Answer
$x+\dfrac{1}{2}x^2+\dfrac{5}{6}x^3+\dfrac{7}{12} x^4+...$
Work Step by Step
We know that the Taylor Series for $ \ln (1+x)$ and $\dfrac{1}{1-x}$ is defined as:
$ \ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$ and $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
Now,
$(\ln (1+x)) \cdot (\dfrac{1}{1-x})=(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-..)\cdot (1+x+x^2+....+x^n)$
or, $=x+\dfrac{1}{2}x^2+\dfrac{5}{6}x^3+\dfrac{7}{12} x^4+...$
