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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?

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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].

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