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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''.

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    @David Mitra: For calculati$n$g the *radius* of co$n$vergence of a power series, one does use root or ratio. For *numerical* series, not so much.2011-11-07

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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.)

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    Even for something like $\sum \frac{n!}{2^n}$ many students have problems with the divergence test.... Yet odly enough, they have no issue with the ratio test.2011-11-07