- #1

twoflower

- 368

- 0

I've little probles with this one:

Analyse uniform and local uniform convergence of this series of functions:

[tex]

\sum_{k = 1}^{\infty} \frac{\cos kx}{k}

[/tex]

I'm trying to solve it using Weierstrass' criterion, ie.

[tex]

\mbox{Let } f_n \mbox{ are defined on } 0 \neq M \subset \mathbb{R}\mbox{, let }

S_n := \sup_{x \in M} \left| f_{n}(x)\right|, n \in \mathbb{N}. \mbox{ If }

\sum_{n=1}^{\infty} S_n < \infty\mbox{, then } \sum_{n=1}^{\infty} f_{n}(x) \rightrightarrows \mbox{ on } M.

[/tex]

I can see from this it won't converge for [itex]x = 2n\pi[/itex], because then

[tex]

\sup_{x \in \mathbb{R}} \left|\frac{\cos kx}{k}\right| = \frac{1}{k} =: S_{k} \mbox{ and thus } \ \sum S_k \ \mbox{diverges} \Rightarrow \sum_{n=1}^{\infty} f_{n} \mbox{ diverges}

[/tex]

So it could converge on [itex]x \in [2n\pi + \epsilon, 2(n+1)\pi - \epsilon], \mbox{where } \epsilon \in (0, \pi)[/itex] but I can't see how could I prove it, at least using this theorem. I also took a look at Abel's criterion, Dirichlet's criterion and theorems about change of sum and derivation\integral, but I didn't see the way to prove that.

Could you help me please?

Thank you.