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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.

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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 $\Bbb R$.

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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 $x$.

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