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?

Know:
\begin{equation}
\frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n} \implies \frac{1}{1-x} - 1 = \sum_{n=1}^{\infty}x^{n}, \label{first series}
\end{equation}
which holds for $|x| < 1$, and
\begin{equation}
\log(1+x) = -\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n} \implies \log(1-x) = -\sum_{n=1}^{\infty}\frac{1}{n}x^{n}, \label{second series}
\end{equation}
also valid for $|x| < 1$. If we now look at our series of interest,
\begin{equation}
f(x) = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n+1},
\end{equation}
we can use an index substitution $k = n+1 \iff n = k-1$ so that
\begin{align}
f(x) &= \sum_{k=1}^{\infty}\frac{k-1}{k}x^{k}\\
&= \sum_{k=1}^{\infty}\left(1-\frac{1}{k}\right)x^{k},\\
\end{align}
and under the condition that our series converges, we can use the linearity of the series operator to show
\begin{equation}
f(x) = \sum_{k=1}^{\infty}x^{k} + \left(-\sum_{k=1}^{\infty}\frac{1}{k}x^{k}\right),
\end{equation}
provided both the individual series converge (which they do). Notice that on the right-hand side of the above equation, the leftmost series is equal to $1/(1-x) - 1$ (from \eqref{first series}), and the rightmost series is equal to $\log(1-x)$ (from \eqref{second series}). Therefore
\begin{align}
f(x) &= \frac{1}{1-x} - 1 + \log(1-x)\\
&= \frac{x}{1-x} + \log(1-x).
\end{align}
