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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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What is the "nth-term" test? –  Phira Nov 7 '11 at 17:34
    
So you want a series in which the terms fail to approach $0$, but where it's not easy to show that they don't approach $0$, but it is easy to show via the ratio test or the root test that the series diverges. I have a question: Has any mathematician ever used the ratio test or the root test to prove divergence of a series, when the question of convergence or divergence was a nontrivial research question? –  Michael Hardy Nov 7 '11 at 17:39
    
+1. Definitely. –  Michael Hardy Nov 7 '11 at 17:40
    
Yes, that's what I want. To answer your question, I would guess ''no''. I'm mostly curious about it. The question occurred to me when I was making a ``decision chart'' for determining whether a series converges or not for my students (in which I demand they apply the nth-term test first). –  David Mitra Nov 7 '11 at 17:45
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@David Mitra: For calculating the radius of convergence of a power series, one does use root or ratio. For numerical series, not so much. –  André Nicolas Nov 7 '11 at 17:52

1 Answer 1

up vote 11 down vote accepted

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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Very nice! Thank you. –  David Mitra Nov 7 '11 at 18:13
    
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. –  N. S. Nov 7 '11 at 22:08

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