Finding the sum $\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \cdots$ Suppose $|x| < 1$. Can you give any ideas on how to find the following sum?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$
 A: Evaluating $$\sum_{q\ge 0} \frac{2^q x^{2^q}}{1+x^{2^q}}$$
we obtain
$$\sum_{q\ge 0} 2^q \sum_{k\ge 0} (-1)^k x^{(k+1)2^q}
= \sum_{n\ge 1} x^n \sum_{2^q|n} 2^q (-1)^{n/2^q-1}.$$
Now observe that
$$\sum_{2^q|n} 2^q (-1)^{n/2^q-1}
= \sum_{p=0}^{v_2(n)} 2^p (-1)^{n/2^p-1}$$
where  $v_2(n)$ is  the  exponent of  the highest  power  of $2$  that
divides $n.$ This is
$$-\sum_{p=0}^{v_2(n)-1} 2^p + 2^{v_2(n)}
= - (2^{v_2(n)}-1) + 2^{v_2(n)} = 1.$$
because $n/2^p$ is even unless $p=v_2(n).$ 
(This also goes through when $n$ is odd and we have one value for $p$, namely zero.)
Hence the end result is
$$\sum_{n\ge 1} x^n  = \frac{x}{1-x}.$$
A: Prove first that for any $x\in(-1,1)$ the identity
$$ \frac{1}{1-x}=\prod_{k\geq 0}\left(1+x^{2^k}\right) \tag{1}$$
follows from a telescoping product or the fact that every $n\in\mathbb{N}^*$ has a unique representation in base-$2$. By considering $\frac{d}{dx}\log(\cdot)$ of both sides of $(1)$, we get:
$$ \frac{1}{1-x} = \sum_{k\geq 0}\frac{2^k x^{2^k-1}}{1+x^{2^k}}\tag{2}$$
from which:
$$\sum_{k\geq 0}\frac{2^k x^{2^k}}{1+x^{2^k}}=\color{red}{\frac{x}{1-x}}.\tag{3}$$

Essentially the same "trick" allows us to derive Euler pentagonal number theorem from Jacobi's triple product, for instance.
