Power Series Expansion for Logarithm of x/Formulation 2

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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

From the corollary to Power Series Expansion for $\map \ln {1 + x}$:

\(\displaystyle \map \ln {1 - x}\) \(=\) \(\displaystyle -\sum_{n \mathop = 1}^\infty \frac {x^n} n\) 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.

$\blacksquare$


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