Uniform convergence of $\sum_{n \to \infty}\frac{\sqrt{x}\cos nx}{n(2nx^2+1)}$

Question

How one can prove that for $x\in(0;+\infty)$ the series $$ \sum_{n=1}^{\infty}\frac{\sqrt{x}\cos nx}{n(2nx^2+1)} $$ converge uniformly?

Answer 1

Let $$f_n(x) = \frac{\sqrt x}{2nx^2+1}, \,\,\,\, x > 0.$$ Notice that \begin{align*} f'_n(x) &= \frac{1}{2\sqrt{x}(2nx^2+1)} - \frac{4n x\sqrt x}{(2nx^2+1)^2} \\ &= \frac{1 - 6nx^2}{2\sqrt x (2nx^2+1)^2}. \end{align*} Thus $f'(x) = 0$ at $x_n^* = \frac{1}{\sqrt{6n}}$. Further $f'(x) > 0$ for $0 < x < x_n^*$ and $f'(x) < 0$ for $x_n^* < x < \infty$, which shows that $x_n^*$ is a global maximum for $f_n$. From this, we can conclude that $$f_n(x) \le f_n(x^*_n) = \frac{(6n)^{-1/4}}{4/3} = \frac{3}{4(6)^{1/4}} \cdot \frac 1 {n^{1/4}}.$$ But then $$\left \lvert \frac{\sqrt x \cos(nx)}{n(2nx^2+1)} \right \rvert \le \frac{C}{n^{5/4}}.$$