TITLE: Zero limit implies zero sequence
QUESTION [11 upvotes]: Let $0 < \alpha_1 < \alpha_2 < \cdots < \alpha_k < 1$, and $a_1,\ldots , a_k \in \mathbb{R}$. Suppose that $$\lim_{n\to + \infty} \sum_{p = 1}^k a_p \sin (n \pi \alpha_p)=0$$
Show that $a_1=a_2=\cdots=a_k=0$.

I tried to use the same method as Cantor Lemma which you can find a related problem on MSE :
Convergence of series of functions: $f_n(x)=u_n\sin(nx)$
But I failed, it seems to be more difficult.

REPLY [3 votes]: There is also a one-line solution through diophantine approximation.
The sequence $\left\{\left(\left\{\frac{n\alpha_1}{2}\right\},\ldots,\left\{\frac{n\alpha_k}{2}\right\}\right)\right\}_{n\in\mathbb{N}}$ is ergodic in $[0,1]^k$, hence is some $a_p$ differs from zero, $\{x_n=\sum_{p=1}^{k}a_p \sin(\pi n \alpha_p)\}_{n\in\mathbb{N}}$ has a non-Cauchy subsequence and cannot converge.