Computing the infinite product $\prod_{n=1}^{\infty}(1-\alpha^n) = \dfrac{1-3\alpha}{1-2\alpha}$ I want to prove that if $\alpha < \dfrac{1}{3}$, then $$\prod_{n=1}^{\infty}(1-\alpha^n) = \dfrac{1-3\alpha}{1-2\alpha}.$$
I've proved (after some calculations) that if
$$P = (1-\alpha)(1-\alpha^2)(1-\alpha^3)(1-\alpha^4)\ldots,$$
then
$$P = 1-\alpha-\alpha^2+\alpha^5+\alpha^7-\alpha^{12}-\alpha^{15}+\alpha^{22}+\alpha^{26}-\alpha^{35}-\alpha^{40}+\alpha^{51}+... $$
where the behaviour of the exponent is given by:

*

*the odd positions from $\alpha^2$, the exponents are given by the sequence of general term $a_n = \dfrac{3n^2+n}{2}$,

*the difference between the exponent of the even possitions on the right side of the pow described in the item before is given by the sequence of odd numbers from $3$,

*the terms before $\alpha^7$ are computed expanding the product,

*the sign of the terms of the sequence change each two positions.

For example, if we want to compute the exponent in the 6th position (which is $12$), which is an even position, has to differ from the previous exponent, which is $7$, in $5$. Then, the exponent has to be $12$. To compute the exponent in the 7th position, which is odd, we use the formula $a_n=\dfrac{3n^2+n}{2}$; this exponent is the third one in the odd positions, therefore it corresponds to $n=3$, i.e., $a_3 = \dfrac{3\cdot 3^2+3}{2}=15$, and so on.
I have my mind open to other procedures or explanations, and I really appreciate any proof.

In fact
\begin{align}
\prod^n_{k=1}(1-\alpha^k)>\frac{1-3\alpha}{1-2\alpha},\qquad 0<\alpha<1/3\tag{0}\label{zero}
\end{align}
Both quantities in the inequality above correspond to the measure of some well known fat Cantor sets $F$ and $K$ that I am constructing below.
Starting with the interval $F_0=K_0=[0,1]$, remove the middle open subinterval of length $\alpha$. This yields a set $F_1=K_1$ consisting of two subintervals of length $\frac{1-\alpha}{2}$ each.  This is the first step of the construction
Construction of $F$: The $n$-th step of the construction yields a set $F_n$ which is the union of $2^n$ disjoint closed subintervals each of length
$$\frac{1}{2^n}(1-\alpha)\cdot\ldots\cdot(1-\alpha^n)$$
From each such subinterval subtract the middle subinterval of length proportional to $\alpha^{n+1}$ of its length, that is, a middle subinterval of length
$$\frac{1}{2^n}(1-\alpha)\cdot\ldots\cdot(1-\alpha^n)\alpha^{n+1}$$
This yields a set $F_{n+1}\subset F_n$ which is the union of $2^{n+1}$ closed subintervals. The length of $F_{n+1}$ (or rather its Lebesgue measure) is
$$\ell(F_{n+1})=(1-\alpha)\cdot\ldots\cdot(1-\alpha^n)(1-\alpha^{n+1})$$
The set $F$ is defined as $F=\bigcap_nF_n$. Its length measure  is
$$\ell(F)=\prod^\infty_{n=1}(1-\alpha^n)$$
Construction of $K$:
ImThe $n$-th step of the construction yields a set $K_n$ consisting of the union of $2^n$ disjoint closed subintervals each of length
$$\frac{1-(\alpha+\ldots +2^{n-1}\alpha^n)}{2^n}$$
From each subinterval, we subtract a middle open subinterval of length $\alpha^{n+1}$. This yields a set $K_{n+1}\subset K_n$ with length measure
$$\ell(K_{n+1})=1-(\alpha+2\alpha^2+\ldots + 2^n\alpha^{n+1})$$
$K$ is defined as $K=\bigcap_nK_n$. Its length measure  is
$$\ell(K)=1-\sum^\infty_{n=1}2^{n-1}\alpha^n=\frac{1-3\alpha}{1-2\alpha}$$

Notice that the length of each of the subintervals in the $n+1$-th step of the construction of $F$ is
$$\frac{(1-\alpha)\cdot\ldots\cdot(1-\alpha^n)\alpha^{n+1}}{2^n}<\alpha^{n+1}$$
The total measure of the subintervals removed from $[0,1]$ in the construction of $F$ and $K$ respectively satisfy
$$1-\ell(F)=\sum^\infty_{n=0}(1-\alpha)\cdot\ldots\cdot(1-\alpha^n)\alpha^{n+1}<\sum^\infty_{n=0}2^n\alpha^{n+1}=1-\ell(K)$$
Inequality \eqref{zero} follows.
