I've stumbled upon this problem, while preparing to my exams. The problem is that I cannot understand which $M_n$ should I finally come to, trying to apply the Weierstrass M-test. I also couldn't have found similar-to questions here. I made several attempts, stating that $\forall n \in\mathbb{N}, \forall x\in E[0,10] : \arctan(nx) \leqslant nx$ or $\arctan(nx) \leqslant \frac{\pi}{2}$, but for some reason nothing works out for me. My best attempt for now was to state that the given $|u_n(x)| \leqslant \frac{\pi}{(n-x)^2}$, but the thing is that it is not true for all x in the given $E$.
Thank you in advance for your help.
If $n>10$ and $x\in[0,10]$, then$$1+(x-n)^3>1+(10-n)^3=1-(n-10)^3,$$and so$$|1+(x-n)^3|>(n-10)^3-1.$$Furthermore, $\arctan(nx)<\frac\pi2$. Therefore,$$\left|\frac{\arctan(nx)}{1+(x-n)^3}\right|<\frac{\pi/2}{(n-10)^3-1}.$$Since the series$$\sum_{n=11}^\infty\frac{\pi/2}{(n-10)^3-1},$$converges, then your series converges uniformly, by the Weierstrass $M$-test.