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).
 A: Like all sums ever, this sum telescopes: we can write
$$
   \frac1{a+1} = \frac1{a-1} - \frac2{a^2-1} 
$$
and more generally
$$
   \frac{2^n}{a^{2^n}+1} = \frac{2^n}{a^{2^n}-1} - \frac{2^{n+1}}{a^{2^{n+1}}-1}
$$
and therefore
$$
   \sum_{n=0}^{N} \frac{2^n}{a^{2^n}+1} = \sum_{n=0}^N \left(\frac{2^n}{a^{2^n}-1} - \frac{2^{n+1}}{a^{2^{n+1}}-1}\right) = \frac1{a-1} - \frac{2^{N+1}}{a^{2^{N+1}}-1}.
$$
If $\lim_{n \to \infty} \frac{n}{a^n-1} = 0$, which happens when $|a|>1$, then the infinite sum simplifies to $\frac1{a-1}$.

Okay, but seriously, what's the deal with telescoping sums?
If you don't notice the identities above, I don't blame you: Mathematica doesn't either, and in fact Mathematica can't tell that this sum is $1$ when $a=2$ unless you ask it to approximate the sum by numerical methods.
However, telescoping identities appear from partial sums, and so if you don't know what to do, the partial sums might be a good place to look. Here, summing the $n=0,1,2$ terms gives us
$$
\frac{a^6+2 a^5+3 a^4+4 a^3+5 a^2+6 a+7}{(a+1) \left(a^2+1\right) \left(a^4+1\right)}
$$
(you might not want to expand the numerator but you should be able to see that this is the denominator) and now it's much less of a stretch to spot that this expression becomes much nicer when we multiply top and bottom by $a-1$. If we do, we get
$$
\frac{a^7+a^6+a^5+a^4+a^3+a^2+a-7}{a^8-1} = \frac1{a-1}-\frac{8}{a^8-1}
$$
which directly leads to the telescoping solution.
A: 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 $2$.
