point of symmetry of the graph of a function the graph of the function $$f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\cdots+\frac{2^{n-1}x^{2^{n-1}-1}}{1+x^{2^{n-1}}}+\cdots$$  where $x \in (-1,1)$ is symmetric about the point---
I tried to integrate the function and could not proceed to find centre of symmetry.I tried it in desmos and i got the following curve.Please help me out.Thanks
 A: Surprisingly $f(x)$ is a very simple function as shown below. To get it, we start from :
$F_0(x)=(1+x)$
$F_1(x)=(1+x)(1+x^2)=(1+x)+x^2(1+x)=1+x+x^2+x^3$
$\begin{cases}
F_2(x)=(1+x)(1+x^2)(1+x^4)=(1+x^2+x^3)+x^4(1+x^2+x^3) \\
F_2(x)=1+x^2+x^3+x^4+x^5+x^6+x^7
\end{cases}$
and so on ...
$\begin{cases}
F_n(x)=(1+x)(1+x^2)...(1+x^{(2^n)}) \\
F_n(x)=1+x+x^2+...+x^{(2^{n+1}-1)}
\end{cases}$
$$F_n(x)=\displaystyle\prod_{k=0}^{n}(1+x^{(2^k)})=\displaystyle\sum_{k=0}^{2^{n+1}-1}x^k=\frac{1-x^{(2^{n+1})}}{1-x}$$
$$\displaystyle\prod_{k=0}^{\infty}(1+x^{(2^k)})=\frac{1}{1-x}$$
$$\ln\left(\displaystyle\prod_{k=0}^{\infty}(1+x^{(2^k)})\right) =\displaystyle\sum_{k=0}^{\infty}\ln(1+x^{(2^k)})=-\ln(1-x)$$ 
The differentiation leads to :
$$\displaystyle\sum_{k=0}^{\infty}\frac{2^kx^{(2^k)-1} }{1+x^{(2^k)}}=\frac{1}{1-x}$$
We recognize $f(x)$ :
$$f(x)=\frac{1}{1-x}$$
All above is valid for $-1<x<1$. So, $f(x)$ is an arc of hyperbola.
Obviously, the symmetry is with respect to the straight line $y=1-x$ with center $(x=0\:,\:y=1)$.
 
This is in good agreement with your own graph (magnified) :

