Representation of irrationals as $\sum_{n\ge 2}\frac{x_n}{n!}$

Question

Prove that every $x\in(0,1)\setminus\mathbb{Q}$ has a unique representation as $x = \sum_{n\ge 2}\frac{x_n}{n!}$, where $x_n\in\mathbb{Z}_n = \{0,1,2,\ldots,n-1\}$.

Probably this is well known, I'd be grateful for a book or article with a proof.

Answer 1

Let $T_n: [0,1) \to [0,1)$ be given by $T_n(x) = nx \pmod{1}$. Also define $U_n(x) = T_nT_{n-1} \cdots T_2T_1(x)$. Let $x_n = nU_{n-1}(x) - U_n(x) = nU_{n-1}(x) - T_n(U_{n-1}(x))$ for $n \geq 2$. Then by definition of $T_n$ it follows that $x_n \in \{0,1,2,\ldots,n-1\}$. We also have $$ \sum_{k=2}^n \frac{x_k}{k!} = \sum_{k=2}^n \frac{kU_{k-1}(x) - U_k(x)}{k!} = \sum_{k=2}^n \left( \frac{U_{k-1}(x)}{(k-1)!} - \frac{U_k(x)}{k!} \right) = U_1(x) - \frac{U_n(x)}{n!} $$ since the sum telescopes. Taking the limit $n \to \infty$ on both sides we find $$ \sum_{k \geq 2} \frac{x_k}{k!} = U_1(x) = x $$ because $\frac{U_n(x)}{n!} \to 0$ for $n \to \infty$ (recall that $U_n(x) \in [0,1)$.). This shows that $x$ can be written in the desired form.

Suppose $x$ is irrational and $x = \sum_{k \geq 2} \frac{x_k}{k!}. $ We have $$ x - \frac{x_2}{2} = \sum_{k \geq 3} \frac{x_k}{k!} \leq \sum_{k \geq 3} \frac{k-1}{k!} = \sum_{k \geq 3} \left( \frac{1}{(k-1)!} - \frac{1}{k!} \right) = \frac12 $$ (a telescoping sum again). This means that $x - \frac{x_2}{2} \in (0, \frac12]$ (note that $0$ is impossible since $x$ is irrational), hence $x_2 \in [2x-1,2x)$. This uniquely determines $x_2$. In the same way one can show that $x_3$, $x_4$, $\ldots$ are uniquely determined. This means that for irrational $x$ the representation is unique.