Showing $\sum x_n\sin(nx)$ converges uniformly

Show the $\sum\limits_{n = 1}^\infty x_n\sin(nx)$ converges uniformly iff $nx_n →0$ as $n →\infty$, where $x_n$ is a decreasing sequence and with $x_n>0$ for all $n=1,2,\cdots$.

My attempt was with M-test. I took $M_n= nx_n$. Since $\sin(x)$ is bounded by $1$, I claim that $M_n >x_n\sin(nx)$ and we have $\sum M_n >\sum x_n\sin(nx)$. Then if $\sum M_n$ converges we have $\sum x_n\sin(nx)$ converges uniformly by theorem.

My problem here is question asking with iff. If $\sum M_n$ converges we know $|M_n|→0$. But I could not understand how we can show the opposite direction in this case. I mean if $|M_n|→0$ we can not be sure if $\sum M_n$ converges or not from the harmonic series case. Could you help me on that?

You cannot do this by directly applying some test for convergence. The proof involves 'summation by parts' and some calculations involving $\sin (nx)$ which are basic to Fourier series. A complete proof can be found here: Theorem 7.2.2 (part 1), p. 112 of Fourier Series: A Modern Introduction by R. E. Edwards. [ If the numbers don't match look for 'The series (C) and (S) as Fourier series' in the contents page].