Does $\sum_{n=1}^{\infty} \frac{1}{n^3 x + \sqrt[3]{n} }$ uniformly converge on $A = (0, + \infty)$
The problem is we don't have minimum for $x$. Consider $A' = [a, + \infty)$, where $a>0$ is some fixed number. Then, by Weierstrass $$|a_n (x)| = \frac{1}{n^3 x +\sqrt[3]{n} } < \frac{1}{n^3 a +\sqrt[3]{n} } < \frac{1}{n^3 a} \to 0, n \to \infty$$ and $\frac{1}{a} \sum_{n=1}^{\infty} \frac{1}{n^3}$ is convergent. Hence, we can conclude that functional sequence $\sum_{n=1}^{\infty} \frac{1}{n^3 x +\sqrt[3]{n} }$ converges uniformly on $A'$.
But what should we do if we don't have min value? How to proceed in such case? I literally have no idea. I don't think we can state that this functional sequence is uniformly convergent on $A$ just because $a$ is actually an arbitrary number.
Let $R_m(x) = \sum_{n=m}^{2m}a_n(x)$
$$\left\|R_m\right\|_{\infty} \ge R_m\left(\frac1{m^3}\right) \ge \sum_{n=m}^{2m} \frac1{8 + \sqrt[3]{2m}} = \frac m{8 + \sqrt[3]{2m}} \to \infty$$
as $m\to \infty$. Can you conclude from this?