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A student asked a very insightful question in my Calculus class this morning. I did not know how to answer him. (Admittedly, I am not an analyst by trade: once I passed my qualifiers I never looked back.) I would like to know if anyone here can give a precise answer, and if that can be massaged into an answer understandable by someone in Calculus II.

The topic of the lecture was the $n$th term test (or "divergence test") for infinite series. I presented it as:

Theorem: If $\sum_n a_n$ converges then $\lim_{n \to \infty} a_n=0$.

I proved this, then had them state the contrapositive:

Divergence Test: If $\lim_{n \to \infty} a_n \neq 0$ then $\sum_n a_n$ diverges.

I then gave a litany of examples. To their credit, they never fell into the false-converse trap. I have been harping on dogs/mammals/horses all semester so they are very good about avoiding that (if you are a dog then you are a mammal but the converse is false if you can find a horse).

My second example was the harmonic series, which of course has $\lim_{n \to \infty} a_n=0$ yet fails to converge. Hence we definitely have horses in my theorem above (and they all spotted this).

Enter the sharp student. He asked if there was an improvement on my first theorem so that the converse becomes true. I had said earlier in my example that although $\lim_{n \to \infty} \frac{1}{n}=0$, it doesn't converge "fast enough" to $0$ to make the harmonic series converge. The student asked for a measure of "fast enough" or at least a precise statement of this. What he is fishing for is something like:

Improved Theorem: If $\sum_n a_n$ converges then $\lim_{n \to \infty} a_n=0$ and (extra-nice condition on the speed of the convergence to $0$).

My instinct is there is no answer in terms of the terms $a_n$. The only answer is that the terms must vanish quickly enough to make the sequence of partial sums convergent (which is an unsatisfying answer to his question). If he only cared about $p$-series then I can be precise ($p > 1$), but he is asking about generic series whose terms vanish in the limit.

I hope my question is clear. Am I correct that all this is much too subtle to have a nice, clean answer on the "rate" of convergence of the terms to $0$?

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  • $\begingroup$ What about using this limit comparison test with the series $\Sigma 1/n^{1+\epsilon}$? $\endgroup$
    – snulty
    Sep 25, 2017 at 15:17
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    $\begingroup$ Note: The proper contrapositive (Divergence Test) should also include the case that $a_n$ does not converge at all, not only the case that it ocnverges to a non-zero limit! $\endgroup$ Sep 25, 2017 at 15:17
  • $\begingroup$ Well, there's the integral test, but I'm not sure that's a particularly "nice" condition. $\endgroup$
    – rogerl
    Sep 25, 2017 at 15:17
  • $\begingroup$ @HagenvonEitzen: yes, I teach that by declaring what I mean to say the limit "isn't 0." $\endgroup$
    – Randall
    Sep 25, 2017 at 15:18
  • $\begingroup$ @rogerl yes, but that would only apply to series with positive, decreasing terms, not a generic series with vanishing temrs. The more I think about this the more I realize my question is ridiculous. $\endgroup$
    – Randall
    Sep 25, 2017 at 15:20

3 Answers 3

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Without additional assumptions on $a_i$, there can be no such theorem. Suppose $x_n$ is a series that, no matter how slowly, converges to zero. Then the sum of the series $\{x_1, -x_1, x_2, -x_2, \ldots \}$ converges to zero, too.

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The answer could depend on how exactly we sum the series, because, conditionally convergent series possibly could be permutated so that terms at the same rate go to zero but one sequence converges and permutated one diverges.

Also it is a question of how to formulate all that as a theorem even in the domain of sequences with all strictly positive terms because there is no sequence with slowest rate of growth among all strictly positive sequences, right?

So, nice question, but certain breakthroughs in the field are needed before we have some major theorem of that kind, maybe that student wil accomplish right that.

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Note that if $a_n = \frac{1}{n^{1 + \epsilon}}$ where $\epsilon > 0$, then the series converges by the integral test (or p-series). Then you can argue via the limit comparison test, e.g. if the terms approach zero in a similar fashion (as $\frac{1}{n^{1 + \epsilon}}$, then the series converges as well.

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