This post is inspired by the Numberphile video 2.920050977316, advertising the paper A Prime-Representing Constant by Dylan Fridman, Juli Garbulsky, Bruno Glecer, James Grime and Massi Tron Florentin, involving an alternative to continued fractions. The goal of this post is to discuss the relevance of this alternative by asking whether it can prove the irrationality of numbers for which it was unknown before.

Let us first recall the notion of continued fraction. For a given number $\alpha>0$, consider the recurrence relation $u_0 = \alpha$ and $$ u_{n+1} = \begin{cases} (u_n - \lfloor u_n \rfloor)^{-1} & \text{ if } u_n \neq \lfloor u_n \rfloor \\ 0 & \text{ otherwise } \end{cases}$$ and let $a_n = \lfloor u_n \rfloor $. Then $$\alpha = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots}}}$$ denoted $[a_0; a_1, a_2, \dotsc]$. It is rational if and only if $a_n = 0$ for $n$ large enough. So it is a great tool to prove the irrationality of some numbers. For example, $\phi = [1;1,1, \dotsc]$ is the golden ratio, because $(\phi-1)^{-1}=\phi$.

Let $p_n$ be the $n$th prime, then we can consider the irrational number $[p_1;p_2,p_3, \dots] = 2.31303673643\ldots$ (A064442), which then compresses the data of all the prime numbers, in a more natural and efficient way than just taking $2.\mathbf{3}5\mathbf{7}11\mathbf{13}17\mathbf{19}\ldots$. The paper mentioned above provides another interesting way to compress the primes numbers, which uses Bertrand’s postulate, i.e. $p_n < p_{n+1} < 2p_n$. This way is a kind of alternative to continued fractions. For a given number $\beta \ge 2$, consider the recurrence relation $u_1=\beta$ and $$u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1).$$ Let $a_n= \lfloor u_n \rfloor $. Then $a_n \le a_{n+1} < 2a_n$ and the mentioned paper proves that $$\beta = \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$$ denoted, let’s say, $(a_1,a_2,a_3, \dots )$.

By the mentioned paper:
Theorem 1: Let $(a_n)$ be a sequence of positive integers such that:

  • $a_n < a_{n+1} < 2a_n$,
  • $\frac{a_{n+1}}{a_n} \to 1$

then $\beta := (a_1,a_2,a_3, \dots ) := \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$ is irrational.

It follows that the number $(p_1,p_2,p_3,\dots) = 2.920050977316\ldots$ is irrational.

Question: Can Theorem 1 be proved by some previously known methods?

Remark: The first point of Theorem 1 can be relaxed to $a_n \le a_{n+1} < 2a_n$, when $(a_n)$ is not eventually-constant.

For a given non-constant polynomial $P \in \mathbb{Z}[X]$ with a positive leading term and $P(n) \neq 0$ for all $n \in \mathbb{N}_{\ge 1}$, consider $a_n=P(n)$. Then it is easy to deduce from Theorem 1 that the number $e_P\mathrel{:=}(a_1,a_2, \dotsc )$ is irrational. For example, take $P(X)=X^k$, with $k \in \mathbb{N}_{\ge 1}$, then $$e_k:= \sum_{n=1}^{\infty} \frac{(n+1)^k-1}{n!^k}$$ is irrational. Note that $e_1 = e$ is Euler’s number.

The following result applies for an alternative proof of the irrationality of $e_k$ for all $k$, and of $e_P$ for many $P$ (not all), but not for $(p_1,p_2,p_3, \dots)$

Theorem 2: Let $(a_n)$ be a sequence of positive integers such that:

  • $a_n \le a_{n+1} < 2a_n$,
  • $\forall k \in \mathbb{N}_{\ge 1}$, $\exists m$ such that $k$ divides $a_m$,

then $\beta := (a_1,a_2,a_3, \dots ) := \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$ is irrational.

proof: Assume that $\beta = \frac{p}{q}$. By assumption, there is $m$ such that $q$ divides $a_m$. By the mentioned paper, if $u_1=\beta$ and $u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1)$, then $a_n= \lfloor u_n \rfloor $. It is easy to see that $u_n$ can always be written with a denominator equal to $q$ (possibly not simplified). It follows that $u_{m+1}=a_m(u_m-a_m+1)$ and that $a_m u_m$ is an integer. So $u_{m+1}$ is an integer. It follows that for all $n>m$ then $u_n=u_{m+1}$, and so $a_n=a_{m+1}$. But the second point of Theorem 2 implies that $a_n \to \infty$, contradiction. $\square$

The following example will show that the condition $\frac{a_{n+1}}{a_n} \to 1$ is not necessary for the irrationality.

Consider $a_n=\lfloor \frac{3^n}{2^n} \rfloor + r_n$, with $0 \le r_n < n$ such that $n$ divides $a_n$. Adjust the sequence for $n$ small so that the first point of Theorem 2 holds. Then $\beta$ is irrational whereas $\frac{a_{n+1}}{a_n} \to \frac{3}{2} \neq 1$.

