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I'm new to this subject and would very much appreciate your help with this question. I'm not really sure about how to approach this.

$$f(x) = \sum_1^\infty\frac{1}{n}x^n$$

  1. If I'm not mistaken the domain of convergence of this function is $x=[-1,1)$.
  2. I need to check if it converges uniformly on $ [0,1)$.
  3. I need to check if it converges uniformly on $(-1,0]$.
  4. To calculate the function which the series converges to within the radius of convergence, please notice that the function should be for the whole radius of convergence and only for $[-a,a]$ for $0\leq a<R$.
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  • $\begingroup$ @AndréNicolas yep, i edited it, thanks $\endgroup$ – Simba Jun 8 '13 at 12:47
  • $\begingroup$ The radius of convergence is a number, not a set. $\endgroup$ – GEdgar Jun 8 '13 at 12:59
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    $\begingroup$ The radius of convergence is a number, not an interval. Here [-1,1) is the domain of convergence. $\endgroup$ – Did Jun 8 '13 at 12:59
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  1. The radius of convergence $R=1$ and the interval of convergence is $[-1,1)$

  2. Since $\lim_{x\to 1} \frac{x^n}{n}=\frac{1}{n}$ and the series $$\sum_{n=1}^\infty \frac{1}{n}$$ is divergent then the given series isn't uniform convergent on $[0,1)$

  3. Let $s=-x$ then we have $$\sum_{n=1}^\infty \frac{x^n}{n}=\sum_{n=1}^\infty \frac{(-1)^ns^n}{n}\quad s\in[0,1)$$ so this series is alternating so we have $$\left|\sum_{k=n+1}^\infty \frac{(-1)^ks^k}{k}\right|\leq\frac{1}{n}\to0\quad \forall s$$ hence we have the uniform convergence on $(-1,0]$ for the given series.
  4. We can prove easily that the series $\sum_n t^n$ is uniformly convergent on every $[-x,x]$ with $0<x<1$ so by integration term by term we have $$-\log(1-x)=\int_0^x\frac{dt}{1-t}=\int_0^x \sum_{n=0}^\infty t^n dt=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=\sum_{n=1}^\infty\frac{x^{n}}{n}=f(x)$$
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