If the integral test applies, use it to determine whether the series converges or diverges. ∑ 1, ∞ (ln(n))2/n
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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.
It is clear that this series diverges because, apart from the first two terms, each term is greater than 1/n.
The series ∑ 1,∞ 1/n diverges.