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

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?

• Couldn't you just note that $\lvert \sin x \rvert \le 1$ and then apply the p test to the resulting expression to show convergence? I would have thought continuity comes from the fact that $\sin nx/(n \sqrt{n})$ is the quotient of continuous functions for $n \in \mathbb{N}$ and $x \in \mathbb{R}$ and $S$ is then the sum of continuous functions. – mattos Nov 19 '19 at 13:27
• @mattos ok thx, I'll think about it. But still, is my reasoning towards convergency correct? – lytkin Nov 19 '19 at 13:51

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 $$x\in\ ]0,2\pi[\$$.
$$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.