Power Series Expansion for Logarithm of 1 + x/Corollary

Corollary to Power Series Expansion for $\map \ln {1 + x}$

$\ds \map \ln {1 - x}$

$=$

$\ds -\sum_{n \mathop = 1}^\infty \frac {x^n} n$

$\ds $

$=$

$\ds -x - \frac {x^2} 2 - \frac {x^3} 3 - \frac {x^4} 4 - \cdots$

valid for $-1 \le x < 1$.

$\ds \map \ln {1 + x} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$

Then:

$\ds \map \ln {1 - x}$

$=$

$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\paren {-x}^n} n$

substituting $x \to -x$

$\ds $

$=$

$\ds -\sum_{n \mathop = 1}^\infty \paren {-1}^{2 n} \frac {x^n} n$

$\ds $

$=$

$\ds -\sum_{n \mathop = 1}^\infty \frac {x^n} n$

$\blacksquare$