# Steps to determine the interval of continuity of $f(x)= \sum_{n=2}^{\infty} \frac{(\sin{nx})^2}{\sqrt{n}}$

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?

• First, the function does not depends on $n$. It should be $f$. Secondly, have you determined the definition domain of your function ? – TheSilverDoe Mar 12 at 20:53
• @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. – E.Nole Mar 12 at 20:58
• Unfortunately, I think that $f$ is not defined on any interval... – TheSilverDoe Mar 12 at 21:02
• @TheSilverDoe you mean the series diverges for all $x$? – 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 $$\sin(x) \ne 0$$.
• Second series needs a $2$ in the denominator. – zhw. Mar 12 at 21:48