Let $\lambda_n=n$ and $A=\overline{\lim}_n |a_n|^{\frac{1}{\lambda_n}}$.
Cauchy's ($n$-th) root test states that
- $A<1 \ \implies \ \sum_n |a_n|< \infty$
- $A>1 \ \implies \ \sum_n |a_n|= \infty$
If we assume only that:
$$\frac{\lambda_n}{\log(n)}\to \infty$$
then we can prove similar claims:
-
If $A<1$ then by the properties of $\lambda_n$ and $\overline{\lim}$ there is a $n_0$ such that if $n\ge n_0$:
$$
|a_n|^{\frac{1}{\lambda_n}}\le \sup_{k\ge n} |a_k|^{\frac{1}{\lambda_k}}< \frac{1+A}{2}=q<1 \\
$$
$$
\log(q)\frac{\lambda_n}{\log(n)}<-2 \\
$$
$$
\implies |a_n| < q^{\lambda_n} =
e^{\log(q)\lambda_n} = e^{\log(n) \log(q)\frac{\lambda_n}{\log(n)}} = n^{\log(q)\frac{\lambda_n}{\log(n)}}< \frac{1}{n^2}
$$ -
If $A>1$ then there is a subsequence $a_{k(n)}$ for which $|a_{k(n)}|>1$, therefore $a_n$ is not a null sequence.
Use cases:
-
1
$$
\sum_n \frac{1}{3^{\sqrt{n}}}<\infty\ \hspace{2cm}
\lambda_n=\sqrt{n}\\
$$ -
2
$$
\sum_n \frac{n}{e^{\sqrt{n}}}<\infty\ \hspace{2cm}
\lambda_n=\sqrt{n}\\
$$ -
[1]+[2]
$$
r>1,\ \alpha, \beta>0\ \ \ \implies \sum_n \frac{n^\beta}{r^{n^\alpha}}<\infty\ \hspace{2cm} \lambda_n=n^{\alpha}\\
$$
Questions:
- Is the derivation correct?
- I think that it is a toy test, but for some sequences it provides a routine way to study the convergence.
What do you think about it?
Best Answer
The derivation of the test is correct. I am not sure how useful the test is in practice, since off the top of my head I can't see examples where this test works and its application is straightforward but an application of the comparison test (or, in case of eventually monotonic sequences like your examples, the integral comparison test or the condensation test) would be considerably more difficult.
Your test is in these examples simpler, but whether the gain in simplicity is large enough to outweigh the burden of remembering yet another test I cannot yet tell.
But even if it turns out to be not very useful in practice, it was good thinking to come up with it.