In this ocassion we consider the followgin series that involve ${n\brace k}$ the Stirling number of the second kind and $(n)_k$ the Pochhammer symbols. I've known from an informative point of view that in the literature was explored an example versus the definition of irrational absolutely abnormal numbers (for example from [1]).

This is the Wikipedia article dedicated to Normal number.

I wondered as curiosity if in the context of these definitions, the definitions and notions concerning normal numbers it is possible to propose some statement or conjecture about the following series

$$\sum_{n=1}^\infty\frac{1}{{2n\brace n}^{{2n\brace n}}} \tag{1}$$ or $$\sum_{n=1}^\infty \frac{1}{(2n)_{n}^{(2n)_{n}}}.\tag{2}$$

Question. Show heuristics/reasonings, or set a proposition or propose a conjecture concerning the series $(1)$ or $(2)$ in the context of the normal numbers. Many thanks.

I hope that my series and question have a good mathematical content and it makes good sense in the context of the theory of number numbers.


[1] Glyn Harman, One Hundred Years of Normal Numbers, proceeding from Number Theory for the Millennium II, A K Peters (2002).

  • 1
    $\begingroup$ "in the context of number numbers"??? $\endgroup$ Sep 6, 2019 at 13:03
  • $\begingroup$ Was a typo, many thanks I hope that my question has a good mathematical content and the series can be studied @GerryMyerson $\endgroup$
    – user142929
    Sep 6, 2019 at 13:54
  • $\begingroup$ It is well-known that (Lebesgue-) almost all real numbers are absolutely normal. So if there is no reason speaking against it, then probably your numbers are absolutely normal. Can you give any reason why you consider these particular numbers, and ask about their normality? $\endgroup$ Sep 9, 2019 at 6:33
  • $\begingroup$ I known the first claim in your comment. I believe that the only reason thus to study numbers that aren't absolutely normal is the purpose to show a specific example. My intention was edit my post since I've created two series after I've known an example that is being studied from the last paragraph of the reference [1]. My belief is that one can to deduce some statement about these reals in my post in the context of normal numbers. Many thanks for your attention @KurisutoAsutora $\endgroup$
    – user142929
    Sep 9, 2019 at 6:56

1 Answer 1


While it is very likely both numbers are absolutely normal, simply by appealing to the idea that there's no obvious reason why they should be abnormal, current proof techniques are very far from being able to prove the normality of such numbers in a given base, let alone in all bases simultaneously.

The closest thing I can think of to your example are the variants of the Korobov-Stoneham construction, which generally look like $$ \sum_{i=1}^\infty \frac{1}{q_i b^{n_i}}, $$ for positive integers $q_i,n_i$ with $q_i$ and $n_i$ increasing. In the right conditions, such numbers are known to be normal in base $b$. For example $$ \sum_{i=1}^\infty \frac{1}{3^i 2^{5^i}} $$ is normal in base $2$.

One generally chooses the $q_i$ in such a way so that the periods of the rational numbers $$ r_K=\sum_{i=1}^K \frac{1}{q_i b^{n_i}} $$ see most short strings to almost the correct frequency (in other words, the period is $(\epsilon,k)$-normal for some small $\epsilon$ and large $k$, in the terminology of Besicovitch). This is usually achieved by demanding that the prime factors of $q_i$ must belong to a finite set. However, in order to show that the full series results in a normal number, the $n_i$ must be chosen to be so swiftly growing that the behavior of the period of $r_K$ can become the dominant contribution to the full series before the next term in the series begins to alter the digits and for some time afterwards too. In particular, $n_i$ usually has to be at least on the order of magnitude of $q_i$. As such, I don't see a way to nicely fit the numbers you have shown into such a framework.

  • $\begingroup$ Many thanks, I'm going to study your answer, of course I accept your words and the words of the other user in previous comment.If in next few weeks there aren't more answers I should to accept your post as a definitively accepted answer. $\endgroup$
    – user142929
    Sep 11, 2019 at 9:22
  • $\begingroup$ I'm going to accept your great answer, convinced that isn't possible to do better it currently. $\endgroup$
    – user142929
    Sep 19, 2019 at 12:41

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