What is the nature of the infinite series expansion in this excerpt from a book? In the image below, I can see that the expression $1/(1-2^{-n})$ has been expanded to several terms that approxaimate it:

My questions are:

*

*What is the mathematical technique by which the series infinite expansion has been done?

*Why was the expansion not just carried out on the $1/(2^n-1)$ itself instead of converting it to $1/[2^n(1-2^{-n})]$
 A: With $x = 2^{-n}$, your first question amounts to showing that
$$
\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^i}) \cdots
\label{prod}
\tag{1}
$$
The geometric series expansion gives
$$
\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots + x^k + \cdots
\label{sum}
\tag{2}
$$
so we have to show that the infinite product (\ref{prod}) agrees with the infinite sum (\ref{sum}).
This amounts to the observation that every non-negative integer has a unique binary expansion:
$$
k = b_0 + 2b_1 + 4b_2 + \cdots + 2^\ell b_\ell, 
\quad \text{where all }
b_i \in \{0, 1\}.
$$
Upon expanding the product (\ref{prod}), each term is obtained by multiplying together either $1$ or $x^{2^i}$ from each binomial, where only finitely many are not $1$. For instance, the $x^{13}$ term in the geometric sum (\ref{sum}) comes about as
$$
(1+\color{blue}{x^1}) 
(1+x^2) 
(1+\color{blue}{x^4}) 
(1+\color{blue}{x^8}) 
(1+x^{16}) \cdots (1+x^{2^i}) \cdots
=
{}\cdots + \color{blue}{x^{1+4+8}} + \cdots{},
$$
and in general
$$
x^k = x^{b_0 + 2b_1 + 4b_2 + \cdots + 2^\ell b_\ell} 
= x^{b_0} \cdot (x^2)^{b_1} \cdot (x^4)^{b_2} \cdots (x^{2^\ell})^{b_\ell}.
$$

The answer to your second question has to do with convergence of the infinite sum/product. With powers of $2^n$, these terms/factors become arbitrarily large.
