Since f(x)=(ln(x))²/x is positive and continuous on [1,∞), the integral test is applicable to the series
∑ n=1∞ (ln(n))²/n.
Consider the improper integral
∫1 ∞(ln(x))²/x dx = limb→∞∫1 b (ln(x))²/x dx
= limb→∞ [(ln(x))³/3]b1
which diverges, so the series ∑ n=1∞ (ln(n))²/n also diverges.