It is true that $\log(y)=2\sum_{k=0}^\infty\frac{((y-1)/(y+1))^{2k+1}}{2k+1}$ for $y>0$?

Question

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.

Answer 1

I am going to use, as you do, the inverse function of

$$f(y)=x=\dfrac{y-1}{y+1}$$

which is $g(x)=y=\dfrac{1+x}{1-x}$

But treat the relationship to be established in a different manner, because it amounts to

$$\ln\left(\dfrac{1+x}{1-x}\right)=2\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}$$

which is a known result (see formula (20) in (http://mathworld.wolfram.com/SeriesExpansion.html)), this formula being a consequence of the substraction of the classical expansions:

$$\cases{\ln(1-x)=\sum_{p=1}^\infty\frac{x^{p}}{p}\\\ln(1+x)=\sum_{p=1}^\infty (-1)^{p+1}\frac{x^{p}}{p}}$$

where only the odd powers of $x$ remain after substraction.