# Absolute convergence and continuty of $\sum_{n=1}^\infty \sin(\frac{x}{n^4})\cos(nx)$

Question: Prove that the series $$\sum_{n=1}^\infty \sin(\frac{x}{n^4})\cos(nx)$$ is converges absolutely, and is continuous on $$\mathbb{R}$$.

Attempt: I can readily see that the series converges absolutely on some closed interval $$[-b,b]$$, since for any $$x\in\mathbb{R}$$ we have $$|\sin(x)|\leq |x|$$ Such that $$\sum_{n=1}^\infty \bigg|\sin\bigg(\frac{x}{n^4}\bigg)\cos(nx)\bigg| \leq \sum_{n=1}^\infty \bigg|\frac{x}{n^4}\cos(nx)\bigg| \leq \sum_{n=1}^\infty \bigg|\frac{x}{n^4}\bigg| \leq \sum_{n=1}^\infty \bigg|\frac{b}{n^4}\bigg|$$

Which converges by absolutely by comparison with $$\sum_{n=1}^\infty \frac{1}{n^2}$$, and I've used $$\cos(nx)\leq 1$$.

However, this is only on some closed interval, I do not think the series converges uniformly on $$\mathbb{R}$$? So how do I show continuity? - and absolute convergence on $$\mathbb{R}$$ in general, rather than just a closed interval?

Can I perhaps say, so long as the series is absolutely convergent on $$[-\pi,\pi]$$, it must be convergent on the rest of $$\mathbb{R}$$, given the periodic behaviour of the functions? Would this also imply continuity? On a closed interval, the series converges uniformly by Weierstrass M-test, and uniform converges preserves continuity.

You have shown that $$\sum_{n=1}^\infty \sin(\frac{x}{n^4})\cos(nx)$$ converges uniformly on every interval $$[-b, b]$$.
It follows that $$f(x) = \sum_{n=1}^\infty \sin(\frac{x}{n^4})\cos(nx)$$ is continuous on $$[-b, b]$$ for arbitrary $$b > 0$$, and therefore continuous on $$\Bbb R$$.
Continuity is a local property: for $$x\in \mathbb{R}$$, there exists $$b$$ such that $$x \in (-b,b)$$. By uniform convergence on $$[-b,b]$$, the limit is continuous on $$[-b,b]$$, hence at $$x$$.