1) I have the series
$$ \sum_{n=1}^{\infty}\frac{1}{2n}x^{2n}$$
, where $x\in\mathbb{R}$ and $0<a\leq1$.
I would like to show that the series converges uniformly on the interval $[-a;a]$ and I was thinking that I can use the Cauchy-Hadamand formula to show that it has a convergens radius of 1 and therefore it is uniformly convergent in a given interval $s<r$ where r is
$$r^{-1} = \lim_{n\to\infty}sup|a_{n}|^{\frac{1}{n}} = \lim_{n\to\infty}sup|\frac{1}{2n}|^{\frac{1}{n}} = \lim_{n\to\infty}sup1=1 $$
according to a theorem in my textbook. But my question is, have I really now shown that is in fact uniformly convergent on the interval [-a;a] or do I need a different approach?
2) From here, how can I show that the function for the series is differentiable and that the following is true?
$$ f'(x) = \frac{x}{1-x^{2}} $$
, for $x\in[-a;a]$
This question I have no idea about other than one of Abel's theorems says that a series like the one I have is continuous in x=r, but I don't know how to use it. Any ideas?
3) To find the function can I just now integrate $ f'(x) = \frac{x}{1-x^{2}} $ or do I need to use a more generel method?
Any help is greatly appreciated!
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