# the series $\sum_{k=1}^\infty a_k$ converges implies the series $\sum_{k=1}^\infty a_k\sin (k\pi x)$ converges for $x$ irrational

Let $\sum_{k=1}^\infty a_k$ be a convergent series. Then can we obtain $\sum_{k=1}^\infty a_k\sin (k\pi x)$ converges for $x$ irrational?

If $\sum_{k=1}^\infty a_k$ converges absolutely, then I can obtain that $\sum_{k=1}^\infty a_k\sin (k\pi x)$ converges for all $x$. But I do not know how to deal with the case $\sum_{k=1}^\infty a_k$ converges conditionally...

• I believe this is not true for conditionnally converging series. Take e.g. $x=\frac{1}{\pi}$ with $a_k=\sin(k)$. – Joce Jun 16 '14 at 8:04
• @Joce, do you mean $a_{k} = (\sin k)/k$? – Sangchul Lee Jun 16 '14 at 8:21
• @sos440 : See math.feld.cvut.cz/mt/txte/2/txe3ec2p.htm – Joce Jun 16 '14 at 8:29
• @Joce : I believe that link you give is mistaken and instead intended to say that $sin(k)$ is bounded rather than convergent. $sin(k)$ diverges by the n'th term test and even by representing as a geometric series as they have done $|e^i|=1$ and hence the geometric series is divergent. – uqtredd1 Jun 16 '14 at 9:12
• @uqtredd1: you're right indeed. But thus $\sin(k)/k$ is conditionally convergent, while $\sin^2(k)/k$ will not be convergent, as we can decompose it into a positive series and a subsample bounded from below by a series in $\alpha/k$, $\alpha>0$. – Joce Jun 16 '14 at 11:55

From this proof, $\sum_{k>0} \sin(k)$ is bounded, and from Dirichlet's test, $\sum_{k>0} \sin(k)/k$ converges conditionally. However, for $x=1/\pi$, $\sum_{k>0} \sin(k\pi x)\sin(k)/k > \sum_{k' \in C} \alpha/k'$ where $0<\alpha<1$ and $C$ is a subsample of $\mathbb{N}$ such that $card(C \cup \{1,...N\})\geq \beta N$, $\beta>0$ depends on $\alpha$. Thus $\sum_{k' \in C} 1/k'$ diverges.