If I have an increasing sequence of integers $a_n$ and $ \sum_{n=1}^{\infty}\frac{\ln(a_n)}{a_n}$ diverges, does $\sum_{n=1}^{\infty}\frac{1}{a_n}$ also diverge
Series convergence help
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sequences-and-series
divergent-series
1 Answers
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For a counter example, consider $a_{n-1} = n \ln^2(n)$ Note that $\displaystyle \sum_{n=1}^{\infty} \dfrac1{a_n}$ converges from here. Whereas $\displaystyle \sum_{n=1}^{\infty} \dfrac{\ln(a_n)}{a_n} = \sum_{n=2}^{\infty} \dfrac{\ln(n \ln^2(n))}{n \ln^2(n)} = \sum_{n=2}^{\infty} \dfrac{\ln(n)+ \ln( \ln^2(n))}{n \ln^2(n)} = \underbrace{\sum_{n=2}^{\infty} \dfrac1{n \ln (n)}}_{\text{Diverges}} + \sum_{n=2}^{\infty} \dfrac{\ln( \ln^2(n))}{n \ln^2(n)}$
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1@BabakSorouh Yes. I too feel the sequence $a_n = \dfrac1{n \log^2(n)}$ is a nice sequence and can help us to analyze/ get an idea for slightly non-trivial convergence/divergence. – 2012-11-24