Prove $\sum_{n=0}^{\infty}a_n \cos{(nx)}$ does not uniformly converge in $\left( 0,2\pi \right)$ with the conditions below

analysisconvergence-divergencereal-analysissequences-and-seriesuniform-convergence

Let $a_n$ be a sequence of real numbers such that:

  • $a_n \ge 0$
  • $\forall n\in \mathbb{N}: a_n \ge a_{n+1}$
  • $\lim_{n \to \infty} a_n =0$
  • $\sum_{n=0}^{\infty } a_n =\infty $

Consider the sum $\sum_{n=0}^{\infty}a_n \cos{(nx)}$ . Prove:

  1. $\sum_{n=0}^{\infty}a_n \cos{(nx)}$ uniformly converges in $\left[ \frac{\pi}{2},\pi \right]$
  2. $\sum_{n=0}^{\infty}a_n \cos{(nx)}$ does not uniformly converge in $\left( 0,2\pi \right)$

I only need help with 2 (I already solved 1). I'm trying to solve it for almost a day now. I'd appreciate if you could explain how to solve it. Thanks.

Best Answer

Since $\sum a_n=\infty$, then we have $$\exists\epsilon>0, \forall N, \exists m, n, (m>n\ge N), \Longrightarrow \sum_{k=n}^{m}a_k\ge\frac \epsilon{\cos(1)}$$

Now, let $x=\frac1{m}$,

$$\sum_{k=n}^{m}a_k\cos(kx)=\sum_{k=n}^{m}a_k\cos\left(\frac k{m}\right)\ge \sum_{k=n}^{m}a_k\cos(1)\ge\epsilon$$

hence, it is not uniformly convergent on $(0,2\pi)$

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