calculus - Show that $sum_{n=1}^{infty}frac{sinfrac{x}{n}sin2nx}{x^2+4n}$ converges uniformly.

How to show that the following series converges uniformly? $$\sum_{n=1}^{\infty}u_n(x),\ \ \ u_n(x)=\frac{\sin\frac{x}{n}\sin2nx}{x^2+4n},\ \ x\in E=(-\infty;+\infty)$$

At first I tried to apply Dirichlet's test. However, I got stuck while trying to prove that $\sum_{n=1}^{\infty}\sin\frac{x}{n}\sin2nx$ is less than some fixed $M$ (multiplying and dividing by $2\sin x$ did not help much). In my other attmepts I also got stuck trying to limit the numerator. So, the problem is with these $\sin$ functions.

Answer

As $f(x)=\sum_{n=1}^{\infty}\frac{\sin\frac{x}{n}\sin2nx}{x^2+4n}$ is even, we can limit the analysis on $[0, \infty)$.

Consider $$v_n(x)=\frac{x}{n(x^2+4n)}.$$

We have $$v_n^\prime(x)=\frac{4n^2-nx^2}{n^2(x^2+4n)^2}$$

Based on that, one can prove that $v_n$ is positive on $[0,\infty)$ and attains its maximum at $x_n = 2\sqrt n$. The maximum having for value $\frac{1}{2n^{3/2}}$. As $\sum \frac{1}{n^{3/2}}$ converges, $\sum v_n(x)$ converges uniformly on $[0, \infty)$ according to Weierstrass M-test.

We then get the uniform convergence of $\sum u_n(x)$ as $\vert u_n(x) \vert \le v_n(x)$ for all $n \in \mathbb N$ for $x \in [0,\infty)$.