I was recently Taylor-expanding `ln` around $(1,0)$. I noticed that this polynomial will have a range of input that converges between $0$ and $2$ regardless of Taylor order. I then found an expansion that did not seem to have this issue, namely:

$\lim\_{n \to +\infty} 2 \cdot \sum\limits\_{i=0}^n (\frac{1}{2i + 1} \cdot {(\frac{x-1}{x+1})}^{2i + 1}) = \ln(x)$

My question is: how is this formula derived (as in created, not the derivative) from `ln`?