Power Series Expansion for Logarithm of 1 + x

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Theorem

The natural logarithm function has a power series expansion:

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

valid for all $x \in \R$ such that $-1 < x \le 1$.


Corollary

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

valid for $-1 < x < 1$.


Proof

From Sum of Infinite Geometric Progression, putting $-x$ for $x$:

$(1): \quad \displaystyle \sum_{n \mathop = 0}^\infty \paren {-x}^n = \frac 1 {1 + x}$

for $-1 < x < 1$.


From Power Series Converges Uniformly within Radius of Convergence, $(1)$ is uniformly convergent on every closed interval within the interval $\left({-1 \,.\,.\,. 1}\right)$.

From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

\(\displaystyle \int_0^x \frac 1 {1 + t} \rd t\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t}^n \rd t\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \ln \paren {1 + x}\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{n + 1} } {n + 1}\) $\quad$ Primitive of Reciprocal and Integral of Power $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \ln \paren {1 + x}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n\) $\quad$ letting $n \to n - 1$ $\quad$

$\blacksquare$


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