Find $\lim_{n \to \infty}\frac1{\ln^2n}\left( \frac{\ln 2}{2} + \frac{\ln 3}{3} +\cdots + \frac{\ln n}{n}\right)$

Define $$ S_n=\frac{\ln 2}{2} + \frac{\ln 3}{3} + ... + \frac{\ln n}{n} $$

The function $\ln(x)/x$ is decreasing for $x\geq 3$, so $$ \frac{\ln 2}{2} + \int_3^n\frac{\ln(x)}{x}\,dx \leq S_n\leq \frac{\ln 2}{2} + \frac{\ln 3}{3} + \int_3^{n}\frac{\ln(x)}{x}\,dx, $$ $$ \frac{\ln(2)}{2} + \frac{\ln(n)^2}{2}-\frac{\ln(3)^2}{2}\leq S_n\leq\frac{\ln(2)}{2}+\frac{\ln(3)}{3} + \frac{\ln(n)^2}{2}-\frac{\ln(3)^2}{2}. $$ Now divide through by $\ln(n)^2$ and apply the squeeze theorem to conclude $$ \lim_{n\to\infty}\frac{S_n}{\ln(n)^2}=\frac{1}{2}. $$