Prove that $S(x) = \sum\limits_{n=1}^\infty \frac{\sin nx}{n\sqrt n}$ is convergent and can be differentiated on $x\in(0, 2\pi)$

Question

my task is to prove that $$S(x) = \sum\limits_{n=1}^\infty \frac{\sin nx}{n\sqrt n}$$ is convergent and is continuous on $(0, 2\pi)$.

I did the following: $$S(x) = \int\limits_{n=1}^\infty \frac{\sin nx}{n\sqrt n}\mathrm{d}x =\frac{\cos(nx)}{\sqrt{n}} $$ Applying Dirichlet Theorem, observe two series: $$a_n = \cos(nx), \quad b_n=\frac{1}{\sqrt{n}}$$ $a_n$ is bounded, $b_n$ approaches zero, therefore the initial series is convergent.

Is it the right way to prove the convergence? And how shall I prove that it is continuous and differentiated on the given interval?

Answer 1

It's unclear what you mean by your line "$S(x)=\int_{n=1}^\infty\ldots$".

Note that $$\left|{\sin(nx)\over n \sqrt{n}}\right|\leq{1\over n^{3/2}}\ .$$ As ${3\over2}>1$ the given series is a uniformly convergent series of continuous functions. Therefore $x\mapsto S(x)$ is a continuous $2\pi$-periodic function.

With differentiation it is different.The derived series $$\sum_{n=1}^\infty{\cos(nx)\over\sqrt{n}}$$ is divergent at $x=0$ (since ${1\over2}<1$), but is convergent at all $x\in\ ]0,2\pi[\ $. To show the latter you need Abel's theorem. Looking at a proof of this theorem you can verify that the derived series is uniformly convergent on every interval $[\delta,\>2\pi-\delta]$, $\delta>0$. This does guarantee that the derived series represents $S'(x)$ for all $x\in\ ]0,2\pi[\ $. The following is a plot of $S(x)$; it corroborates that $S(x)$ is differentiable at all $x\in\ ]0,2\pi[\ $.

Answer 2

$S(x)$ is not differentiable at the origin: let $x=\frac{1}{N}$ for some large $N$. We have $$ \sum_{n=1}^{2N}\frac{\sin(nx)}{n\sqrt{n}}\geq \frac{4}{5N}\sum_{n=1}^{2N}\frac{1}{\sqrt{n}}\geq \sqrt{\frac2N} $$ while the partial sums of $\sin(n x)$ are bounded by $\frac{1}{2}\cot\frac{x}{4}\leq 2N$, so by summation by parts $$ \left|\sum_{n>2N}\frac{\sin(nx)}{n\sqrt{n}}\right|\leq 2N\sum_{n>2N}\left(\frac{1}{n\sqrt{n}}-\frac{1}{(n+1)\sqrt{n+1}}\right)\leq \sqrt{\frac1N} $$ and $\frac{S(x)}{x}$ is unbounded as $x\to 0^+$. This implies that $S$ is not differentiable at the origin.