 # 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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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.