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.