Prove that $\sum_{n=0}^{\infty} \frac{n}{n+1}x^{n+1}=\frac{x}{1-x} + \ln(1-x)$ Prove that $\sum_{n=0}^{\infty} \frac{n}{n+1}x^{n+1}=\frac{x}{1-x} + \ln(1-x)$
Right off the bat, I noticed that $\frac{x}{1-x}=x\frac{1}{1-x}= x \sum_{n=0}^{\infty}x^n$, and I know that $\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}$. If somehow I can express $\ln(1-x)$ in terms of $\ln(1+x)$, then I'm set. Can I do something like $\ln(1-x)=\ln(1+(-x))=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x)^n}{n}$?
What's next? Should I add these 2 together? Can someone give me an exact answer? Also, the radius of convergence is supposed to be $(-1, 1)$ right?

\begin{equation}
f(x) = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n+1}
\end{equation}
$$f'(x)=\sum_{n=0}^{\infty} n x^n=x\sum_{n=0}^{\infty} n x^{n-1}=x\left(\sum_{n=0}^{\infty} x^{n}\right)'=\frac{x}{(x-1)^2}$$
$$f'(x)=\frac{x}{(x-1)^2}=\frac{x-1+1}{(x-1)^2}=\frac 1{x-1}+\frac 1 {(x-1)^2}$$ Just integrate.
