8

Is the following number demonstrably irrational? Is it demonstrably transcendental?

$$ \prod_{n=0}^{\infty} \left(1 + \frac{1}{n!} \right) = 7.36430827236725725637277250963105\dots = 7 + \frac{1}{2 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{11 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1+\ddots }}}}}}}}} $$

This number appears in the OEIS as A238695, with the note:

Conjectured to be irrational, transcendental and normal, none have been shown.

It was discussed on Math Stack Exchange: How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$?

A Bound on the Tail

Let the partial product be

$$ f_N := \prod_{n=0}^N \left(1 + \frac{1}{n!} \right) $$

Then the tail is

$$ f - f_N = f_N \left( \prod_{n=N+1}^{\infty} \left(1 + \frac{1}{n!} \right) - 1 \right) $$

We bound the tail using the inequality $ \log(1 + x) < x $:

$$ \log \left( \prod_{n=N+1}^\infty \left(1 + \frac{1}{n!} \right) \right) < \sum_{n=N+1}^\infty \frac{1}{n!} < \frac{2}{(N+1)!} $$

$$ \Rightarrow \prod_{n=N+1}^\infty \left(1 + \frac{1}{n!} \right) < \exp\left( \frac{2}{(N+1)!} \right) $$

$$ \Rightarrow f - f_N < f_N \left( \exp\left( \frac{2}{(N+1)!} \right) - 1 \right) $$

Since $ f_N < f $ and $ \exp(x) - 1 < 2x $ for small $ x $, we get:

$$ f - f_N < \frac{4}{(N+1)!} $$

Open Question

  • Can anything rigorous be concluded from this bound? Are there stronger bounds we can construct?
  • Are irrationality or transcendence results known?
  • Are continued fraction properties known that could help?

References or ideas are appreciated.

Tag Justifications

Does this have a differential annihilator that helps our cause? $$\prod_{n=0}^{\infty} \left(1 + \frac{x}{n!} \right)-\prod_{n=0}^{\infty} \left(1 + \frac{x^n}{n!} \right)$$

Mason
  • 4,489
  • Have you seen the proof of $\sum 1/n!$ being irrational? Basically, follow the same idea. – Salcio Jun 02 '25 at 22:09
  • 2
    And multiply through by the superfactorial? It didn't click through when I tried this. – Mason Jun 02 '25 at 22:34
  • 2
    @Mason one has to be careful in invoking divisibility arguments for infinite products or limits more generally; for example with $a_n=\frac{2^{2n}-1}{3}$ odd integer we have $\frac{a_n}{2^{2n}}\to \frac{1}{3}$ so if we write $\frac{a_n}{2^{2n}}=\Pi_{k \le n}q_k$ with rational $q_k$ it is easy to see that $2$ divides the denominator of each $q_n$ but it doesn't divide the denominator of the limit which is $1/3$ – Conrad Jun 03 '25 at 15:56
  • 1
    what do you mean by a transcendental function in this context? entire functions are not algebraic unless they are polynomials but this doesn't mean that their value at $1$ cannot be rational even if they have rational coefficients - just take $(z-1)e^z+P(z)$ where $P$ is any rational coefficients polynomial – Conrad Jun 04 '25 at 15:57
  • note that entire functions are never algebraic unless they are polynomials - even for $G$ functions (for example Taylor series with rational coefficients with denominators not increasing superexponentially and satisfying linear ODE's over $C[z]$) and the discussion of their values at rational or algebraic points is quite tricky; – Conrad Jun 04 '25 at 16:34
  • Let us continue this discussion in chat. – Mason Jun 04 '25 at 16:35
  • On topic but not good enough to get the result: https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0004972711002309 – Mason Jun 09 '25 at 23:17

0 Answers0