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 $\alpha$.

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 $.
