2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ "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
1
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.