Let $G(x)=\sum _{n=1}^{\infty }\frac{nx}{1+n^6x^2}$ Prove that $G(x)$ is bounded 
Let $G:\mathbb{R}\to\mathbb{R}$
$$G(x)=\sum _{n=1}^{\infty }\frac{nx}{1+n^6x^2},x\in\mathbb{R}$$ Prove that $G(x)$ is bounded


by AGM inequality  :
$$\frac{1+n^6x^2}{2}\ge \sqrt{n^6x^2}$$
$$\iff \frac{1}{2} \ge \frac{\sqrt{n^6x^2}}{1+n^6x^2}$$
$$ \iff \frac{1}{2n^2} \ge \frac{n|x|}{1+n^6x^2}$$
so G(x) is bounded , $M=\frac{1}{2n^2}$
is this correct ? and if i want to prove that it's Pointwise convergence i can say that because $\frac{1}{2n^2}$ is convergence we get also that $\frac{n|x|}{1+n^6x^2}$ is convergence for every $x\in R$?
thanks
 A: Easy trick by differentiating:
Let $$g_n(x)=\frac{nx}{1+n^6x^2},\implies g'_n(x)=\frac{n-n^7x^2}{(1+n^6x^2)^2}$$ 
$$g'_n(x)=0 \Longleftrightarrow x= \pm\frac{1}{n^3}$$ and 
$$ \lim_{x\to\infty}g_n(x)=0=\lim_{x\to0}g_n(x)$$
Observing that $g_n(x)$ is odd function one easily get that $|g_n|$  attains its maximum at $x= \frac{1}{n^3}$ therefore for all $x$,
$$|g_n(x)|\le |g_n(\frac{1}{n^3})| = \frac{1}{n^2(1+n^3)}$$
And hence,  $$|G(x)|\le\sum _{n=1}^{\infty }|g_n(x)|\le \sum _{n=1}^{\infty }\frac{1}{n^2(1+n^3)} \le \sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi^2}{6}$$ 
A: edit: I was confused by post before edits, so I decided to offer a solution.
Note that if 
$$
y\geq 1
$$
then 
$$
\sqrt{y}\leq y
$$
so, 
$$
1+n^6x^2\geq \sqrt{1+n^6x^2}\geq n^3|x|\\
\implies\frac{n|x|}{1+n^6x^2}\leq\frac{1}{n^2}
$$
so 
$$
|G(x)|=\left|\sum_{n=1}^\infty\frac{nx}{1+n^6x^2}\right|\leq
\sum_{n=1}^\infty\left|\frac{nx}{1+n^6x^2}\right|\\
=\sum_{n=1}^\infty\frac{n|x|}{1+n^6x^2}\leq\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}
$$
As for your second question, we may conclude that the series converges uniformly by the Weirstrass M test.
