Can you apply the root test with the $\ln(n)$th root? I have been arguing with several people online, in school and with my GSI and no one can seem to agree on this point. Are you allowed to take the $\ln(n)$-th root of a sequence of partial sums and test for convergence the same way as the $n$th root?
 A: Sort of....
Let $\{a_n\}$ be a sequence of real numbers such that $\lim_{n\to\infty} |a_n|^{1/\ln n} = L$. Then:


*

*If $L>1$, then the series $\sum_{n=1}^\infty a_n$ diverges.

*If $L<\frac1e$, then the series $\sum_{n=1}^\infty a_n$ converges absolutely.

*If $\frac1e<L\le1$, then the series $\sum_{n=1}^\infty a_n$ does not converge absolutely; it might or might not converge conditionally.

*If $L=\frac1e$, then the series might converge absolutely, might converge conditionally, or might diverge.


Proof: 1 is trivial, since $L>1$ means that $|a_n|$ tends to infinity.
To prove 2: choose $\alpha$ between $L$ and $\frac1e$. All but finitely many $|a_n|^{1/\ln n}$ are less than $\alpha$, which means that all but finitely many $|a_n|$ are less than $\alpha^{\ln n} = n^{\ln \alpha}$. Since $\alpha<\frac1e$, we have ${\ln \alpha} < {-1}$, and so $\sum|a_n|$ converges by the comparison test.
To prove the first part of 3: choose $\beta$ between $\frac1e$ and $L$. All but finitely many $|a_n|^{1/\ln n}$ are greater than $\beta$, which means that all but finitely many $|a_n|$ are greater than $\beta^{\ln n} = n^{\ln \beta}$. Since $\beta>\frac1e$, we have ${\ln \beta} > {-1}$, and so $\sum|a_n|$ diverges by the comparison test.
To prove the second part of 3: given $\frac1e<L<1$, the examples $a_n = n^{\ln L}$ and $a_n = (-1)^n n^{\ln L}$, for which $\lim_{n\to\infty} |a_n|^{1/\ln n} = \lim_{n\to\infty} (n^{\ln L})^{1/\ln n} = \lim_{n\to\infty} L = L$, demonstrate that the series might diverge but also might converge conditionally. For $L=1$, consider the series $a_n = 1/\ln n$ and $a_n = (-1)^n/\ln n$.
To prove 4, consider the three series $a_n = 1/(n \ln^2 n)$, $a_n = (-1)^n/n$, and $a_n = 1/n$, respectively.
If I may editorialize: any belief that one could just replace $\sqrt[n]{|a_n|}$ by $\sqrt[\ln n]{|a_n|}$ without changing the conclusion of the root test betrays a lack of deep understanding about why the root test is true. $\lim_{n\to\infty} \sqrt[n]{|a_n|}=L$ means that the sequence $|a_n|$ acts enough like $L^n$ for the series's convergence or divergence to match that of the geometric series $\sum L^n$. Similarly, we see in the above proof that $\lim_{n\to\infty} \sqrt[\ln n]{|a_n|}=L$ means that the sequence $|a_n|$ acts enough like $L^{\ln n} = n^{\ln L}$ for the series's convergence or divergence to match that of the power-of-$n$ series $\sum n^{\ln L}$. In the proof of 2, for example, the upper bound $|a_n|^{1/\ln n} < n^{\ln \alpha}$ was the rigorous way of demonstrating that $|a_n|$ acts "enough like" $n^{\ln L}$.
Final note: it's easy to show that if $\lim_{n\to\infty} \sqrt[\ln n]{|a_n|}$ is finite and nonzero, then $\lim_{n\to\infty} \sqrt[n]{|a_n|}=1$, since the hypothesis is equivalent to $\lim_{n\to\infty} \ln \sqrt[\ln n]{|a_n|}$ being finite, the conclusion is equivalent to $\lim_{n\to\infty} \ln \sqrt[n]{|a_n|}=0$, and $\ln \sqrt[n]{|a_n|} = \frac{\ln n}n \ln\sqrt[\ln n]{|a_n|}$.
