Converging product

This is an infinite product which has close similarities to the infinite geometric series. It is well known that if $| x |< 1$ then $$1+x+x^2+\cdots +x^n=\frac{1-x^{n+1}}{1-x}$$ and hence (taking limits) we have the sum of the infinite geometric series $$1+x+x^2+\cdots +x^n+\cdots =\frac{1}{1-x}$$

 

Graeme from Madras College obtained a similar formula for an infinite product.

$$(1+x)(1+x^2)(1+x^4)(1+x^8)\dots (1+x^{2^n})\cdots =\frac{1}{1-x}$$ The first step is to evaluate the product of a few terms and then to prove a general result. $$\begin{array}{rll}(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)& =& (1-x^2)(1+x^2)(1+x^4)(1+x^8)\\ & =& (1-x^4)(1+x^4)(1+x^8)\\ & =& (1-x^8)(1+x^8)\\ & =& (1-x^{16})\end{array}$$ The next step is to use the axiom of mathematical induction to prove the following result: $$(1+x)(1+x^2)\dots (1+x^{2^n})=\frac{1-x^{2^{n+1}}}{1-x}$$ For $n = 1$, $$\begin{array}{rll}\text{LHS}& =& (1+x)(1+x^2)\\ & =& 1+x+x^2+x^3\\ \text{RHS}& =& \frac{1-x^{2^{1+1}}}{1-x}\\ & =& \frac{1-x^4}{1-x}\\ & =& \frac{(1-x)(1+x+x^2+x^3)}{1-x}\\ & =& 1+x+x^2+x^3\end{array}$$ So the statement is true for $n = 1$.

Now assume it is true for $n = k$ where $k$ is an integer. $$(1+x)(1+x^2)\dots (1+x^{2^k})=\frac{1-x^{2^{k+1}}}{1-x}$$ It follows that $$\begin{array}{rll}(1+x)(1+x^2)\dots (1+x^{2^{k+1}})& =& \frac{1-x^{2^{k+1}}}{1-x}(1+x^{2^{k+1}})\\ & =& \frac{1-x^{2\times 2^{k+1}}}{1-x}\\ & =& \frac{1-x^{2^{k+1+1}}}{1-x}\end{array}$$ Hence, by mathematical induction the statement holds for any positive integer value of $n$. Taking limits, where $| x | < 1$ $$\underset{n\to \mathrm{\infty }}{lim}{x}^{2^n}=0$$ so the formula for the infinite product follows, namely: $$(1+x)(1+x^2)(1+x^4)(1+x^8)\dots (1+x^{2^n})\cdots =\frac{1}{1-x}$$