# Which series converges the most slowly?

We say $a_n$ converges slower than $b_n$ 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.

Ignoring constant factors, which type of function converges the slowest?

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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

There is not such a function! Walter Rudin in his "Principles of Mathematical Analysis" has a series of exercises trying to indicate that there is no function at the "boundary" between convergence and divergence.

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.

I recommend that you take a look at the answers at MO and at the references they suggest, for more subtle examples.

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What happens if you apply that trick recursively, b_n->a_n/(r_n)^0.5, c_n->b_n/(r(b)_n)^0.5, then it converges for a,b,c,d,e.., if we had inf letters, and took the limit, it should also converge? What function does it converge to? – TROLLKILLER Feb 4 '11 at 11:08
@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. – Andrés E. 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