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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.
Converging Product
Age 16 to 18
Challenge Level 
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.