Why $\sum_{n=1}^\infty {\ln(n^2) \over n^2 }$ converges?

Prove that $0\le \ln n \le \sqrt{n}$ first, and $$ 0\le \frac{\ln n^2}{n^2} =\frac{2\ln n}{n^2} \le \frac{2}{n\sqrt{n}}. $$ $\sum_{n=1}^{\infty}\frac{2}{n\sqrt{n}}$ converges.

Consider $a_n:=\frac{\ln (n^2)}{n^2}$, $b_n:=\frac{1}{n^{3/2}}$. We have that

$$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=\lim\limits_{n \to \infty}\frac{\ln (n^2)}{n^{1/2}}=\lim\limits_{n \to \infty}\big(2\cdot \frac{\ln (n)}{n^{1/2}}\big)=0.$$

Therefore, there exists $N$ such that $n>N$ implies $\frac{a_n}{b_n}<1$. Hence, $n> N$ implies $a_n< b_n$, and the comparison test yields convergence.