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Does $\sum\limits_{n=3}^{\infty} \dfrac{1}{(\log\log n)^{\log n}}$ converge or diverge ? I tried some tests , but nothing conclusive is coming . Pleas help

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closed as off-topic by Did, B. Mehta, Chris Custer, Siong Thye Goh, paul garrett May 6 '18 at 22:50

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Try the Cauchy condensation principle (http://en.m.wikipedia.org/wiki/Cauchy_condensation_test) - I leave the verification of the prerequisites to you.

This shows that the convergence of the series is equivalent to that of

$$ \sum_n 2^n \cdot \frac{1}{(\log (n \cdot \log 2))^{n \cdot \log 2}}. $$

Now, for $n$ large, you have $(\log(n \cdot \log 2))^{\log 2}> 3$, which should allow you to conclude convergence.

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  • $\begingroup$ Further comment: For series containing $\log$ terms, this condensation principle is often a good idea. $\endgroup$ – PhoemueX Nov 23 '14 at 8:58