What is the derivative of $\sum_{n=1}^\infty\frac{x^n}{n}$?

Question

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

$$f'(x)=?$$

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?

Answer 1

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.

Answer 2

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

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