Find $\alpha \ge 0$ such that following is uniformly convergent on (1) $[0,1]$ and (2) $[0,\infty)$ $$\sum \frac{x^\alpha}{1+n^2x^2}$$
I thought about using $1+n^2x^2\ge 2nx$ but cant conclude anything from here as we only get $\frac{x^\alpha}{1+n^2x^2} \leq \frac{x^\alpha}{2nx}$. But RHS is divergent.
By observing here, it appears that $\alpha \geq 1$ for case 1.
For $\alpha >0$ the series obviously converges pointwise to $0$ for $x=0$, but fails to converge uniformly for $x \in [0,b]$ or $x\in[0,\infty)$ if $\alpha \in (0,1]$ since,
$$\sup_{x \in (0,b]}\left|\sum_{k = n+1}^\infty \frac{x^\alpha}{1 + k^2x^2}\right| = \sup_{x \in (0,b]}\sum_{k = n+1}^\infty \frac{x^\alpha}{1 + k^2x^2} \\ >\sup_{x \in (0,b]}\sum_{k = n+1}^{2n}\frac{x^\alpha}{1 + k^2x^2} \\> \frac{n x^\alpha}{1 + 4n^2x^2}$$
Taking $x = 1/n$ we have if $\alpha \leqslant 1$,
$$\sup_{x \in (0,b]}\left|\sum_{k = n+1}^\infty \frac{x^\alpha}{1 + n^2x^2}\right| > \frac{n^{1-\alpha}}{5} \not\to 0$$
I'll leave it to you to show when the series converges uniformly.