Suppose |x|<1. Can you give any ideas on how to find the following sum? xx+1+2x2x2+1+4x4x4+1+8x8x8+1+…

Prove first that for any ${\displaystyle x\in (-1,1)}$ the identity
${\displaystyle \frac{1}{1-x}=\prod _{k\ge 0}(1+x^{2^k})}$ (1)
follows from a telescoping product or the fact that every ${\displaystyle n\in {\mathbb{N}}^{+}}$ has a unique representation in base-2. By considering ${\displaystyle \frac{d}{dx}\log (.)}$ of both sides of (1), we get
${\displaystyle \frac{1}{1-x}=\sum _{k\ge 0}\frac{2^k{x}^{2^k-1}}{1+x^{2^k}}}$
from which:
${\displaystyle \sum _{k\ge 0}\frac{2^k{x}^{2^k}}{1+x^{2^k}}=\frac{x}{1-x}}$

Evaluating
${\displaystyle \sum _{q\ge 0}\frac{2^q{x}^{2^q}}{1+x^{2^q}}}$
we obtain
${\displaystyle \sum _{q\ge 0}{2}^q\sum _{k\ge 0}{(-1)}^k{x}^{(k+1)2^q}=\sum _{n\ge 1}{x}^n\sum _{2^1\mid n}{2}^q{(-1)}^{\frac{n}{2^q}-1}}$
Now observe that
${\displaystyle \sum _{2^q\mid n}{2}^q{(-1)}^{\frac{n}{2^q}-1}=\sum _{p=0}^{v_2\left(n\right)}{2}^p{(-1)}^{\frac{n}{2^q}-1}}$
where ${\displaystyle v_2\left(n\right)}$ is the exponent of the highest power of 2 that divides n. This is
${\displaystyle \sum _{p=0}^{v_2\left(n\right)-1}{2}^p+v^{2^{v_2\left(n\right)}}-1)+2_{v_2\left(n\right)}=1}$
because ${\displaystyle \frac{n}{2^p}}$ is even unless ${\displaystyle p=v_2\left(n\right)}$
(This also goes through when n is is odd and we have one value for p, namelly zero.)
Hence the end result is
${\displaystyle \sum _{n\ge 1}{x}^n=\frac{x}{1-x}}$