Does $\sum_{n=1}^{\infty} a_n$ is absolutely convergent $\Rightarrow$ $\sum_{n=1}^{\infty} a_n\sin(nx)$ is absolutely and uniformly convergent?

Question

I'm proving that if $\sum_{n=1}^{\infty} a_n$ is absolutely convergent $\Rightarrow$ $\sum_{n=1}^{\infty} a_n\sin(nx)$ is absolutely and uniformly convergent.

I defined for a set $\mathbb{D} \subset \mathbb{R}$ the following sequence of functions: \begin{equation*} f_n: \mathbb{D} \to \mathbb{R}, \end{equation*}

\begin{equation*} \phantom{1000}x \mapsto a_n\sin(nx) \end{equation*}

Since $-1 \le \sin(nx) \le 1, $

then $\forall n \in \mathbb{N}, \forall x \in \mathbb{R} : |a_n\sin(nx)| \le |a_n| $

and $\sum_{n=1}^{\infty} a_n$ is absolutely convergent

Therefore apliying M-Weierstrass test we got that $\sum_{n=1}^{\infty} a_n\sin(nx)$ is absolutely and uniformly convergent.

Is my reasoning correct?

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct.