$\sum_{n=1}^\infty \frac{\sin nx}{n+1}$ is not continuous at $0$ and $2\pi$?

Question

$\sum_{n=1}^\infty \frac{\sin nx}{n+1}$ is not continuous at $0$ and $2\pi$? It is easy by Dirichlet test that the above sum is continuous on $(0,2\pi)$. What about the endpoint case $0$ and $2\pi$? Computation show it is not continuous, but what is the rigourous proof?

Answer 1

The sum can be computed explicitly. Let $\color{blue}{0<x<2\pi}$. By Abel's theorem $$\sum_{n=1}^\infty\frac{\sin nx}{n+1}=\Im\lim_{r\to1^-}\sum_{n=0}^\infty\frac{r^{n+1}}{n+1}e^{inx}=\Im\lim_{r\to1^-}\big(-e^{-ix}\log(1-re^{ix})\big),$$ and since $\lim\limits_{r\to1^-}\log(1-re^{ix})=l(x)-is(x)$ with $$l(x)=\log\left|2\sin\frac x2\right|,\qquad s(x)=\frac{\pi-x}{2},$$ we get $$\sum_{n=1}^\infty\frac{\sin nx}{n+1}=s(x)\cos x+l(x)\sin x.$$ This holds for all $x$ if we assume the RHS at $x=0$ to be $0$ (that is, if we let $s(0)=0$ and extend $l(x)\sin x$ by continuity) and if we let $s(x+2\pi)=s(x)$ for all $x$.

Now the discontinuity is apparent: the sum tends to $\pm\pi/2$ as $x\to\pm 0$.