Taking the derivative, you find . Integrating both sides, we get . Since , this becomes (after you reindex to start at one instead of zero).
Let $g(x)=\ln(1+x)$
$f(x)=g(-x)$
We know (or can easily find out) that
$g(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{x^k}{k}$
$f(x)=g(-x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{(-x)^k}{k}$
$f(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}(-1)^k\dfrac{x^k}{k}$
$f(x)=\displaystyle{\sum_{k=1}^\infty}-\dfrac{x^k}{k}$