$$\sum\limits_{n=1}^{\infty}\left(1- \cos\sqrt{\frac{x^5}{n}}\right)\arctan \frac{e^x}{\sqrt{n}}$$
For $x$ in:
$1)\ E_1 = (0;1)$
$2)\ E_2 = (1,+\infty)$
Attempt:
1) $E_1 = (0;1)$
Let $x = \frac{1}{n}, n > 1$. Then $x \in (0, 1)$.
Therefore $\exists n_0 \in \mathbb{N}$ such that
$$\left(1- \cos \frac{1}{n^3}\right)\arctan \frac{e^{\frac{1}{n}}}{\sqrt{n}} \leq \left(1 - \left[1-\frac{1}{n^6}\right]\right)\arctan \frac{1 + \frac{1}{n}}{\sqrt{n}} = \frac{1}{n^6}\arctan \frac{1 + \frac{1}{n}}{\sqrt{n}}$$
Since $\frac{1}{n^6}\arctan \frac{1 + \frac{1}{n}}{\sqrt{n}} = 0$ as $n \to \infty$, it follows that the leftmost series converges uniformly.
Question:
Is this approach correct? I feel that the choice of $x = \frac{1}{n}$ was somewhat arbitrary. How can I be sure that there is no an $x \in (0,1)$ not of the form $x = \frac{1}{n}$ for which the series non-uniformly converges (or even divergences)
2) $E_2 = (1,+\infty)$
Here I made the guess that the series non-uniformly convergence. Let $e^x = \sqrt{n}$. Then $x = \frac{1}{2}\ln n$, and we have the inequality:
$$\left(1-\sqrt{\frac{\ln^5 n}{32n}} \right)\frac{\pi}{4} \leq \left(1-\cos \sqrt{\frac{\ln^5 n}{32n}} \right)\frac{\pi}{4} \leq \frac{\pi}{4}$$
Since $\left(1-\sqrt{\frac{\ln^5 n}{32n}} \right)\frac{\pi}{4} = \frac{\pi}{4}$ as $n \to \infty$, we have that $\left(1-\cos \sqrt{\frac{\ln^5 n}{32n}} \right)\frac{\pi}{4} = \frac{\pi}{4}\neq 0$. Therefore the series non-uniformly converges.
Your first part is wrong. Note for $x\in E_1$, $$ \sin x\le x, \arctan x\le x. $$ So for $x\in E_1$, $$ 0<\left(1- \cos \sqrt{\frac{x^5}{n}}\right)\arctan \frac{e^x}{\sqrt{n}}=2\sin^2\left(\frac12\sqrt{\frac{x^5}{n}}\right)\arctan \frac{e^x}{\sqrt{n}}\le 2\left(\frac12\sqrt{\frac{x^5}{n}}\right)^2\frac{e^x}{\sqrt{n}}\le\frac{e}{2}\frac{1}{n^{3/2}}.$$ Since $\sum\frac{1}{n^{3/2}}$ converges, $\sum\limits_{n=1}^{\infty}\left(1- \cos \sqrt{\frac{x^5}{n}}\right)\arctan \frac{e^x}{\sqrt{n}}$ converges uniformly.