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How do I show the series $\sum{\dfrac{1}{\log(n)^{\log(n)}}}$ converges?

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  • $\begingroup$ Is that what you meant? $\endgroup$
    – Em.
    Apr 11, 2016 at 19:33
  • $\begingroup$ Yes it is. I know it converges but don't know how to prove. Thanks for the fast reply and clarification. $\endgroup$
    – user330591
    Apr 11, 2016 at 19:35
  • $\begingroup$ Comparison test: en.wikipedia.org/wiki/Direct_comparison_test $\endgroup$
    – JohnWO
    Apr 11, 2016 at 19:36
  • $\begingroup$ Thanks. Which series should I use to compare? $\endgroup$
    – user330591
    Apr 11, 2016 at 19:37

3 Answers 3

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We have

$$(\ln n)^{\ln n}=\exp(\ln n(\ln(\ln n))=n^{\ln(\ln n)}\ge n^2\;\text{for sufficiently large}\; n$$ so by comparison with $\sum\frac1{n^2}$ the given series is convergent.

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Hint: Try Cauchy's condensation test, which says that if $f$ is non-increasing and non-negative, then $\sum f(n)$ converges if and only if $\sum 2^{n} f(2^{n})$ converges. This test gives a pretty neat comparison.

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You may also use the Cauchy Condensation Test pretty easily here.

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