Show the series converges uniformly $\sum_{n =1}^\infty \frac{n^2x^2}{n^4+x^4} $ 
Show that the following series converges uniformly: $$\sum_{n =1}^\infty \frac{n^2x^2}{n^4+x^4} $$

I approached the problem by finding maximum value of $$ f_n(x )$$
It turned out to be  $$ \frac {1}{2}$$So Weirerstrass test. Where did I go wrong ?
 A: Your observation implies that the series does not converge uniformly on the entire real line. Below I am guessing what the "real question" probably was meant to be :-)
The maximum of a single term $n^2x^2/(n^4+x^4)$ is achieved at $x=n$. As these drift further and further out, we can take advantage, and still prove that the series sums to an everywhere continuous function.
A way to do that with Weierstrass $M$-test is the following. Let's fix a bounded interval $[-M,M]$ where $M>0$ is some (large) number. We set as our goal to prove that the series converges uniformly in this interval. Now, with  $x\in[-M,M]$, we have
$$
\frac{n^2x^2}{n^4+x^4}\le \frac{M^2n^2}{n^4}=\frac{M^2}{n^2}.
$$
The series $\sum_{n=1}^\infty M^2/n^2$ converges, and Weierstrass kicks in.
Consequently, the series converges uniformly in $[-M,M]$ and has a continuous sum there. As $M$ was arbitrary, the series converges everywhere, and the sum function is continuous everywhere.
A: Just added for your curiosity.
I have been surprised to see your problem since, two weeks ago, in a problem from physics, I faced the same one and found that 
$$f(x)=\sum_{n =1}^\infty \frac{n^2x^2}{n^4+x^4}=\frac{\pi  x}{2 \sqrt{2}}\frac{\sinh\left(\sqrt{2} \pi  x\right)-\sin \left(\sqrt{2} \pi  x\right)}{\cosh
   \left(\sqrt{2} \pi  x\right)-\cos \left(\sqrt{2} \pi  x\right)}$$ which is quite nice $V$ shaped function.
