Evaluating infinite series with $a^{(2^n)} + 1$ form denominator Evaluate: 
$$\sum_{n = 0}^\infty \frac{2^n}{a^ \left( 2^n \right) + 1 } $$
This doesn't seem to have factors in the denominator. I tried stuff like expanding this series and then multiplying by a factors like $ \frac{1}{x-1} $ and then shifting the terms, subtracting the series.
(Edit: Sorry for not telling that a $ \gt $ 1).

Alternative approach: for any $a>0$ we have
$$ 1-a^2 = (1+a)(1-a),\quad 1-a^4=(1+a^2)(1+a)(1-a),\quad 1-a^8 = (1+a^4)(1+a^2)(1+a+1)(1-a),\qquad \frac{1-a^{2^N}}{1-a}=\prod_{k=0}^{N-1}\left(1+a^{2^k} \right)\tag{1}$$
and
$$ \log(1-a^{2^N})-\log(1-a) = \sum_{k=0}^{N-1} \log\left(a^{2^k}+1\right)\tag{2} $$
so by applying $\frac{d}{da}$ to both sides of $(2)$, then multiplying by $a$:
$$ \frac{2^N a^{2^N}}{a^{2^N}-1}-\frac{a}{a-1}=\sum_{k=0}^{N-1}\frac{2^k a^{2^{k}}}{a^{2^k}+1}\tag{3}$$
then mapping $a$ into $\frac{1}{a}$:
$$ \frac{2^N}{1-a^{2^N}}+\frac{1}{a-1}=\sum_{k=0}^{N-1}\frac{2^k}{a^{2^k}+1}.\tag{4}$$
If $a>1$, by considering the limit of both sides of $(4)$ as $N\to +\infty$ we get
$$ \sum_{k=0}^{N-1}\frac{2^k}{a^{2^k}+1} = \frac{1}{a-1}.\tag{5} $$
Notice that this is equivalent to the fact that every natural number has a unique representation in base $2$.
