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$$f(x)=\sum_{n=1}^\infty\frac{x^n}n,\quad -1<x<1$$

$$f'(x)=?$$

I think it is a Riemann sum as upper limit is infinity. I tried to modify the function so as to represent it through defined integral and solve but x to the power of n is confusing me. Can someone help me to solve the problem?

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closed as off-topic by Simply Beautiful Art, Namaste, Shailesh, Daniel W. Farlow, Trevor Gunn Jul 15 '17 at 2:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Simply Beautiful Art, Namaste, Shailesh, Daniel W. Farlow, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Why not just take the derivative with respect to x? $\endgroup$ – kingW3 Jul 14 '17 at 21:32
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    $\begingroup$ It's easier than you think. Differentiate the series term by term, and see whether you recognize the result. $\endgroup$ – Arthur Jul 14 '17 at 21:32
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    $\begingroup$ Please take the time to write out formulas using MathJax. Pictures may not be legible, and secondly, they do not appear in searches. $\endgroup$ – Simply Beautiful Art Jul 14 '17 at 21:37
  • $\begingroup$ That said, you say you "tried to modify the function so as to represent it through [a definite] integral" but I don't see any attempt to do so. To avoid having your questions closed, we recommend showing us what you've tried. Shouldn't be too hard if you say you did try. $\endgroup$ – Simply Beautiful Art Jul 14 '17 at 21:41
  • $\begingroup$ @SimplyBeautifulArt I take your feedback into consideration. Thank you for informing me $\endgroup$ – R.Temur Jul 14 '17 at 21:44
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Since the function $f$ converges absolutely for $|x|<1$, the function $f$ is differentiable in $(-1,1)$, and its derivative can be computed by differentiating term by term: $$ f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n} = \sum_{n=0}^\infty x^{n} = \frac{1}{1-x}. $$ This function $f$ is an example of an analytic function.

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For $x\in (-1,1) $, $$f (x)=x+\frac {x^2}{2}+\frac {x^3}{3}+... $$ $f $ is differentiable at the open disc of convergence and

$$f'(x)=1+x+x^2+...=\frac {1}{1-x} $$

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