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Determine whether ∑ 1 / (log n)log n converges

Test the following series for convergence:

    \[ \sum_{n=2}^{\infty} \frac{1}{(\log n)^{\log n}}. \]

The series converges.

Proof. First, we use the definition of the exponential to rewrite the terms of the series,

    \[ \frac{1}{(\log n)^{\log n}} = \frac{1}{e^{(\log n)(\log (\log  n))}} = \frac{1}{n^{\log (\log n)}}. \]

But then for sufficiently large n we know \log (\log n) > 2 (since \log  (\log n) is strictly increasing and unbounded above). Therefore, for all sufficiently large n we have \frac{1}{(\log n)^{\log n}} < \frac{1}{n^2}; hence, by the comparison test the series

    \[ \sum_{n=2}^{\infty} \frac{1}{(\log n)^{\log n}} \]

converges. \qquad \blacksquare


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