Claim in Rudin 3.36 On p68 he claims:

More precisely: Whenever the ratio test shows convergence, the root test does too; whenever the root test is inconclusive, the ratio 
  test is  too.  

I'm having trouble seeing why this is true. Consider the series $a_n = 1$. Then by theorem 3.33 (c) on p65 the root test is inconclusive. But by theorem 3.34 (b) on p66 the ratio test shows divergence. Isn't this a counterexample?
 A: Your observation is correct. The fact that 
\begin{equation}
\limsup_{n\to\infty} \sqrt[n]{a_n} = 1
\end{equation}
does imply that the root test gives no information, according to Theorem 3.35 (c). 
Since
\begin{equation}
\left\lvert \frac{a_{n+1}}{a_n} \right\rvert \geq 1,
\quad \forall n\in \mathbb{N},
\end{equation}
which implies
\begin{equation}
\liminf_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert \geq 1,
\end{equation}
Theorem 3.34 (b) indeed concludes that $\sum a_n$ diverges. One can also easily see this from $\lim_{n\to\infty}a_n\neq 0$.
So the issue is about interpretation of Rudin's Remarks 3.36 on page 68. I think that he meant to say that based on $\limsup_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n} \right\rvert\geq 1$ alone, the ratio test as in Theorem 3.34 says nothing about the convergence or divergence of $\sum a_n$. One can confirm this from context, specifically, the examples that he gave in 3.35. 
A: In your example both tests are inconclusive:
$$\lim_{n\to\infty}\root n\of a_n=\lim_{n\to\infty}\root n\of 1=1,$$
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{1}{1}=1.$$
A: As usual Rudin (requiescat in pace) is a bit cryptic and expects a lot from his readers.  The series of exercises Ex  3.6.7-3.6.16 in  Elementary Real Analysis elaborates on this theme (but we flagged some of them as "advanced"--if you are studying Rudin  then you are advanced!).  If Rudin annoys you occasionally then perhaps you might find the equivalent material here a bit less so.
