- Prove that if \(| z | < 1\) , then
\[ (1 + z) (1 + z ^ {2}) (1 + z ^ {4}) (1 + z ^ {8}) \dots = \prod_ {k = 0} ^ {\infty} (1 + z ^ {2 ^ {k}}) = \frac {1}{1 - z}. \]
Solution
For every \(N\ge0\), repeated use of \[ 1-w^2=(1-w)(1+w) \] gives the finite identity \[ (1-z)\prod_{k=0}^{N}(1+z^{2^k}) =1-z^{2^{N+1}}. \tag{1} \] Indeed, each factorization successively replaces \(1-z^{2^j}\) by \[ (1-z^{2^{j-1}})(1+z^{2^{j-1}}). \] If \(|z|<1\), then \[ z^{2^{N+1}}\longrightarrow0. \] Since \(1-z\ne0\), taking \(N\to\infty\) in (1) gives \[ \boxed{ \prod_{k=0}^{\infty}(1+z^{2^k})=\frac1{1-z}}. \] Thus the product converges and has the stated value throughout the open unit disc.