# Power Series Expansion for Logarithm of 1 + x/Corollary

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

 $\displaystyle \ln \paren {1 - x}$ $=$ $\displaystyle -\sum_{n \mathop = 1}^\infty \frac {x^n} n$ $\displaystyle$ $=$ $\displaystyle -x - \frac {x^2} 2 - \frac {x^3} 3 - \frac {x^4} 4 - \cdots$

valid for $-1 < x < 1$.

## Proof

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

Then:

 $\displaystyle \ln \paren {1 - x}$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\paren {-x}^n} n$ substituting $x \to -x$ $\displaystyle$ $=$ $\displaystyle -\sum_{n \mathop = 1}^\infty \paren {-1}^{2 n} \frac {x^n} n$ $\displaystyle$ $=$ $\displaystyle -\sum_{n \mathop = 1}^\infty \frac {x^n} n$

$\blacksquare$