Examine whether the series ∞∑ 1/(logn)^logn∑ is convergent.n=2

kuCAu 2021-08-18 Answered

Examine whether the series \(\sum_{n=2}^\infty \frac{1}{(\log n)^{\log n}}\) is convergent. \(\displaystyle{n}={2}\)

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Expert Answer

aprovard
Answered 2021-08-19 Author has 11370 answers

Notice that, for all \(\displaystyle{n}\Rightarrow{2}\)
\((\log n)^{\log n}=e^{\log((\log n)^{\log^n})}=(e^{\log n})^{\log(\log n)}=n^{\log(\log n)}\)
Also note that \(\displaystyle{\log{{\left({\log{{n}}}\right)}}}{>}{2}\) for all \(n>e^{{e}^2}\). Choose n0 such that \(\displaystyle{n}_{0}{>}{e}^{{{e}}^{{2}}}\). Then for all \(\displaystyle{n}\Rightarrow{n}{0}\) we have \(\displaystyle\frac{{1}}{{{\left({\log{{n}}}\right)}^{{\log{{n}}}}}}=\frac{{1}}{{{n}^{{\log}}}}{\left({\log{{n}}}\right)}\le\frac{{1}}{{n}^{{2}}}\)
The given series is convergent by comparison test.

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