My attempt included finding the maximum of the sequence of functions to find the maximum of the function and then using the Wierstrauss $M$-Test. However, I couldn't find a suitable $M$ that worked for comparison to the original function. I looked at this very similar problem but I am not sure it would work for this version with an exponent alpha greater than $\frac{1}{2}$.
Here is my proof:
WTS: For every $\epsilon$ greater than 0, $\exists M_\epsilon.\mid{f(x)} - \sum_{n=1}^\infty{f_n}(x)\mid\leq \mid \sum_{n=1}^\infty{f_n}(x)\mid \leq M_\epsilon$ where the $M$ is an arbitrary function chosen by finding the maximum of the given sequence.
$\mid \sum_{n=1}^\infty \frac{x}{n^\alpha (1+nx^2)}\mid \leq \sum_{n=1}^\infty \mid \frac{1}{n^\alpha} \mid$. This is as far as I got since when $\alpha = 1$ the series diverges. I eventually would like to come to the conclusion that
$\mid \sup_{x\in R[a,b]}\sum_{n=1}^\infty{f_n} \mid \leq \sup_{x\in R[a,b]}\sum_{n=1}^\infty \mid{f_n}\mid < M_\epsilon$
