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?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$

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.
