TITLE: Show that this limit is positive,
QUESTION [3 upvotes]: 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,

REPLY [3 votes]: 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,
$$
\pi-2x\le\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{1/2}x}\le\pi
$$
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