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

OlmekinjP 2021-06-23 Answered

Examine whether the series \(\sum_1^∞=(\log n)^{\log n}\) is convergent.

\(\displaystyle{n}={2}\)

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Expert Answer

Derrick
Answered 2021-06-24 Author has 15485 answers

Notice that, for all \(\displaystyle{n}\Rightarrow{2}\)
\(\displaystyle{\left({\log{{n}}}\right)}^{{\log{{n}}}}={e}^{{\log{{\left({\left({\log{{n}}}\right)}^{{\log^{n}}}\right)}}}}={\left({e}^{{\log{{n}}}}\right)}^{{\log{{\left({\log{{n}}}\right)}}}}={n}^{{\log{{\left({\log{{n}}}\right)}}}}\)
Also note that \(\displaystyle{\log{{\left({\log{{n}}}\right)}}}{>}{2}\) for all \(\displaystyle{n}{>}{e}^{{e^{2}}}\).

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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