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...

  • $\begingroup$ I believe this is not true for conditionnally converging series. Take e.g. $x=\frac{1}{\pi}$ with $a_k=\sin(k)$. $\endgroup$ – Joce Jun 16 '14 at 8:04
  • $\begingroup$ @Joce, do you mean $a_{k} = (\sin k)/k$? $\endgroup$ – Sangchul Lee Jun 16 '14 at 8:21
  • $\begingroup$ @sos440 : See math.feld.cvut.cz/mt/txte/2/txe3ec2p.htm $\endgroup$ – Joce Jun 16 '14 at 8:29
  • $\begingroup$ @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. $\endgroup$ – uqtredd1 Jun 16 '14 at 9:12
  • $\begingroup$ @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 $\alpha/k$, $\alpha>0$. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.