Sum Representation of log(1 + x)

Question

$\log(1+x) = \sum_{k=1}^{\infty} \left(\dfrac{x}{1+x}\right)^{k} \dfrac{1}{k} = \sum_{k=1}^{\infty} \left(1 - \dfrac{1}{1+x}\right)^k \dfrac{1}{k}$

Why is this true? The most sum representation of $\log(1+x)$ (in this case a MacLaurin series) is $\sum_{k=1}^{\infty} (-1)^{k+1} \dfrac{x^k}{k}$

Answer 1

From \begin{align} \ln(1+x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{k}}{k} \end{align} let $x = -u$ for which \begin{align} \ln(1-u) = - \sum_{k=1}^{\infty} \frac{u^{k}}{k} \end{align} Now let $u = \frac{x}{1+x}$ to obtain \begin{align} \ln(1+x) = \sum_{k=1}^{\infty} \frac{1}{k} \left( \frac{x}{1+x} \right)^{k} \end{align}

Answer 2

Hint: Differentiate the right-hand side term by term. We get a familiar series. Note that there are restrictions on $x$.