# Uniform convergence of $\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^{8}x^{3}}$

Consider $$\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^{8}x^{3}}\quad \text{with x \in \left(0,\frac{\pi}{2}\right).}$$ We want to determine convergence of such series. I showed that for $x \in [\delta,\frac{\pi}{2})$ this series convergent uniformly (it's easy to see by Weierstrass-test). But what about interval.

First of all , I've tried to use opposite of Cauchy-criteria, but I hadn't found any point to show that $|\sum_{n=p}^{M}\frac{\sin(nx)}{1+n^{8}x^{3}}| \ge \epsilon$, for some constant $\epsilon > 0$.

Any hints?

## 1 Answer

Hint. Note that for $x\in (0,\pi/2)$, $$\frac{|\sin(nx)|}{1+n^{8}x^{3}}\leq g_n(x):=\frac{nx}{1+n^{8}x^{3}}\leq\frac{nx_n}{1+n^{8}x_n^{3}}=\frac{2^{2/3}}{3n^{5/3}}$$ where $x_n=1/(2^{1/3}n^{8/3})$ is such that $g_n'(x_n)=0$.

• And because of $a_{n}(x) \le \frac{const}{n^{5/3}}$ series is converge? – openspace Dec 6 '17 at 13:03
• @openspace Yes, exactly! – Robert Z Dec 6 '17 at 13:04
• thanks for hint! – openspace Dec 6 '17 at 13:05