Let us consider absolute convergent series $\ell^{1^+}$ ordered under eventual dominance (mod finite) $<^*$. T. Bartoszynski proved that unbounded number ${\frak b}(\ell^{1^+}, <^*)$ equals additivity of Lebesgue measure. In P. Vojtas. The strength of the comparison test versus gaps between convergent and divergent series, In Prague Topological Symposium 1986, Helderman Verlag, Berlin, 1988, 617-622 (http://www.ksi.mff.cuni.cz/~vojtas/MathPub/1986_TheStrengthOfTheComparisonGaps.pdf) we proved that there is in ZFC a Hausdorff gap $\{a_\alpha, b_\alpha, \alpha < \omega_1\}$ such that $\{a_\alpha, \alpha < \omega_1\}\subseteq\ell^{1^+}$ and $\{b_\alpha, \alpha < \omega_1\}\subseteq c_0^+\setminus\ell^{1^+}$.

So approaching the border between absolute convergence and divergence of series from one side is consistently harder than from both sides simultaneously by a gap. Nevertheless you need uncountable many steps always.

This is contrast with the countable increasing sequence of spaces $\bigcup_{p=2}^{\infty}\ell^{1-\frac{1}{p}}$. Moreover $\ell^{1^+}\setminus \left(\bigcup_{p=2}^{\infty}\ell^{1-\frac{1}{p}}\right)$ is nonempty - this is an external description of set of series which are "close to border". Is there an internal characterization - at least of some series described by an explicit formula - such that series converges but no power $1-\frac{1}{p}$ of it does?

Such an internal explicit description can show that our attempt to reach the border by a countable language is hopeless and in asymptotics one has to work with infinitary methods either in ZFC or in PA. Are there any papers in this direction?

  • 2
    $\begingroup$ Math Stackexchange is more appropriate place for asking this type of questions . $\endgroup$ – Piotr Hajlasz May 10 '18 at 13:02
  • 4
    $\begingroup$ And on MathSE, as a title you should write something informative instead of a useless comment. $\endgroup$ – YCor May 10 '18 at 13:05
  • 4
    $\begingroup$ How about $1/(n(\log n)^2)$? $\endgroup$ – Emil Jeřábek May 10 '18 at 13:21
  • $\begingroup$ I have changed whole question on May 12 - whole title, description in context of a research challenge - sorry for useless title and too low level question $\endgroup$ – Peter Vojtas May 17 '18 at 7:22

Your Answer

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

Browse other questions tagged or ask your own question.