Uniform convergence of $\sum_{n=1}^{\infty}\left(\frac{nx}{1+n^{2}x^{2}}\right)^{n}$ 
Prove with Weierstrass M-Test, that given series $$\sum_{n=1}^{\infty}\left(\frac{nx}{1+n^{2}x^{2}}\right)^{n} ,\hspace{6mm} |x| < +\infty$$ is uniformly convergent

First thing I tried to do, is to find $\sup\left \| \left(\frac{nx}{1+n^2x^2}\right)^n  \right \| = \frac{1}{2^n}$ when we choose $x=\frac{1}{n}$.
So the series $\sum_{n=1}^{\infty}\frac{1}{2^n}$ converges to $1$. And we can conclude that original series converges uniformly.
The problem is, I dont get the intuition that I'm really done here.
And the questions I'm left with:
Am I on the right track and Is this solution correct?
How can I obviously find the supremum? In this example I guessed.
Any other approaches apply here?
 A: Notice that by Am-Gm
$n|x| \leq \frac{1 +n^{2}x^{2}}{2}$
Hence
$(\frac{n|x|}{1 + n^{2}x^{2}})^{n} \leq \frac{1}{2^{n}}$ 
Which justifies your estimate
A: If you look at the conditions of the Weierstrass M test, you should notice yourself that you are indeed done. Hence your solution is correct (once you rigorously show the claim about the supremum).
Actually, in this special case we do not even need to compute the supremum explicitly if we employ a little trick:
The function $$f(x)=\frac{x}{1+x^2}$$ is continuous and converges to $0$ as $x\to\pm\infty$. Therefore $M:=\sup |f(x)|<\infty$. In fact the supremum is attained somewhere because the restriction of $f$ to a compact interval $I$ such that $|f(x)|\le\frac M2$ for $x\notin I$ attains its maximum. 
This general property of such functions should also be clear "intuitively" from looking at a "typical" graph of such a function.
The only more specific property of $f$ that we need to show is that $M<1$.
As $f$ is odd and $f(0)=0$, we may concentrate on $0< x<\infty$. Here we have $$1-x+x^2> 1-2x+x^2=(1-x)^2\ge 0,$$ hence $x<1+x^2$ and $0\le f(x)<1$.
As there exists some $x$ with $f(x)=M$, we conclude $M<1$.
Now by substitution $x\leftarrow nx$ we clearly have
$\sup\frac{nx}{1+n^2x^2}=M$ and then $$\sup\left(\frac{nx}{1+n^2x^2}\right)^n=M^n.$$ As $0<M<1$, the desired condition $\sum_n M^n<\infty$ follows from the convergence of the geometric series.
