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 $\sum_{n=2}^\infty \frac{\ln n}{1+n^2 x}$ &quot;diverges&quot; 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 $\alpha$.
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) &amp;= \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\\
&amp;= 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)\\
&amp;= 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

$$0 &lt; \int_2^\infty \frac{\sqrt{x}\ln \frac{t}{\lfloor t\rfloor}}{1+t^2x}\,dt &lt; \ln \frac{3}{2} \int_0^\infty \frac{du}{1+u^2}.$$

For the other term, noting that $n \geqslant \frac{2}{3}t$ and $t^2 - n^2 &lt; 2t+1 &lt; 3t$, we can estimate

\begin{align}
0 &amp;&lt; \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\\
&amp;&lt; \int_2^\infty \frac{x^{3/2}3t\ln t}{\bigl(1+\frac{4}{9}t^2x\bigr)(1+t^2x)}\,dt\\
&amp;= 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

$$\sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2x} = -\frac{\pi}{4}\ln x + O(1).$$
I know we already have an answer, but I would argue this way:

for all $x$ and $n\ge 2$ we have

$$
\frac{\sqrt{x}\ln{n}}{1+n^2x} &gt; \frac{\sqrt{x}\ln{n}}{n^2(1+x)}
$$

Then 

$$
\sum_{n \ge 2} \frac{\sqrt{x}\ln{n}}{1+n^2x} &gt; \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{\ln{n}}{n^2} &gt; \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{1}{n^2} = \frac{\sqrt{x}}{1+x}c
$$

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 $.