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Does the series $$\sum_2^\infty\frac{\log n}{n(\log \log n)^2}$$ converge?

The root and ratio tests are inconclusive, the integral test may be too difficult to apply. I've tried the limit comparison test with $\log n/n$ but that is also inconclusive since $$\frac{\frac{\log n}{n(\log \log n)^2}}{\frac{\log n}{n}}=\frac{1}{(\log \log n)^2}\to 0.$$

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    $\begingroup$ $\sum_n \frac{1}{n}$ diverges thus so does your series $\endgroup$ – reuns Dec 18 '17 at 10:27
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Using the Cauchy condensation test, (cf Wikipedia), your series diverges. The transformed series reads : $$ \sum_{n} 2^n \frac{n}{2^n \log^2 n} = \sum_{n}\frac{n}{\log^2 n}$$

Remark : I interpreted $\log$ as a base two logarithm, which does not affect the result

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