Power Series Expansion for Logarithm of x/Formulation 1

Theorem
valid for all $x \in \R$ such that $-1 < x < 1$.

Proof
From Power Series Expansion for $\dfrac 1 2 \ln \left({\dfrac {1 + x} {1 - x} }\right)$:
 * $(1): \quad \displaystyle \frac 1 2 \ln \left({\frac {1 + x} {1 - x} }\right) = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}$

for $-1 < x < 1$.

Let $z = \dfrac {1 + x} {1 - x}$

Then:

Then we have:
 * $\displaystyle \lim_{x \mathop \to 1^-} \dfrac {1 + x} {1 - x} \to +\infty$

and:
 * $\displaystyle \lim_{x \mathop \to -1^+} \dfrac {1 + x} {1 - x} \to 0$

Thus when $x \in \left({-1 \,.\,.\, 1}\right)$ we have that $z \in \left({0 \,.\,.\, \to +\infty}\right)$.

Thus, substituting $z$ for $\dfrac {1 + x} {1 - x}$ in $(1)$ gives the result.