Show that this limit is positive, I want to show that, for $\alpha = \frac{1}{2}$,
$$\lim_{x \to 0} \sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{\alpha}}>0$$
Any ideas are welcome.
(In a previous question, I considered the case for $\alpha > \frac{1}{2}$, from which we were able to derive a uniform upper bound and use Weierstrass M-test and Dominated Convergence Theorem to establish the continuity of the series and evaluate the limit as $x \to 0$)  
Thanks,
 A: For small $x$ we can distinguish the initial part of the series with $nx^2\leq 1$ and the rest.  For computational convenience assume that $x=1/m^{1/2}$ for some integer $m$.  The initial part of the series then goes from $n=1$ up to $n=1/x^2$. We then have $1+nx^2\leq 2$ in that part of the series, and so the sum is $\geq (x/2)\sum_{n=1}^{1/x^2} n^{-1/2}$, which is $$\geq {x\over 2}\int_1^{1/x^2} y^{-1/2}\, dy= xy^{1/2}|_1^{1/x^2}=x(1/x-1)=1-x.$$
A: Multiplying the numerator and denominator by $x$, it can be seen that this is a Riemann Sum:
$$
\begin{align}
\lim_{x\to0}\sum_{n=1}^\infty\frac{x}{(1+nx^2)n^{1/2}}
&=\lim_{x\to0}\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{1/2}x}\\
&=\int_0^\infty\frac{\mathrm{d}t}{(1+t)\sqrt{t}}\\
&=\int_0^\infty\frac{2\mathrm{d}u}{1+u^2}\\[6pt]
&=\pi
\end{align}
$$

Convergence of the Riemann Sum
In comparison to the curve $\frac1{(1+t)\sqrt{t}}$, the sum above is the sum of the area of the rectangles below whose widths are $x^2$.

Thus,
$$
\begin{align}
\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{1/2}x}
&\le\int_0^\infty\frac{\mathrm{d}t}{(1+t)\sqrt{t}}\\
&=\pi
\end{align}
$$
However, if we shift the rectangles right by $x^2$,

we see that
$$
\begin{align}
\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{1/2}x}
&\ge\int_{x^2}^\infty\frac{\mathrm{d}t}{(1+t)\sqrt{t}}\\
&\ge\int_0^\infty\frac{\mathrm{d}t}{(1+t)\sqrt{t}}
-\int_0^{x^2}\frac{\mathrm{d}t}{\sqrt{t}}\\[6pt]
&=\pi-2x
\end{align}
$$
Therefore,
$$
\pi-2x\le\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{1/2}x}\le\pi
$$
A: Just to supplement ForgotALot's nice answer, here is a proof that
$$\sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{\alpha}}$$
converges when $\alpha = 1/2$, so that the question makes sense. Note that $1 + nx^2 > nx^2$, so for $x \neq 0$ we have
$$\frac{1}{1+nx^2} < \frac{1}{nx^2}$$
Consequently,
$$\left|\frac{x}{(1+nx^2)n^{\alpha}}\right| < \frac{|x|}{n^{1 + \alpha}x^2} = \frac{1}{n^{1+\alpha}|x|}$$
For fixed $x \neq 0$, the sum of the right-hand side is finite provided that $\alpha > 0$, and this includes $\alpha = 1/2$, so this shows that
$$F(x) = \sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{1/2}}$$
converges absolutely for all $x \neq 0$. Also, since the terms are all zero at $x=0$, we also have $F(0) = 0$.
Since ForgotALot's answer shows that $F(x) \geq 1-x$ for $x > 0$, we can certainly rule out $\lim_{x \to 0}F(x) = 0$. 
However, this doesn't necessarily imply that $\lim_{x \to 0}F(x) > 0$, because (as far as I can tell) it hasn't yet been established that $\lim_{x \to 0}F(x)$ exists. I'll edit if I can prove this.
