# Which series converges the most slowly?

a_n converges more slowly then b_n if there exist an x such that for all m>x, a_m>b_m, and both sum a_n and sum b_n converges for n=1 to n=inf.

Ignoring constant factors, which type of function converges the most slowly?

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See also Qiaochu Yuan's answer to this related question: math.stackexchange.com/questions/9350/… –  Jonas Meyer Feb 4 '11 at 13:05

A quick way of seeing that there is no "slowest" convergent series is Rudin's exercise 12.b, mentioned at the link above: If $\sum a_n$ converges, and the $a_n$ are positive, then $\sum a_n/\sqrt {r_n}$ converges, where $r_n$ is the tail $\sum_{i\ge n}a_i$. Note $r_n\to 0$, so $a_n/\sqrt{r_n}>a_n$ for $n$ large enough.
@solomoan: If you have such a countable sequence of series, you can always diagonalize to get a larger one (carefully, to ensure convergence), and re-start the process. If you know about ordinals, you see that this will lead us to a "sequence" of length at least $\omega_1$. I'm pretty sure how far one can go quickly leads to independence questions in set theory. –  Andres Caicedo Feb 4 '11 at 14:53
Andres: I seem to remember something about the 'Gowers series' mentioned in that answer being the 'primitive recursive' boundary: If we define $f(n) = n*\mathrm{log}(n)*...*\mathrm{log}^{(k)}(n)$, iterated to the last positive term, then the series $\Sigma {1\over f(n)}$ diverges, but for any unbounded PR function $g(n)$, the series $\Sigma {1\over f(n)g(n)}$ converges. Maybe this was in Companion To Concrete Mathematics somewhere? I don't have my copy to hand... –  Steven Stadnicki Feb 4 '11 at 18:25