All right, managed to get \displaystyle ln(ln(n))^{ln(n)} = n^{ln(ln(ln(n)))}. Guess using n^{2} as the comparison function would work, right? Because at infinity, n^{ln(ln(ln(n)))} ≈ n^{n}, therefore I would it should converge.
Homework Statement
Analyze the convergence of the following series, describing the criteria used:
\displaystyle\sum_{n=9}^{\infty}\frac{1}{(ln(ln(n)))^{ln(n)}}
Homework Equations
None
The Attempt at a Solution
Wolfram Alpha says it converges due to comparison test, however I can't...
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Title.
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