The log test

Thinking about how the ratio and root tests work while teaching, I realized that a similar test could be devised for comparing against $p$-series rather than geometric series. Written how it would appear in Stewart:

Theorem. Let ${\sum }_{n=1}^{\infty }{a}_n$ be a series of positive terms. Consider ${lim}_{n\to \infty }{\log }_n{a}_n^{-1}$. Then:

1. If this limit is $L>1$ or diverges to $\infty$, then the series converges;
2. If this limit is $L<1$ or diverges to $-\infty$, then the series diverges; and
3. Otherwise the test is inconclusive.

It can be easily strengthened to:

Theorem. Let ${\sum }_{n=1}^{\infty }{a}_n$ be a series of positive terms. If the sequence ${\log }_n{a}^{-1}$ is eventually bounded below by a number $L>1$, then the series converges, and if the sequence is eventually bounded above by a number $L<1$, then the series diverges.

Examples

1. ${\sum }_{n=2}^{\infty }(\ln n{)}^{-\ln n}$. Since ${\log }_n\left(\right(\ln n{)}^{-\ln n})=\ln (\ln n)$ tends to $\infty$ as $n\to \infty$, the series converges.
2. For which $p$ does ${\sum }_{n=2}^{\infty }(n^p\ln n{)}^{-1}$ converge? ${\log }_n\left(n^p\ln n\right)=p+\frac{\ln \left(\ln n\right)}{\ln n}$, which tends to $p$ as $n\to \infty$, so the series converges if $p>1$ and diverges if $p<1$. When $p=1$, then we may instead use the integral test.