Why does ln n use root n as a direct comparison? $$
\sum^{\infty}_{n=1}\frac{(\ln n)}{n^2}
$$

Use the direct comparison test

My teachers solution is:

$\frac{(\ln n)}{n^2} &lt; \frac{\sqrt n}{n^2} = \frac{1}{n^\frac{3}{2}}$ and so it converges because p-series > 1

How does the direct comparison go from $\ln n$ to $\sqrt n$?
hint

For $x\geq 4$, put

$$f (x)=\ln (x)-\sqrt{x}$$

$$f'(x)=\frac {2-\sqrt {x}}{2x}\leq 0$$

thus $f $ is decreasing at $[4,+\infty). $

For   enough great $n $, we have

$$f (n)\leq f (4)=\ln (4)-2\leq 0$$
and
$$\ln (n)\leq \sqrt {n} .$$

we also use this to prove that

$$\lim_{n\to\infty}\frac {\ln (n)}{n}=0.$$
It is a well known fact that $y=\ln x$ is always growing at a slower rate than $f(x)=x^\alpha$ for all values of $\alpha\gt0$ as $x\to+\infty$. It is straightforward to prove this fact.

Thus, your teacher could have chosen $\alpha = 0.0001$, for instance. The choice of $\alpha = 1/2$ is just a convenience because she could then obtain an inequality for all values of $n\ge 1$.