The question arise from the Taylor expansion around $x=0$ of $f(x):=\log((1+x)/(1-x))$ for $x\in(-1,1)$. If Im not wrong
$$\mathcal T(f,0)=2\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1},\quad\text{for }|x|<1$$
Because the series converges for $|x|<1$ I assumed that
$$\log((1+x)/(1-x))=\mathcal T(f,0),\quad\text{whenever }|x|<1$$
If I define $y:=(1+x)/(1-x)$ then $x=(y-1)/(y+1)$, if I substitute this in $\mathcal T(f,0)$ I get the expression
$$\mathcal T(f,0)=2\sum_{k=0}^\infty \frac{((y-1)/(y+1))^{2k+1}}{2k+1},\quad\text{for }y>0$$
then the question of the title, it is true that
$$\log(y)=2\sum_{k=0}^\infty\frac{((y-1)/(y+1))^{2k+1}}{2k+1}$$ for $y>0$? Can you confirm or disprove this hypothesis? Thank you.