$\sum\left|\sin\alpha n+\sin\beta n\right|/n=\infty$ when $\alpha\neq-\beta$ Is there an elementary way of showing that the series $$\sum\frac{\sin\alpha n+\sin\beta n}{n}$$ is not absolutely convergent?
Assuming $\alpha,\beta$ are not $0$ or $\pi$, I can show that there exists a subsequence $n_k=O\left(k\right)$ such that $\sin\alpha n_k$ is always $\geq 1/2$ (actually, I have shown that $n_{k+1}-n_k\in\left\{p,p+1,p+2,p+3\right\}$ for some $p$), and then another subsequence $m_k=O\left(k\right)$ with $\sin\beta m_k\geq 1/2$. Next, I'd like to show that these sequences overlap often enough, so that the series above is comparable to $\sum 1/n$. I think this is justified by some kind of orthogonality between $\sin\alpha n$ and $\sin\beta n$, but I can't prove it.
I would like to find an elementary way of doing this that does not involve results like the equidistribution theorem (I don't even know if it's relevant, but looks like). Is it possible?
 A: Just take a slightly weaker lower bound. We have to show that
$$ \lim_{N\to +\infty}\sum_{n=1}^{N}\frac{2}{n}\left|\,\sin\left(\frac{\alpha-\beta}{2}\,n\right)\right|\cdot\left|\,\cos\left(\frac{\alpha+\beta}{2}\,n\right)\right|=+\infty$$
but for any interval $I=[A,A+M]$, assuming that $M$ is big enough, we have strictly more than $\frac{M}{2}$ integers for which $\left|\,\sin\left(\frac{\alpha-\beta}{2}\,n\right)\right|>\frac{1}{4}$ and strictly more than $\frac{M}{2}$ integers for which $\left|\,\cos\left(\frac{\alpha+\beta}{2}\,n\right)\right|>\frac{1}{4}$, so for at least one integer in the given interval, both the inequalities hold and the series is so divergent by comparison with the harmonic series.
Thanks to Roberto Pagaria, we also have a nice proof by contradiction. Assuming that the given series is absolutely convergent, any rearrangement of its terms must lead to the same limit. However, by exchanging the terms corresponding to even values of $n$ and odd values, it is easy to check that it is not the case, so the original series is not absolutely convergent.
