Prove that with $0<\delta <1$
$\sum_{n=0}^{\infty} \frac{n^2x}{1+n^4x^2}$
converges uniformly on $[\delta, 1]$
Prove that the series does not converge on $(o,1]$.
The issue is that for series I have been abusing the Weierstrass M-Test. I do not see how it is applicable in this scenario. I think that I want to just directly prove that a function $f$ defined by $f(x)=\sum_{n=0}^{\infty} \frac{n^2x}{1+n^4x^2}$
For the first part, we can do exactly as Santiago suggested: Let $\delta\in]0,1[$. Then for $x\in[\delta,1]$, we have $f_n(x):=\dfrac{n^2x}{1+n^4x^2}\leq \dfrac{n^2x}{n^4x^2}=\dfrac{1}{n^2x}\leq \dfrac{1}{n^2\delta}$. Set $\forall n: M_n:=\dfrac{1}{n^2\delta}$. Then as $\sum M_n=\dfrac{1}{\delta}\sum{\dfrac{1}{n^2}}<\infty$, by Weierstrass $M$-test $\sum{f_n(x)}$ is uniformly convergent on $[\delta,1]$.
For the second part, note that the use of picking a positive $\delta$ is simply that we don't evaluate the sum near $0$. Set $\{x_n\}_n:=\left\{\frac{1}{n^2}\right\}_n$. Then $x_n\downarrow 0$, and $\forall n: f_n(x_n)=\dfrac{1}{2}$. Consequently $\sum{f_n(x_n)}=\infty$.
As for finding the limit function $f$, I don't see it explicitly. However, recall that the uniform limit of a sequence (or series) of functions is not different than its pointwise limit. So that the common procedure in finding a uniform limit is to first find the pointwise limit and then test it for uniform convergence. Hence, after proving that the series actually converges uniformly, it suffices to find a function $g$ that gives the exact values as the sum at each point in $[\delta,1]$. $g$ and $f$ the same functions on $[\delta,1]$ (why?). Finding explicit formulas are usually cumbersome, hence we have all these tests and theorems and what have you.