Is the root test stronger than the ratio test In Rudin, principles of mathematical analysis https://web.math.ucsb.edu/~agboola/teaching/2021/winter/122A/rudin.pdf p.68, he claims that whenever the root test is inconclusive then the ratio test is aswell. But lets consider the sequence
$1 + 1 + ...$
Clearly $\left|\frac{a_{n+1}}{a_n}\right| = 1 \quad \forall n$ therefore the ratio test implies divergence but
$$\limsup_{n \to \infty} |a_n|^{1/n} = 1$$ so the root test gives no information. Is this a counter example to Rudin's claim or am I missing something?
Edit: It seems that there is cofusion here around the definition of the ratio test. Rudin claims (p.66) that a sequence diverges if
$$\exists n_0 \text{ such that } n>n_0\text{ implies } \left|\frac{a_{n+1}}{a_n}\right| \geq 1.$$
There may be other definitions of the ratio test but with this definition is the example given a counter example?
 A: The ratio test is inconclusive in the case where $$\left|\frac{a_{n+1}}{a_n}\right| = 1.$$  There are both convergent and divergent series for which this ratio equals $1$.  The link illustrates examples of such cases.  Therefore, your example does not furnish a counterexample to the claim.
A: The ratio test giving you a $1$ is not divergence, but rather says that it is inconclusive. Also the ratio test looks at the limit, not simply the ratio for a fixed $n$.
In fact your example is one of the examples on the Wikipedia page.
Wikipedia seems to say that the root test actually implies divergence if the $\limsup$ is $1$ and it approaches strictly from above; I wasn't aware of this, I'm pretty sure Rudin doesn't mention this form of the root test either, but if this is true, this actually shows that your example diverges.
The root test is stronger than the ratio test for the simple reason that the limit of the expression in the root test is less than or equal to that of the ratio test; so if the limit exists in the latter, the limit must exist in the former. This answer gives a formal proof.
