the uniform convergence of series (M-test). $\sum_{n=1}^{\infty} \dfrac{\arctan n + \sqrt {|x|}}{n^2}$. 
Does the series converge uniformly on $R$. ?
Let $f_k(x) = \dfrac{\arctan k + \sqrt {|x|}}{k^2}$. Then, I have to find $M_k$ such that $|f_k(x)| \le M_k$ for all $x \in R$. Concurrently, $\sum_{k=1}^{\infty} M_k$ has to converge. I think such $M_k$ does not exist because $\sqrt{|x|}$ can approach to infinity. 
If I am correct, I have to disprove this, but I am struggling to find a way to disprove it. Could you give some hint?  
Thank you in advance. 
 A: Let
$$R_n(x)=\sum_{k=n}^{\infty}\dfrac{\arctan k+\sqrt{|x|}}{k^2}$$
This is the rest of the series.
Uniform convergence would mean that
$$\forall \varepsilon>0,\exists N,\forall n>N,\forall x\in\Bbb R, |R_n(x)|<\varepsilon$$
If you suspect that the series does not converge uniformly, then you have to prove the contrary:
$$\exists \varepsilon>0,\forall N,\exists n>N, \exists x\in\Bbb R, |R_n(x)|\ge\varepsilon$$
Now, let $n\ge1$, you have
$$R_n(x)>\sqrt{|x|}\sum_{k=n}^\infty\dfrac{1}{k(k+1)}=\frac{\sqrt{|x|}}{n}$$
It should not be difficult to conclude.
A: It's not uniformly Cauchy: Fix $N\in \mathbb{N}$, then for any $k\geq 1$ we ought to be able to keep 
$$
\sup_{x\in \mathbb{R}}\left|\sum_{n=N}^{N+k} \frac{\arctan(n)+\sqrt{|x|}}{n^2}\right|
$$ 
small by making $N$ large. However,
$$
\sup_{x\in \mathbb{R}}\left|\sum_{n=N}^{N+k} \frac{\arctan(n)+\sqrt{|x|}}{n^2}\right|=
\sup_{x\in \mathbb{R}}\sum_{n=N}^{N+k} \frac{\arctan(n)+\sqrt{|x|}}{n^2}\\
\geq \sup_{x\in\mathbb{R}}
\sum_{n=N}^{N+k} \frac{\sqrt{|x|}}{n^2}\\
\stackrel{|x|=(N+k)^4}{\geq} (N+k)^2\sum_{n=N}^{N+k} \frac{1}{n^2}\\
\geq (N+k)^2 k\left(\frac{1}{(N+k)^2}\right)\\
=k\not\to0
$$
as $N\to \infty$.
