Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=2}^{\infty} \frac{1}{(\log n)^{\frac{1}{n}}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9ab3a5916e50e1d210b58d716553922c_l3.png "Rendered by QuickLaTeX.com")
The series divergence by the comparison test. For 
we have
![\begin{align*} \log n > 1 && \implies && \log n &> \left( \log n )^{\frac{1}{n}} \\[9pt] && \implies && \frac{1}{\log n} &< \frac{1}{(\log n)^{\frac{1}{n}}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d002d1d3ddadff63b9c16db1f4b4e643_l3.png "Rendered by QuickLaTeX.com")
Thus,
![\[ \sum_{n=2}^{\infty} \frac{1}{(\log n)^{\frac{1}{n}}} > \sum_{n=2}^{\infty} \frac{1}{\log n} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5eee76d425fce2d8c933bba30e364360_l3.png "Rendered by QuickLaTeX.com")
which we know diverges. Hence, the given series diverges as well.
Determine if the following series converges or diverges and justify your decision.
The series divergence by the comparison test. For we have
Thus,
which we know diverges. Hence, the given series diverges as well.