5
$\begingroup$

I lost my baby Rudin book on real analysis book but I recall a pair of results in homework exercises that he seemed to indicate that there is no "boundary" between convergent and divergent series of positive decreasing terms. One result was that if $a_n$ is positive decreasing, and $\sum_n a_n$ is divergent, then $\sum_n a_n/s_n$ is also divergent where $s_n$ is the $n$th partial sum of the $a_i$. So, since the original series diverges, we can keep repeating this construction to get series that diverge slower and slower, since $\lim_{n \to \infty} s_n = \infty$.

Rudin paired this with another homework exercise result about convergent series, something showing that given any convergent series (possibly of positive decreasing terms) you could construct a new series that converged "slower" in some obvious sense. Can someone recall that result?

$\endgroup$
  • $\begingroup$ See this answer for links to literature on this topic, and excerpts, (du Bois-Reymond, Hardy, etc). $\endgroup$ – Bill Dubuque Apr 14 '15 at 19:05
3
$\begingroup$

If $\sum a_n = \infty,$ then we can find $n_1 < n_2 < \dots $ such that $\sum_{n_k\le n < n_{k+1}} a_n > 1$ for all $k.$ Define $b_n = a_n/k$ for $n_k\le n < n_{k+1}.$ Then $\sum b_n =\infty,$ and $b_n/a_n \to 0.$

If $\sum a_n < \infty,$ then we can find $n_1 < n_2< \dots $ such that $\sum_{n_k\le n < n_{k+1}} a_n < 1/2^k$ for all $k.$ Define $b_n = ka_n$ for $n_k\le n < n_{k+1}.$ Then $\sum b_n <\infty,$ and $b_n/a_n \to \infty.$

$\endgroup$
0
$\begingroup$

Never seen it, but I can repeat the standard collection of examples that exhibit this behavior. The series with terms

$$ \frac{1}{n}, $$ $$ \frac{1}{n \log n}, $$ $$ \frac{1}{n \log n \log \log n}, $$ $$ \frac{1}{n \log n \log \log n \log \log \log n}, $$ all diverge (we need to start at $n = n_0$ large enough to get all terms defined)

The series with terms $$ \frac{1}{n^2}, $$ $$ \frac{1}{n \log^2 n}, $$ $$ \frac{1}{n \log n \left( \log \log n \right)^2 }, $$ $$ \frac{1}{n \log n \log \log n \left( \log \log \log n \right)^2}, $$ all converge.

In both families we are comparing with easy explicit integrals. For the first bunch, divergent, what is the derivative of $\log \log \log \log x?$ Trying to remember second bunch.

There we go, what is the derivative of $$ \frac{-1}{\log \log \log x}? $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.