How to find the power series representation of $(x+1)\ln(1-x)$ using the geometric series?
I thought i could multiply the $\ln(1-x)$ through the brackets and than add up the series using manipulation of the geometric series but it didnt quite work?
The answer is $\displaystyle -x-\sum_{n=1}^\infty \frac{2n+1}{n(n+1)}x^{n+1}$.
Best Answer
Hint: $\displaystyle (\ln (1-x))'=-\frac{1}{1-x}$
Follow up: It's easy to find $\displaystyle -(x+1)\ln (1-x)=(x+1)\sum \limits_{n=0}^{+\infty}\frac{x^{n+1}}{n+1}$.
Then $$\begin{align} \displaystyle (x+1)\sum \limits_{n=0}^{+\infty}\frac{x^{n+1}}{n+1}&=\sum \limits_{n=0}^{+\infty}\left(\frac{x^{n+2}}{n+1}\right)+\sum \limits_{n=0}^{+\infty}\left(\frac{x^{n+1}}{n+1}\right)\\ &=\sum \limits_{\color{blue}{n=1}}^{+\infty}\left(\frac{x^{n+1}}{n}\right)+\left[\sum \limits _{\color{blue}{n=1}}^{+\infty}\left(\frac{x^{n+1}}{n+1}\right)+x\right]\\ &=\sum \limits_{n=1}^{+\infty}\left(\frac{x^{n+1}}{n}+\frac{x^{n+1}}{n+1}\right)+x\\ &\, \vdots\end{align}$$
(Of course $x$ ranges over $(-1,1)$ throughout).