Prove functional series $\sum\limits_{n=1}^\infty\frac{\cos nx}{\sqrt{n}}$ doesn't converge uniformly using Cauchy's convergence test 
$$\sum\limits_{n=1}^\infty\frac{\cos nx}{\sqrt{n}}$$ where $x\in (0;2\pi)$.

My attempt: 
$$|\frac{\cos(n+1)x}{\sqrt{n+1}}+\frac{\cos(n+2)x}{\sqrt{n+2}}+\ldots+\frac{\cos(n+p)x}{\sqrt{n+p}}|\ge p\cdot\frac{\cos^2 x}{\sqrt{n+p}}=\{p=n\} = \frac{n\cos^2 x}{\sqrt{2n}}\ge\frac{n\cdot\cos^2 x}{2n}=\frac{\cos^2 x}{2}=\{x=\frac{\pi}{4}\}=\frac{1}{4}$$
Let $\varepsilon = \frac{1}{4}$. Am I right? Just doing it the first time
 A: Your first inequality seems dubious to me. But I think the idea of showing this series is not uniformly Cauchy is a good one.
Let $S_m(x)$ denote the $m$th partial sum of the series. We'll show that for any $m,$ there exists $x_m\in (0,2\pi)$ such that $S_{2m}(x_m) - S_{m-1}(x_m) \to \infty$ as $m \to \infty.$ That shows the uniform Cauchy criterion fails in a big way.
In fact we can let $x_m = 1/2m.$ Then 
$$S_{2m}(x_m) - S_{m-1}(x_m)=\sum_{n=m}^{2m}\frac{\cos (n/2m)}{\sqrt n}$$ $$ \ge \cos 1\cdot  \sum_{n=m}^{2m}\frac{1}{\sqrt n} \ge  \cos 1\cdot m \cdot\frac{1}{\sqrt {2m}}$$
As $m\to \infty,$ the last expression $\to \infty,$ and we're done.
A: I do not get your very first inequality, in particular how it may account for a lot of cancellation in the LHS, due to the fact that $\cos(nx)$ changes its sign pretty often. 
For sure, $\left|\sum_{n=1}^{N}\cos(nx)\right|\leq \frac{1}{\sin(x/2)}$ for any $N\in\mathbb{N}^+$ and any $x\in(0,2\pi)$, hence the given series is pointwise convergent to some function $f(x)$ by Dirichlet's test. On the other hand, by using approximations of the identity,
$$ \lim_{x\to 0^+}f(x) = \lim_{m\to +\infty}\sum_{n\geq 1}\frac{1}{\sqrt{n}}\int_{0}^{+\infty} m\cos(nx)e^{-mx}\,dx = \lim_{m\to +\infty}\sum_{n\geq 1}\frac{m^2}{\sqrt{n}(m^2+n^2)} $$
equals $+\infty$. So, how can the convergence to $f(x)$ be uniform? It simply cannot.
