# Power Series Expansion for Logarithm of x/Formulation 2

## Theorem

 $\displaystyle \ln x$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \paren {\frac {x - 1} x}^n$ $\displaystyle$ $=$ $\displaystyle \frac {x - 1} x + \frac 1 2 \paren {\frac {x - 1} x}^2 + \frac 1 3 \paren {\frac {x - 1} x}^3 + \cdots$

valid for all $x \in \R$ such that $x \ge \dfrac 1 2$.

## Proof

 $\displaystyle \map \ln {1 - x}$ $=$ $\displaystyle -\sum_{n \mathop = 1}^\infty \frac {x^n} n$ for $-1 < x \ge 1$ $\text {(1)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle \map \ln {\frac 1 {1 - x} }$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {x^n} n$ Logarithm of Reciprocal

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

Then:

 $\displaystyle z$ $=$ $\displaystyle \dfrac 1 {1 - x}$ $\displaystyle \leadsto \ \$ $\displaystyle z \paren {1 - x}$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle z - z x$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle z x$ $=$ $\displaystyle z - 1$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \frac {z - 1} z$

Then we have:

$\displaystyle \lim_{x \mathop \to 1^-} \dfrac 1 {1 - x} \to +\infty$

and:

$\displaystyle \lim_{x \mathop \to -1^+} \dfrac 1 {1 - x} \to \frac 1 2$

Thus when $x \in \openint {-1} 1$ we have that $z \in \hointr {\dfrac 1 2} \to$.

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

$\blacksquare$