Show that series converges uniformly

convergence-divergencepower seriessequences-and-seriesuniform-convergence

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!

Best Answer

  1. It is simpler to use the Weierstrass $M$ test: the series $\sum_{n=0}^\infty\frac1{2n}a^{2n}$ (by, say, the ratio test) and, since$$(\forall x\in[-a,a]):\left|\frac1{2n}x^{2n}\right|\leqslant\frac1{2n}a^{2n},$$your series converges uniformly (and absolutely).
  2. Since $\left(\frac1{2n}x^{2n}\right)'=x^{2n-1}$ and since the series $\sum_{n=1}^\infty x^{2n-1}$ converges uniformly on $[-a,a]$ to $\frac x{1-x^2}$ (by the same argument), you have $f'(x)=\frac x{1-x^2}$.
  3. Since $f(0)=0$,$$f(x)=\int_0^x\frac t{1-t^2}\,\mathrm dt=\log\left(\frac1{\sqrt{1-x^2}}\right).$$
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