I am trying I determine the interval of continuity of $$f_n(x)= \sum_{n=2}^{\infty} \frac{(\sin{nx})^2}{\sqrt{n}}$$ I tried to find the domain by Dirichlet's test, but the sum $\sum _{n=1}^{\infty }(\sin (nx)^2)$ does not converge. I also tried the ratio test on which I got stuck on. The root test was inconclusive. What other tests can I apply?

  • $\begingroup$ First, the function does not depends on $n$. It should be $f$. Secondly, have you determined the definition domain of your function ? $\endgroup$ – TheSilverDoe Mar 12 at 20:53
  • $\begingroup$ @TheSilverDoe I was trying to find the domain using Dirichlet's test, the root test and the ratio test but didn't get anything helpful. I initially guessed that the domain is all real $x$ except $0$, but I can't back my argument yet. $\endgroup$ – E.Nole Mar 12 at 20:58
  • $\begingroup$ Unfortunately, I think that $f$ is not defined on any interval... $\endgroup$ – TheSilverDoe Mar 12 at 21:02
  • $\begingroup$ @TheSilverDoe you mean the series diverges for all $x$? $\endgroup$ – E.Nole Mar 12 at 21:06

$$\sum_{n=1}^\infty \frac{\sin^2(nx)}{\sqrt{n}} = \sum_{n=1}^\infty \frac{1 - \cos(2nx)}{2\sqrt{n}}$$

Of course, $\sum_n 1/\sqrt{n}$ diverges, while $\sum_n \cos(2nx)/\sqrt{n}$ converges by Dirichlet's test whenever $\sin(x) \ne 0$. Therefore your series must diverge whenever $\sin(x) \ne 0$.

  • 1
    $\begingroup$ Second series needs a $2$ in the denominator. $\endgroup$ – zhw. Mar 12 at 21:48
  • $\begingroup$ Oops, yes. Editing. $\endgroup$ – Robert Israel Mar 13 at 1:46

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