TITLE: Do we need to use the Ratio/Root test to determine divergence of a series?
QUESTION [12 upvotes]: From the proofs of the Root and Ratio tests for a series, one deduces that if one of these tests shows divergence, then the terms of the series in question do not tend to zero. 
I am therefore interested in finding  an example of a divergent series (accessible to Calc II students) for which the Ratio or Root test is substantially easier to apply that the $n^{\rm th}$-term test (the Divergence Test). Does anyone know of one?
Thank you for any help, and I apologize in advance for the vague requirement ``substantially easier''.

REPLY [12 votes]: A series like this perhaps:
$$\sum_{n=1}^\infty \frac {3^n n!}{n^n}$$
Although the limit of this sequence is indeed not zero, I don't think most Calc I or II students would be able to prove it easily without resorting to a very tailored approach for this problem. On the other hand, the ratio test handles this one easily.
That is, provided they are not commonly aware that $$\lim_{n\to\infty} \frac {(n!)^{\frac 1 n}} n=\frac 1 e$$
(I wasn't when I took Calc I and II.)