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Converge/Diverge?

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2 Answers

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.