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.