Power Series Expansion for Logarithm of 1 + x/Corollary

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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

By Power Series Expansion for $\ln \paren {1 + x}$:

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

Then:

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

$\blacksquare$