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I've seen an example that the root test is stronger than the ratio test here, and tried to deal with it myself but it didn't go well;

Here's my attempt:

$ 1\\ +1\\ +0.5+0.5\\ +0.25+0.25+0.25+0.25\\ +\dots\\ =4\\$

$ a_1 = a_2 = 1,\ a_n = \left(\frac{1}{2}\right)^{\lceil\log_2 n\rceil-1}\ n\geq 3\\$

where $\lceil 3.5\rceil=4$ is the ceiling function

$ \text{root test : }\limsup \sqrt[n]{a_n} = \limsup \sqrt[n]{\left(\frac{1}{2}\right)^{\lceil\log_2 n\rceil-1}} =\limsup \left(\frac{1}{2}\right)^{\frac{\lceil\log_2 n\rceil-1}{n}}\\ \leq \limsup \left(\frac{1}{2}\right)^{\frac{\log_2 n-1}{n}} =\limsup 2^{\frac{1}{n}}\cdot \left(\frac{1}{2}\right)^{\log_2{n^{\frac{1}{n}}}} = 1\cdot \left(\frac{1}{2}\right)^{\log_2{\limsup n^{\frac{1}{n}}}}\\ =1\cdot \left(\frac{1}{2}\right)^{\log_2 1}=1, $

so it's inconclusive.

I guess I did a mistake when I found the general term $a_n$ or in the middle of the root test(after $\leq$).

I know this series is inconclusive when using the ratio test. So I just need to fix this root test thing, but I got stuck.

(And this is just a counterexample of (root test) $\subset$ (ratio test). How do we know that (root test) $\supset$ (ratio test)?)

user159234
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1 Answers1

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It can be shown that

$$\liminf \frac{a_{n+1}}{a_n}\le \liminf \sqrt[n]{a_n}\le \limsup \sqrt[n]{a_n}\le \limsup\frac{a_{n+1}}{a_n}$$

therefore convergence by ratio test implies convergence by root test but not viceversa.

Refer also to the related:

For the series we can use that

$$\sum_{n=3}^\infty \left(\frac{1}{2}\right)^{\lceil\log_2 n\rceil-1} =\sum_{n=1}^\infty \,\sum_{k=2^{n}+1}^{2^{n+1}} \left(\frac{1}{2}\right)^{n} =\sum_{n=3}^\infty(2^{n}-1)\left(\frac{1}{2}\right)^{n}$$

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