Answer to Question #114836 in Real Analysis for Sheela John

Question

let {an} be a decreasing sequence of positive terms .prove that the series summation an sin (nX) converges uniformily on R if and only if nan tends to 0 as n tends to infinity

Answer 1

Let $\{a_n\}$ be the decreasing sequence of positive terms. Suppose $\sum_{n=1}^{\infty} a_n \sin(nx)$  converges uniformly and let the first $n$ partial sum of the given series be $S_{n}(x)=\sum_{r=1}^{n} a_r \sin(rx)$ , By the hypothesis $S_{n}(x)$ converges uniformly.Now by Cauchy's criteria of uniform convergence, for every $\epsilon >0,\exist N\in \mathbb{N}$ such that $n\geq N , m \geq N$ for sufficiently large $m,n \implies$

$|S_{n}(x)-S_{m}(x)|<\epsilon , \forall x \in \mathbb{R}$ $\iff$ $|\sum_{r=m+1}^{n} a_r \sin(rx)|<\sum_{r=m+1}^{n}|a_r \sin(rx)|$ (since, from general triangle inequality),thus

$|\sum_{r=m+1}^{n} a_r \sin(rx)|<\sum_{r=m+1}^{n}|a_r \sin(rx)|=\sum_{r=m+1}^{n}|a_r|| \sin(rx)|$ .We also know that, always $|\sin(rx)|<1$ $\iff \sum_{r=m+1}^{n} |a_r||\sin(rx)|<\sum_{r=m+1}^{n} |a_r|=\sum_{r=m+1}^{n} a_r$ (since, $a_r >0 ,\forall r \in \mathbb{N}$ .Hence choose $\epsilon = \sum_{r=m+1}^{n} a_r \iff \epsilon = \sum_{r=m+1}^{n} a_r> a_n (n-m)$ ($\because \{a_n\}$ is decreasing sequence),thus $a_n \rightarrow 0 ,$ as $n \rightarrow \infty$ (since, $\epsilon$ is finite positive real number).

Hence, we are done.