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

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

It can be easily strengthened to:

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