# Power Series Expansion for Logarithm of 1 + x over 1 + x

## Theorem

 $\displaystyle \frac {\ln \left({1 + x}\right)} {1 + x}$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n + 1} H_n x^n$ $\displaystyle$ $=$ $\displaystyle x - H_2 x^2 + H_3 x^3 - H_4 x^4 + \cdots$

where $H_n$ denotes the $n$th harmonic number:

$H_n = \displaystyle \sum_{r \mathop = 1}^n \dfrac 1 r = 1 + \dfrac 1 2 + \dfrac 1 3 \cdots + \dfrac 1 r$

valid for all $x \in \R$ such that $\left\lvert{x}\right\rvert < 1$.

## Proof

Let $f \left({x}\right) = \dfrac {\ln \left({1 + x}\right)} {1 + x}$.

By definition of Maclaurin series:

$(1): \quad f \left({x}\right) \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} f^{\left({n}\right)} \left({0}\right)$

where $f^{\left({n}\right)} \left({0}\right)$ denotes the $n$th derivative of $f$ with respect to $x$ evaluated at $x = 0$.

$\dfrac {\d^n} {\d x^n} \dfrac {\ln \left({1 + x}\right)} {1 + x} = \left({-1}\right)^{n + 1} n! \dfrac {H_n - \ln \left({1 + x}\right)} {\left({1 + x}\right)^{n + 1} }$

The result follows by setting $x = 0$ and substituting for $f^{\left({n}\right)} \left({0}\right)$ in $(1)$.

$\blacksquare$