Rudin PMA p.63

$\sum 1/n$ diverges, but $\sum 1/{n^2}$ converges. Likewise, $\sum \frac 1{n \log n}$ diverges, but $\sum \frac 1{n (\log n)^2}$ converges.

This might lead us to the conjecture that there is a limiting situation of some sort "boundary".

However, the conjecture is false. The book says one can check this by showing following two theorems.

Theorem 1;

If $a_n >0$, $s_n=a_1 + \cdots + a_n$ and $\sum a_n$ diverges, then $\sum a_n / s_n$ diverges.

Theorem 2;

If $a_n>0$, $r_n=\sum_{m=n}^\infty a_m$ and $\sum a_n$ converges, then $\sum a_n / \sqrt r_n$ converges.

I have proved those two theorems, but still don't understand how are these two theorems related to "boundary conjecture"..

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  • $\begingroup$ I have never liked this kind of phrasing in textbooks (exercises of this sort are common in analysis books). In my view, it is more appropriate to say that these theorems show that one way (or perhaps even a class of ways) of attempting to formalize a "boundary" between convergent and divergent series will not work. (The fact that one could likely come up with examples dooming any way of trying to do this is interesting, but almost "meta-mathematical," and not really the point.) Someone will surely give more detail about the relevance of this example to a "boundary" in an answer. $\endgroup$ – leslie townes Oct 1 '12 at 15:54
  • $\begingroup$ The "correct" statement of Theorem 2 should have as conclusion that $\sum a_n/\sqrt{r_n}$ converges. Use Exercise 3.12(b) instead of 3.12(a)! $\endgroup$ – Per Manne Oct 1 '12 at 16:04
  • $\begingroup$ @Per My mistake! Edited $\endgroup$ – Katlus Oct 1 '12 at 16:13
  • $\begingroup$ Parentheses, please $1/n \log n$ can be read as $(1/n)\log n$ and similarly for $1/n(\log n)^2$ or use stacked fractions with \frac {numerator}{denominator}. $\endgroup$ – Ross Millikan Oct 1 '12 at 17:05

Theorem 1 says that if $\sum a_n$ is a divergent series with positive terms, there is always another divergent series $\sum b_n$ whose terms increase more slowly than $a_n$. (More slowly means that $b_n/a_n \to 0$ as $n\to \infty$.)

Theorem 2 (with the correction in my comment) says that if $\sum a_n$ is a convergent series with positive terms, there is always another convergent series $\sum c_n$ whose terms increase more rapidly than $a_n$. (More rapidly means that $c_n/a_n\to\infty$ as $n\to\infty$.)

In other words, no single series $\sum a_n$ can function as a boundary between convergent and divergent series.

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