Showing that $\sum_{n=1}^{\infty} \frac{1}{n+1} \sin n\theta$ is not continuous I have been asked to show that the series
$$
f(\theta) = \sum_{n=1}^{\infty} \frac{1}{n+1} \sin n\theta 
$$
is point-wise convergent. I have been able to show this, but after this I have also been asked the smoothness of the function. That is, whether $f(\theta)$ is $C^0 C^1 C^2, \dots,$ or $C^{\infty}$.
I think that it is probably $C^0$, because if I plot it, it looks sort of like a sawtooth wave:
https://www.desmos.com/calculator/mzgk0knaaq
it seems to be discontinuous as $2\pi n$, and therefore it should be $C^0$. But I'm not really sure how I can actually show that it is discontinuous at these points. May I receive a hint of how I can do this?
 A: Summation by part yields $$f(x)=\sum_{n=1}^\infty \frac{1}{(n+1)(n+2)}\frac{\sin^2((n+\frac 12)x)}{\sin(\frac x2)}-\frac 14\tan(\frac x4)$$
Suppose $f$ is continuous at $0$. Then $\lim_{x\to 0}\sum_{n=1}^\infty \frac{1}{(n+1)(n+2)}\frac{\sin^2((n+\frac 12)x)}{\sin(\frac x2)}=0$.
Take $x=\frac 1N$. Since all the terms are $\geq 0$ and $(n+\frac 12)\frac 1N\leq \frac {\pi}2$, 
$$ \begin{align}
\sum_{n=1}^\infty \frac{1}{(n+1)(n+2)}\frac{\sin^2((n+\frac 12)x)}{\sin(\frac x2)} 
&\geq \sum_{n=1}^N \frac{1}{(n+1)(n+2)}\frac{\sin^2((n+\frac 12)x)}{\sin(\frac x2)}\\
&\geq \frac{8}{\pi^2}\frac 1N \sum_{n=1}^N \frac{1}{(n+1)(n+2)}n^2
\end{align}$$
hence $\frac 1N \sum_{n=1}^N \frac{1}{(n+1)(n+2)}n^2$ goes to $0$ as $N\to \infty$.
But $\sum_{n=1}^N \frac{1}{(n+1)(n+2)}n^2\geq \frac{1}{(N+1)(N+2)}\sum_{n=1}^N n^2=\frac{N(2N+1)}{6(N+2)}$, hence $$\frac 1N\sum_{n=1}^N \frac{1}{(n+1)(n+2)}n^2\geq \frac{2N+1}{6(N+2)}\geq \frac{1}{12}$$
a contradiction.

A similar reasoning may be used to prove the following theorems:

Let $a_n$ a sequence of real numbers that decreases to $0$.
  
  
*
  
*$\sum_{n\geq 1} a_n \sin(nx)$ converges uniformly iff $a_n=o\left( \frac 1n \right)$
  
*$\sum_{n\geq 1} a_n \sin(nx)$ is continuous at $0$ iff $a_n=o\left( \frac 1n \right)$

