I will try again asking my question: I have $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$, for x$\in$R. Then I have used wolframalpha finding the sum function: https://www.wolframalpha.com/input/?i=sum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%5Cfrac%7Bn*x%5E%7Bn%7D%7D%7Bn%2B1%7D And with Maple I have reduced the sum function to: $\frac{1}{1-x}+\frac{ln(1-x)}{x}$ for |x|<1. But how can I show it formally that this is the sum function?

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    $\begingroup$ "For $x \in \Bbb{R}$" your power series does not converge for $x > 1$. $\endgroup$ – Clement Yung May 28 at 22:01
  • $\begingroup$ Are you asking how to derive the closed form for the series, or are you asking how to determine for which real numbers it converges? $\endgroup$ – Brian M. Scott May 28 at 22:08

Note that for $|x|<1$

$$\begin{align} \sum_{n=1}^\infty \frac{nx^n}{n+1}&=\sum_{n=1}^\infty x^n-\frac1x\sum_{n=1}^\infty \frac{x^{n+1}}{n+1}\\\\ &=\frac x{1-x}-\frac1x\int_0^x \sum_{n=1}^\infty t^n\,dt\\\\ &=\frac x{1-x}-\frac1x\int_0^x \frac{t}{1-t}\,dt \end{align}$$

Can you finish now?

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  • $\begingroup$ Thank you it makes sence $\endgroup$ – Lifeni May 28 at 22:21
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola May 28 at 22:22
  • $\begingroup$ Hi Mark ! I know that you enjoy playing with bounds of logarithms. Yesterday, just for the fun ot it, I used some of them. Would you please have a look at math.stackexchange.com/questions/3695080/… and tell me what you think. Thanks & cheers. $\endgroup$ – Claude Leibovici May 29 at 5:11

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