find the maclaurin series for f(x)=ln(1-x) and its radius of convergence. how do i do this? is it similar to that of ln(1+x)?
Taking the derivative, you find $\displaystyle f'(x) = -\dfrac{1}{1-x} = -\sum_{n\ge 0}x^n$. Integrating both sides, we get $\displaystyle f(x) = -\sum_{n\ge 0}\dfrac{x^{n+1}}{n+1}+C$. Since $\displaystyle f(0) = 0 = C$, this becomes $\displaystyle \ln(1-x) = -\sum_{n\ge 1}\dfrac{x^n}{n}$ (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
$g(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{x^k}{k}$
$f(x)=g(-x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{(-x)^k}{k}$
$f(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}(-1)^k\dfrac{x^k}{k}$
$f(x)=\displaystyle{\sum_{k=1}^\infty}-\dfrac{x^k}{k}$