}) =2 \coth (2) - 2^{-N} \coth (2^{-N})\tag3
$$ and, using $\displaystyle 2^{-N}\coth(2^{-N}) \to 1$, as $N \to +\infty$, leads to 

$$
\sum_{n=0}^{\infty}2^{-n}\tanh(2^{-n}) =\color{blue}{2 \coth (2) - 1}.\tag4
$$