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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

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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 It is a magic sequence $a_n$. I was searching in my old examples but didn't find such this key-like sequence. –  Babak S. Nov 24 '12 at 7:59
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@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. –  user17762 Nov 24 '12 at 8:01

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