Using the Weierstrass M-test properly, I am slightly confused by the last part of my work here.  For $x>0$ and for $\alpha > \frac{1}{2}$,
$$\left |\frac{x}{(1+nx^2)n^{\alpha}}\right |$$
$$=\left |\frac{1}{(\frac{1}{x}+nx)n^{\alpha}}\right |$$
$$\le\left |\frac{1}{(nx)n^{\alpha}}\right |$$
$$\le \frac{1}{xn^{1+\alpha}}$$
and now I claim that 
$$\sum_{n=1}^{\infty}\frac{1}{xn^{1+\alpha}}<\infty$$
by the p-series / integral tests, and so 
$$\sum\frac{x}{(1+nx^2)n^{\alpha}}$$ converges uniformly (to a continuous function of $x$ >0) by the Weierstrass M-test.
Is my claim correct?  The $\large \frac{1}{x}$ factor in the summand confuses me a little bit; but, it is fixed, so I don't think it affects the convergence of the p-series.
Any comments are welcome.
Thanks,
 A: Edited to point out that we actually have uniform convergence on all of $\mathbb R$, not just $(0,\infty)$.

Observe that
$$0 \leq (\sqrt{n}|x| - 1)^2 = nx^2 - 2\sqrt{n}|x| + 1$$
so
$$1 + nx^2 \geq 2\sqrt{n}|x|$$
and therefore,
$$\frac{|x|}{1 + nx^2} \leq \frac{1}{2\sqrt{n}}$$
Consequently, for all $x \in \mathbb R$ we have
$$\left|\frac{x}{(1+nx^2)n^{\alpha}}\right| = \frac{|x|}{(1+nx^2)n^{\alpha}}
\leq \frac{1}{2n^{\alpha + 1/2}}$$
Now, observe that the right hand side does not depend on $x$, and its sum converges since $\alpha > 1/2$. So you can conclude from the Weierstrass $M$-test that
$$\sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{\alpha}}$$
converges uniformly for all $x \in \mathbb{R}$. Among other things, this allows us to conclude that the limit function
$$F(x) = \sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{\alpha}}$$
is continuous everywhere.
A: You cannot keep any $x$ in the upper bound -- you are looking for a uniform upperbound on $f_n(x) = \frac{x}{n^\alpha(1+nx^2)}$, i.e. something of the form
$$
\exists (M_n)_{n\geq 1} \text{ s.t. }\forall n\geq 1, \forall x > 0,\ \lvert f_n(x)\rvert \leq M_n; \text{ and } \sum_{n=1}^\infty M_n < \infty
$$
Here, your $M_n$ depends on $x$: this is not legit. A way to solve your problem would be to see if, for every $n\geq 1$, $\lvert f_n\rvert$ can be maximized by some value $x_n >0$; and set $M_n = \lvert f_n(x_n)\lvert$.
For instance, studying the function $f_n$ (and differentiating it), it appears that $f_n=\lvert f_n\rvert$ is maximized at $x_n = \frac{1}{\sqrt{n}}$. If you take $M_n = \lvert f_n(\frac{1}{\sqrt{n}})\lvert$, you do get (by construction!) $\lvert f_n(x)\rvert \leq M_n$ for all $x>0$. But then, it only remains to check whether $\sum_{n=1}^\infty M_n < \infty$.
A: In a similar fashion to this answer, we see, using the Beta Function, that
$$
\begin{align}
\lim_{x\to0}\,x^{1-2\alpha}\sum_{n=1}^\infty\frac{x}{(1+nx^2)n^{\alpha}}
&=\lim_{x\to0}\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{\alpha}x^{2\alpha}}\\
&=\int_0^\infty\frac{\mathrm{d}t}{(1+t)t^\alpha}\\[6pt]
&=\Gamma(1-\alpha)\Gamma(\alpha)\\[9pt]
&=\frac{\pi}{\sin(\pi\alpha)}
\end{align}
$$
and we can bound
$$
\frac{\pi x^{2\alpha-1}}{\sin(\pi\alpha)}-\frac{x}{1-\alpha}\le \sum_{n=1}^\infty\frac{x}{(1+nx^2)n^{\alpha}}\le\frac{\pi x^{2\alpha-1}}{\sin(\pi\alpha)}
$$
