4
$\begingroup$

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"..

  • 0
    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}.2012-10-01

1 Answers 1

4

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.