0
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ The power series does converge uniformly in any segment strictly inside of the interval of convergence. If you are allowed to use that, you're done. Otherwise, you can start by proving that if $\sum \sup_{x \in [-a,a]} |f_n(x)|$ converges (we say that $f_n$ converges normally) then $\sum f_n$ converges uniformly on [-a,a] and conclude. For the derivative, consider the series of the derivatives then take the integral. $\endgroup$ – Hashem Ben Abdelbaki May 16 '20 at 14:29
1
$\begingroup$
  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 $f(0)=0$,$$f(x)=\int_0^x\frac t{1-t^2}\,\mathrm dt=\log\left(\frac1{\sqrt{1-x^2}}\right).$$
$\endgroup$
9
  • $\begingroup$ José, thanks for the info. When I apply the ratio test to $\frac{1}{2n}a^{2n}$ I get that the limit goes to $a^{2}$, is that correct? and does that prove that the series converge for all $|x|<1$ or what was your thoughts about that? $\endgroup$ – Eod Enaj May 16 '20 at 17:34
  • $\begingroup$ The limit is $a^2$ and therefore the limit is smaller than $1$ of $0\leqslant a<1$. Note that you cannot take $a=1$, since your series doesn't converge when $x=1$. But, by the Weierstrass $M$ test, it converges uniformly on any interval $[-a,a]$ when $0\leqslant a<1$. Si, given $x\in(-1,1)$, take some $a$ such that $|x|\leqslant a<1$ and then, since the series converes uniformly on $[-a,a]$, in particular it converges at the point $x$. $\endgroup$ – José Carlos Santos May 16 '20 at 17:50
  • $\begingroup$ That makes sense, thanks a lot! $\endgroup$ – Eod Enaj May 16 '20 at 18:05
  • $\begingroup$ Yes of course. Last question, do you have any idea how we can use the result from c) to show that $\sum_{n=1}^{\infty}\frac{1}{2^{n}n} = log(2)$? I can't seem to get it right. $\endgroup$ – Eod Enaj May 16 '20 at 19:33
  • $\begingroup$ Sure:\begin{align}\sum_{n=1}^\infty\frac1{2^nn}&=2\sum_{n=1}^\infty\frac1{2n}\left(\frac1{\sqrt2}\right)^{2n}\\&=2\log\left(\frac1{\sqrt{1-\left(\frac1{\sqrt2}\right)^2}}\right)\\&=2\log\left(\sqrt2\right)\\&=\log(2).\end{align} $\endgroup$ – José Carlos Santos May 16 '20 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.