Uniform convergence of $\sum_{n = 1}^\infty {\arctan{\frac{2x}{x^2+n^2}}}$ on $\mathbb R$ 
Check if $\sum_\limits{n = 1}^\infty{\arctan{\dfrac{2x}{x^2+n^2}}}$ is uniformly convergent series of function on $\mathbb{R}$.

"Uniform convergence" really messes up my mind.
Since $\dfrac{2x}{x^2+n^2} \approx \arctan{\dfrac{2x}{x^2+n^2}}$ when $n\to\infty$, could I use the uniform convergence of $\sum_\limits{n = 1}^\infty{\dfrac{2x}{x^2+n^2}}$ to check the uniform convergence of $\sum_\limits{n = 1}^\infty{\arctan{\dfrac{2x}{x^2+n^2}}}$?
Thank you in advance.
 A: Your heuristics may give some idea, but any such heuristics should be justified in order to yield a valid proof.
Let me demonstrate a possible way. It follows from AM-GM inequality that
$$ \left|\frac{2x}{x^2 + n^2}\right| \leq \frac{1}{n} $$
uniformly in $x \in \Bbb{R}$. Together with the inequality $\frac{1}{2}|x| \leq |\arctan x| \leq |x|$ which holds for $|x| \leq 1$, we find that
$$ \frac{1}{2} \left\| \sum_{n=M}^{N} \frac{2x}{x^2 + n^2} \right\|_{\sup}
\leq \left\| \sum_{n=M}^{N} \arctan\left( \frac{2x}{x^2 + n^2} \right) \right\|_{\sup}
\leq \left\| \sum_{n=M}^{N} \frac{2x}{x^2 + n^2} \right\|_{\sup}. $$
That is, $\sum_{n=1}^{\infty} \arctan( \frac{2x}{x^2 + n^2} )$ is uniformly Cauchy if and only if $\sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2}$ is uniformly Cauchy. This is enough to conclude that we can investigate the uniform convergence of $\sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2}$ .

Now let us prove that the series does not converge uniformly. @Did already provided a short and elegant solution for this, but let me articulate a little bit by showing that, loosely speaking, 'positive mass' concentrates near the infinity.
Let $S(x) = \sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2}$ and we write $S(x)$ as
$$ S(x) = \sum_{n = 1}^{\infty} \frac{2}{1 + (\frac{n}{x})^2} \frac{1}{x}. $$
We can recognize $S(x)$ as a Riemann sum of the function $f(t) = \frac{2}{1+t^2}$. To prove that this is indeed true, let $x > 0$ and notice that
$$ \int_{\frac{n}{x}}^{\frac{n+1}{x}} \frac{2}{1+t^2} \, dt
\leq \frac{2}{1 + (\frac{n}{x})^2} \frac{1}{x}
\leq \int_{\frac{n-1}{x}}^{\frac{n}{x}} \frac{2}{1+t^2} \, dt $$
for all $n \geq 1$. Summing over $n$ and taking limit as $x \to \infty$, this proves that
$$ \lim_{x\to\infty} S(x) = \int_{0}^{\infty} \frac{2}{1+t^2} \, dt = \pi. $$
This is enough to conclude that $S(x)$ does not converge uniformly, since any partial sum $S_N(x) = \sum_{n=1}^{N} \frac{2x}{x^2 + n^2}$ satisfies
$$ \lim_{x\to\infty} S_N(x) = \sum_{n=1}^{N} \lim_{x\to\infty} \frac{2x}{x^2 + n^2} = 0 $$
and hence there is no way we have $\| S_N - S \|_{\sup} \to 0$ as $N \to \infty$.
