# Converging Product

It is well known that if $| x | < 1$ then $$1 + x + x^2 + \dots + x^n =\frac{1 - x^{n+1}}{1 - x}$$ and hence (taking limits) we have the sum of the infinite geometric series $$1 + x + x^2 + \dots + x^n + \dots = \frac{1}{1 - x}$$ We are now going to obtain a similar formula for an infinite product, namely $$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\dots(1+x^{2^n})\dots = \frac{1}{1 - x}$$ Evaluate the product $$(1 - x)(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)$$ Show, by induction, that $$(1 + x)(1 + x^2)\dots(1+x^{2^n}) = \frac{1 - x^{2^{n+1}}}{1 - x}$$ and hence (taking limits) the given formula for the infinite product follows.

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

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

$$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\dots(1 + x^{2^n})\dots = \frac{1}{1 - x}$$ The first step is to evaluate the product of a few terms and then to prove a general result. $$\eqalign{ (1 - x)(1 + x)(1 + x^2)(1 + x^4)(1 + x^8) &=& (1 - x^2)(1 + x^2)(1 + x^4)(1 + x^8) \\ \; &=& (1 - x^4)(1 + x^4)(1 + x^8) \\ \; &=& (1 - x^8)(1 + x^8) \\ \; &=& (1 - x^{16})}$$ 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 $n = 1$, $$\eqalign{ \mbox{LHS} &=& (1 + x)(1 + x^2) \\ \; &=& 1 + x + x^2 + x^3 \\ \mbox{RHS} &=& \frac{1 - x^{2^{1+1}}}{1-x} \\ \; &=& \frac{1 - x^4}{1 - x} \\ \; &=& \frac{(1 - x)(1 + x + x^2 +
x^3)}{1 - x} \\ \; &=& 1 + x + x^2 + x^3}$$ 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 $$\eqalign{ (1 + x)(1 + x^2)\dots(1 + x^{2^{k+1}}) &=& \frac{1 - x^{2^{k+1}}}{1 - x}(1 + x^{2^{k+1}}) \\ \; &=& \frac{1 - x^{2\times 2^{k+1}}}{1 - x} \\ \; &=& \frac{1 - x^{2^{k+1+1}}}{1 - x}}$$ Hence, by mathematical induction
the statement holds for any positive integer value of $n$. Taking limits, where $| x | < 1$ $$\lim_{n\rightarrow\infty}x^{2^n}=0$$ so the formula for the infinite product follows, namely: $$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\dots(1 + x^{2^n})\dots = \frac{1}{1 - x}$$