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