Non-equivalence of D'Alembert's and Cauchy's criterion? Is there a simple example where D'Alembert's and Cauchy's criterion (the root test) for convergence of infinite series don't agree, i.e. one of them proves inconclusive?
Can you explain why that happens? Intuitive explanations along with rigorous reasoning are more than welcome!
 A: Notice that Cauchy's root test uses $\limsup\sqrt[n]{|a_n|}$, which always exists, while d'Alembert ratio test requires $\lim\left|\frac{a_{n+1}}{a_n}\right|$ to exist. So one can find examples where Cauchy's test works, but d'Alembert doesn't. But if the the limit exists, then $$\limsup\sqrt[n]{|a_n|}=\lim\left|\frac{a_{n+1}}{a_n}\right|$$
So they both give the same result.
A: A comparison between (D'Alembert's) Ratio Test and (Cauchy's) Root Test can be found here or here.  In particular, the folk adage that "The root test is stronger than the ratio test" is given a precise meaning in terms of limsups and liminfs.  Like most expositions of undergraduate analysis, it is borrowed from Rudin's Principles of Mathematical Analysis (in this case, Theorem 3.37).
(There is no need to look at both: it's the same discussion, just taking place inside highly overlapping lecture notes for two different undergraduate courses.  In the unlikely event that anyone cares this much: the second set of notes is called a "source book" on sequences and series, so is meant to be a more exhaustive reference for that topic, whereas the first set of notes is an honors calculus text treating plenty of other things besides sequences and series.  So the first set of notes has some things on sequences and series that the second doesn't, and aspirationally it should contain much more on those topics.)
A: The Cauchy criterion is always conclusive, by definition of convergence. 
The d'Alambert criterion has several ways to be inconclusive: The limit may not be defined, not exist or equal 1. The latter is the case for $\sum_{n=1}^\infty \frac1{n^s}$ for arbitrary $s$ (we have convergence for $s>1$, divergence for $0<s\le 1$).
