Evaluating $\sum_{n=0}^{\infty } 2^{-n} \tanh (2^{-n})$ 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.
 A: 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 

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

A: We have:
$$ \tanh(x) = \sum_{m\geq 0}\frac{8x}{\pi^2(2m+1)^2+4x^2}\tag{1}$$
so:
$$\begin{eqnarray*} \sum_{n\geq 0} 2^{-n}\tanh(2^{-n})&=&2\sum_{n\geq 0}\sum_{m\geq 0}\frac{1}{1+4^n \pi^2(m+1/2)^2}\\&=&2\sum_{n\geq 0}\sum_{m\geq 0}\left(\frac{1}{4^n \pi^2(m+1/2)^2}-\frac{1}{4^{2n} \pi^4(m+1/2)^4}+\ldots\right)\\&=&2\sum_{m\geq 0}\left(\frac{4}{(4-1) \pi^2(m+1/2)^2}-\frac{4^{2}}{(4^{2}-1) \pi^4(m+1/2)^4}+\ldots\right)\\&=&2\sum_{k\geq 1}\sum_{m\geq 0}\frac{4^k(-1)^{k+1}}{(4^k-1)\pi^{2k}(m+1/2)^{2k}}\\&=&2\sum_{k\geq 1}\frac{4^k \zeta(2k)(-1)^{k+1}}{\pi^{2k}}\tag{2}\end{eqnarray*}$$
and at last one just needs to exploit:
$$ \sum_{k\geq 1}\zeta(2k)(-1)^k x^{2k} = \frac{1-\pi x \coth(\pi x)}{2}\tag{3}$$ 
to get:

$$ \sum_{n\geq 0}2^{-n}\tanh(2^{-n}) = \color{red}{2\coth(2) -1} .\tag{4}$$

