$\sum_{n=1}^{\infty} \frac{\sin(nx)}{\sqrt{n}}, [0, 2\pi]$
I have been thinking for hours and I do not know how to approach whether it uniformly converges or not. Of course, I did search, but there was no desired result, and even if I found similar contents, it was difficult or unintelligible.
Is it possible to show that $\sum_{n=1}^{\infty} \frac{\sin(nx)}{\sqrt{n}}, [0, 2\pi]$ is whether uniform converges or not by using Cauchy's test or Weierstrass-M test?
I wait for an answer that will be clear and detailed. Please.
Thank you for reading so far and hope that today's good work will be full.
Let $S_n(x)$ denote partial sums of this infinite sum at $x$. $$S_n(1/n) = \sum_{i=1}^n \frac{\sin(i/n)}{\sqrt{i}} = \sqrt{n}\cdot\frac{1}{n}\sum_{i=1}^n\frac{\sin(i/n)}{\sqrt{i/n}} $$ Now $$0\neq\int_0^1 \frac{\sin(x)}{\sqrt{x}}dx = \lim_{n\to\infty}\frac{1}{n} \sum_{i=1}^n\frac{\sin(i/n)}{\sqrt{i/n}} $$ where we used definition of Riemann integral.
Hence $$\lim_{n\to\infty} S_n(1/n) = \infty \neq S(0) = 0 $$ This means that the convergence can't be uniform. See here for more details.
This approach also shows, that we shouldn't expect uniform convergence of $$ \sum_{n=1}^\infty \frac{\sin(nx)}{n^a}$$ for $a<1$, $x$ is as earlier.
Indeed, for $0<a<1$ we have that $$S_n(1/n) = n^{1-a}\cdot \frac{1}{n}\sum_{i=1}^n\frac{\sin(i/n)}{(i/n)^a}$$ And $$\lim_{n\to\infty}\frac{1}{n} \sum_{i=1}^n\frac{\sin(i/n)}{(i/n)^a} = \int_0^1 \frac{\sin(x)}{x^a}dx>0 $$ Hence $$\lim_{n\to\infty} S_n(1/n) = \infty\neq S(0) = 0 $$ as a multiple of sequence converging to $\infty$ and to a finite non-zero number. So in this case, the sum isn't uniform convergent.