Bonus question: What is a necessary and sufficent condition for irrationality?

Joel Moreira suggested in this comment that it may be rational if and only if $(a_n)$ is eventually-constant. See the new post Do these rational sequences always reach an integer? dedicated to this question.

FYI, it is easy to compute that $$\pi = (3, 3, 4, 5, 5, 7, 10, 10, 13, 17, 31, 35, 67, 123, 223, 305, 414, 822, 1550, 2224, ...) $$

  • $\begingroup$ but at least for $e_k$, there is a proof of irrationality of it which is very elementary,we can directly multiple $((n-1)!)^k$ on both side. $\endgroup$
    – katago
    Nov 29, 2020 at 21:04
  • 1
    $\begingroup$ $e_k=\sum_{n}\frac{f_k(n)}{(n!)^k}, f_k(n)=(n+1)^k-1=\sum_{i=0}^ka_in^i$, so $\forall m\in \mathbb{N}^*$, the fraction part of $(m!)^ke_k$ is $\{A_m\}$, where $A_m=\sum_{n\geq m+1}\sum_{i=0}^n \frac{1}{(n(n-1)...(m+1))^k}a_in^i$, the point is we can drop the $(n,i)=(m+1,k)$ term and get bound $0<|A_m|<\sum_{i=1}^{k-1}\frac{B_i}{(m+1)^i}+\sum_{n\geq m+2}\frac{(m!)^k}{((n-1)!)^k}<1$ hold for $m$ suffice large, where $B_i$ is unrelated to $m$. $\endgroup$
    – katago
    Nov 30, 2020 at 7:30
  • 2
    $\begingroup$ Regarding the Bonus question, is there a counter-example to the (naïve, but natural) conjecture that $\alpha$ is rational if and only if the sequence $(a_n)$ is eventually constant? (The less interesting "if" direction isn't difficult to prove.) $\endgroup$ Nov 30, 2020 at 12:33
  • 2
    $\begingroup$ @JoelMoreira: see this new post dedicated to your suggestion. $\endgroup$ Dec 2, 2020 at 3:47
  • $\begingroup$ What Joel called $\alpha$ is now called $\beta$. $\endgroup$ Dec 16, 2020 at 11:48

1 Answer 1


I am sorry If the comment is misleading, and welcome to point out any mistakes in the following proof. This is a clarification of the previous comment.

And this is only a proof of the irrationality of $e_k$.

And the proof strategy is an imitation of Fourier's proof of the irrationality of Euler's number $e$.

if $\forall n=\mathbb{N}^{*} \quad n$, $n$ suffice large, $$ \left(n!\right) \cdot a \notin \mathbb{Z} \quad \text { then } a \notin \mathbb{Q} \hspace{1cm}(1) $$

WLOG, in the following calculation we don't distinguish $x,y$ if $x-y\in \mathbb{Z}$. And we write $x=y+\mathbb{Z}$ iff $x-y\in \mathbb{Z}$.

$\begin{aligned} m ! e_{k} +\mathbb{Z}&=\sum_{n \geq m+1} \frac{(n+1)^{k}-1}{(m+1) \cdots(n-1) n)^{k}}+\mathbb{Z} \\ &=\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((n-1) \cdots(n-1) n)^{k}}+\frac{(m+2)^{k}-1}{(m+1)^{k}}+\mathbb{Z}\\ &=\sum_{n \geq m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i} \cdot(m+1)^{i}}{(m+1)^{k}}+1 +\mathbb{Z}\\ &=\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{( m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}+\mathbb{Z}\hspace{1cm}(*) \end{aligned}$

In fact in $(*)$ we have $\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}= O(\frac{1}{m^{k}})$, $\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}=O(\frac{1}{m})$.

Now take $m$ suffice large, in fact $m=10000\cdot k^{100}$ is ok, then $$0< \sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}< 1$$

So $(*)\neq \mathbb{Z}$, so $(1)$ is true, $ e_{k}$ is not rational.

  • 1
    $\begingroup$ Ok, I see, thanks! Regarding my (now deleted) comment, it is true that $\frac{(n+1)^k-1}{n^k}$ is (in general) neither integer nor less than $1$, but $\frac{(n+1)^k-1}{n^k} = 1 + \sum_{s=1}^{k-1} {k\choose s} n^{s-k}$, so $1$ can go to the integral part, and the rest should be less than $1$. It remains to prove in more details your inequality $0 < \cdot < 1$. $\endgroup$ Dec 17, 2020 at 6:01
  • $\begingroup$ @ Sebastien Palcoux, yes this is the crux of the original proof, and btw the question related to $u_{n+1}=\left\{\begin{array}{ll}\left(u_{n}-\left\lfloor u_{n}\right\rfloor\right)^{-1} & \text {if } u_{n} \neq\left\lfloor u_{n}\right\rfloor \\ 0 & \text { otherwise }\end{array}\right.$ is also interesting, I have not figure out how to prove it. $\endgroup$
    – katago
    Dec 17, 2020 at 6:10

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