Studing representation of real number, i have found (2.13) in The Real Numbers - A Survey of Constructions by Ittay Weiss a Cantor theorem: Every real number $a > 1$ can be written uniquely as $a =\displaystyle\prod\limits_{k=1}^∞\biggl(1+\dfrac{1}{a_k}\biggr)$ where $a_k > 0$ is an integer for all $k \ge 1$, with $a_k > 1$ from some point onwards, and further $a_{k+1} \ge a^2_k$ for all $k ≥ 1$.
I have found a proof in 'Irrationalzahlen' book of Perron, pages 134-136 but i'm a little struggling with German.
I therefore kindly ask if anyone can give me a best reference (possibly in English) of the Cantor's representation. Thanks
As best I can glean from Perron's book, to construct such a product, we start with a number $a_0 > 1$ (denoted $a$ in your question), and define, for $n \ge 0$, \begin{align*} q_n &:= \left\lfloor \frac{a_n}{a_n - 1}\right\rfloor \\ a_{n+1} &:= a_n\left(1 + \frac{1}{q_{n}} \right)^{-1}. \end{align*} The resulting sequence of $q_n$s yields $$a_0 = \prod_{n=0}^\infty \left(1 + \frac{1}{q_n}\right). \tag{1}$$ I'm not going to go into this, because this is not what the question at hand is about. What we need to do is show that there could be no other sequence $q_n$ that works. If we can show that the choice of $q_0$ is the only possibility, then we are essentially done: the same logic can be applied recursively to show that $q_1, q_2, \ldots$ are all fixed as well.
So, suppose $(1)$ holds, assuming only that $q_n$ is an integer sequence, $q_n \ge 1$, $q_{n+1} \ge q_n^2$, and $q_n$ is eventually at least $2$. We wish to show that $$q_0 \le \frac{a_0}{a_0 - 1} < q_0 + 1 \tag{2}.$$
We start by tackling the strict inequality in $(2)$. Consider, $a_2 \ge 1$, and $$a_1 = \left(1 + \frac{1}{q_1}\right)a_2 \ge 1 + \frac{1}{q_1} > 1.$$ Thus, following similar logic with now strict inequalities, $$a_0 > 1 + \frac{1}{q_0}.$$ Standard manipulations show that this equivalent to our strict inequality from $(2)$: $$\frac{a_0}{a_0 - 1} < q_0 + 1.$$
Next, we tackle the non-strict inequality in $(2)$, and to do so, we need a sharp upper bound on $\prod_{n=1}^\infty \left(1 + \frac{1}{q_n}\right)$. We can get such an upper bound by minimising the $q_n$ sequence for $n \ge 1$, which we would get by setting $q_{n+1} = q_n^2$. Given the value of $q_0$, the minimum possible value would be achieved by $$q_n = q_0^{2^n}.$$ Thus, $$\prod_{n=1}^\infty \left(1 + \frac{1}{q_n}\right) \le \prod_{n=1}^\infty \left(1 + q_0^{-2^n}\right).$$ We have, for any positive integer $N$, $$(1 - q_0^{-2})\prod_{n=1}^N \left(1 + q_0^{-2^n}\right) = 1 - q_0^{-2^{N+1}},$$ so $$\prod_{n=1}^N \left(1 + q_0^{-2^n}\right) = \frac{1 - q_0^{-2^{N+1}}}{1 - q_0^{-2}} \to \frac{1}{1 - q_0^{-2}} = \frac{q_0^2}{q_0^2 - 1}.$$ Thus, \begin{align*} &\prod_{n=1}^\infty \left(1 + \frac{1}{q_n}\right) \le \frac{q_0^2}{q_0^2 - 1} \\ \implies \, &a_0 = \prod_{n=0}^\infty \left(1 + \frac{1}{q_n}\right) \le \frac{q_0^2}{q_0^2 - 1} \cdot \frac{q_0 + 1}{q_0} = \frac{q_0}{q_0 - 1} \\ \iff \, &\frac{1}{a_0} + \frac{1}{q_0} \ge 1 \\ \iff \, &q_0 \le \frac{a_0}{a_0 - 1}. \end{align*} This establishes the non-strict inequality in $(2)$, and hence $(2)$ is true, and hence only a single integer $q_0$ exists that will satisfy our requirements. Applying the same logic to $q_1$, when building $a_1$, will show that a single choice for $q_1$ exists. Using induction, we can show that the entire sequence must be fixed in this way.