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?

  • $\begingroup$ 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

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 $b_n/a_n \to \infty.$


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 $$ \frac{-1}{\log \log \log x}? $$


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