What is $\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$? 
What is $\displaystyle\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ ?
Find an asymptotic expansion of $\displaystyle \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ as $x\to 0$

One the one hand, $\sqrt{x}\to 0$, but $\displaystyle\frac{\ln n}{1+n^2 x}\sim \ln n$ which suggests that $\sum_{n=2}^\infty \frac{\ln n}{1+n^2 x}$ "diverges" to $\infty$.
I haven't been able to tell which term dominates here, let alone the asymptotic expansion.
Here's the graph of $x\to \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ for $x\in (0,0.5)$
This suggests the limit is $\infty$ and the estimate should be $1/x^\alpha$ for some positive $\alpha$.
 A: If we look at the corresponding integral, we get
$$\int_2^\infty \frac{\sqrt{x}\ln t}{1 + t^2x}\,dt = \int_{2\sqrt{x}}^\infty \frac{\ln u - \ln \sqrt{x}}{1+u^2}\,du.$$
Both functions,
$$\frac{\ln u}{1+u^2} \quad\text{and}\quad  \frac{1}{1+u^2}$$
are Lebesgue integrable over $(0,\infty)$, so for the integral, we have the asymptotic development
\begin{align}
I(x) &= \int_0^\infty \frac{\ln u}{1+u^2}\,du - \frac{\pi}{4}\ln x - \int_0^{2\sqrt{x}}\frac{\ln u}{1+u^2}\,du + \frac{1}{2}\ln x\int_0^{2\sqrt{x}} \frac{1}{1+u^2}\,du\\
&= C - \frac{\pi}{4}\ln x + \frac{1}{2}\arctan (2\sqrt{x})\ln x - 2\sqrt{x}\ln (2\sqrt{x}) + 2\sqrt{x} + O(x)\\
&= C - \frac{\pi}{4}\ln x + 2(1-\ln 2)\sqrt{x} + O(x).
\end{align}
Now let's look at the difference between the series and the integral.
$$\frac{\sqrt{x}\ln n}{1+n^2x} - \int_{n}^{n+1} \frac{\sqrt{x}\ln t}{1+t^2x}\,dt = \sqrt{x}\int_n^{n+1} \frac{(t^2-n^2)x\ln n}{(1+n^2x)(1+t^2x)} - \frac{\ln \frac{t}{n}}{1+t^2x}\,dt.$$
We always have $0 \leqslant \ln \frac{t}{n} \leqslant \ln \frac{3}{2}$ on $[n,n+1]$, so we can estimate
$$0 < \int_2^\infty \frac{\sqrt{x}\ln \frac{t}{\lfloor t\rfloor}}{1+t^2x}\,dt < \ln \frac{3}{2} \int_0^\infty \frac{du}{1+u^2}.$$
For the other term, noting that $n \geqslant \frac{2}{3}t$ and $t^2 - n^2 < 2t+1 < 3t$, we can estimate
\begin{align}
0 &< \int_2^\infty \frac{x^{3/2}(t^2-\lfloor t\rfloor^2)\ln \lfloor t\rfloor}{(1+\lfloor t\rfloor^2x)(1+t^2x)}\,dt\\
&< \int_2^\infty \frac{x^{3/2}3t\ln t}{\bigl(1+\frac{4}{9}t^2x\bigr)(1+t^2x)}\,dt\\
&= 3\sqrt{x}\int_{2\sqrt{x}}^\infty \frac{u\ln \frac{u}{\sqrt{x}}}{\bigl(1+\frac{4}{9}u^2\bigr)(1+u^2)}\,du.
\end{align}
The latter tends to $0$ when $x\to 0$, due to the $\sqrt{x}$ factor, so the difference between the series and the integral remains bounded.
Overall
$$\sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2x} = -\frac{\pi}{4}\ln x + O(1).$$
A: Substitute $x=y^2$, then use the Riemann Integral:
$$
\begin{align}
\lim_{x\to 0}\sum_{n=2}^\infty\frac{\sqrt{x}\log(n)}{1+n^2x}
&=\lim_{y\to 0}\sum_{n=2}^\infty\frac{\log(ny)-\log(y)}{1+n^2y^2}\,y\\
&=\int_0^\infty\frac{\log(u)}{1+u^2}\,\mathrm{d}u-\log(y)\int_0^\infty\frac1{1+u^2}\,\mathrm{d}u+O(y\log(y))\\[6pt]
&=-\frac\pi2\log(y)+O(y\log(y))\\[9pt]
&=-\frac\pi4\log(x)+O(\sqrt{x}\,\log(x))\\[9pt]
&\to\infty
\end{align}
$$
A: I know we already have an answer, but I would argue this way:
for all $x$ and $n\ge 2$ we have
$$
\frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}\ln{n}}{n^2(1+x)}
$$
Then 
$$
\sum_{n \ge 2} \frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{\ln{n}}{n^2} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{1}{n^2} = \frac{\sqrt{x}}{1+x}c
$$
Once the series $\sum_{n \ge 2} \frac{1}{n^2}$ is convergent. 
This shows you that goes to infinity faster than $\frac{1}{\sqrt{x}}$, when $x \rightarrow 0 $.
