Is my analysis of the series $\sum_{n=1}^\infty{\frac{\ln n}{n}}$ correct?

Although your method is correct and works, the method to determine convergence that seems most obvious to me (and the easiest to do) is the direct comparison test with $\frac{1}{n}$. Specifically $$\frac{\ln(n)}{n} > \frac{1}{n}$$ for $n \ge 3$. Therefore $$\sum_{n=1}^\infty \frac{\ln(n)}{n} > \sum_{n=1}^\infty \frac{1}{n}$$ which is known to diverge.

Your answer to this question is perfect .

$$ \sum_{n = 1}^{N}\frac{\log n}{n} > \sum_{n = 1}^{N}\frac{1}{n} = \log N + \gamma + O(1/N) $$ Hence it is divergent by comparison with the harmonic series $1/n$.