Prove that the series $\sum_{n=0}^{\infty}{\frac{n^2 x}{1+n^4 x^2}}$ is uniformly convergent. Show that
$\sum_{n=0}^{\infty}{\frac{n^2 x}{1+n^4 x^2}}$
converges uniformly on $[\delta,1]$ where $0<\delta<1$.

I tried to use $M$ test.
$$\frac{n^2 x}{1+n^4 x^2} \leqslant \frac{n^2 x}{n^4 x^2} = \frac{1}{n^2 x}$$
Then, can I say that: $$\sum_{n = 1}^{\infty}\frac{1}{n^2x}$$ converges if I prove that: $$\sum_{n = 1}^{\infty}\frac{1}{n^2}$$ is convergent??
If so, it would prove the theorem right?
So here, do I treat $x$ in this case as a constant??
I am confused...
 A: Since $x \in [\delta ,1]$ $1/x \le 1/ \delta$. Hence $\sum_{n=1}^\infty\frac 1 {n^2x} \le \frac 1 \delta\sum_{n=1}^\infty \frac 1  {n^2}$ and you get your result by the M-test.
A: For almost esthetical reasons, I enjoy infinite sums of series and I always wonder if they have a closed form. Your series looked very appealing to me and, surprizingly or not, using a CAS, I found that, for the infinite sum, the result is $$\frac{1}{4} \pi  \Re\left(\frac{\sqrt[4]{-1} \left(\cot \left(\frac{(-1)^{3/4} \pi
   }{\sqrt{x}}\right)+i \cot \left(\frac{\sqrt[4]{-1} \pi
   }{\sqrt{x}}\right)\right)}{\sqrt{x}}\right)$$ which I have not been able to simplify further (the complex part of this expression is zero). I suppose that developing the $cot$ would probably give simpler things but I am too lazy to do it. 
Added later
Thanks to b.gatessucks on Mathematica StackExchange, this simplifies nicely to
$$\frac{\pi  \left(\sin \left(\frac{\sqrt{2} \pi }{\sqrt{x}}\right)-\sinh
   \left(\frac{\sqrt{2} \pi }{\sqrt{x}}\right)\right)}{2 \sqrt{2} \sqrt{x} \left(\cos
   \left(\frac{\sqrt{2} \pi }{\sqrt{x}}\right)-\cosh \left(\frac{\sqrt{2} \pi
   }{\sqrt{x}}\right)\right)}$$  
What is also interesting is that, for large values of $x$, the corresponding Taylor expansion gives $$\frac{\pi ^2}{6 x}-\frac{\pi ^6}{945
   x^3}+\frac{\pi ^{10}}{93555 x^5}-\frac{2 \pi ^{14}}{18243225 x^7}$$
