Question

show that the series ∑∞n=1 sin n theta/n does not converge uniformly on the interval ]0,2π[.

Answer 1

$\sum_{n=1}^{\infty} \frac{sin n\theta}{n} , \theta\in]0,2\pi[.\\ Then,\\ f_n=\frac{sin n\theta}{n} \to0as n\to \infty\\ f'_n=cosn\theta\\ f'_n(\pi)=(-1)^n\text{ does not converge as } n\to \infty \\ \text{Thus, the given series is not uniformly covergent.}\\$