# Closed form for $\prod_{i=0}^{\infty}(1+x^{2^i})$

I've recently come across the infinite product $\prod_{i=0}^{\infty}(1+x^{2^i})$ and I was wondering if there is a closed form expression for this, or even if it diverges for all non-zero $x$.

Thanks!

• In addition to Jack D'Aurizio's answer, convergence can be shown for all $-1\lt x \lt 1$ by using the standard theorems on convergence of infinite series (or alternately, by finding the expression for the finite product and showing that the limit exists) – Steven Stadnicki May 3 '16 at 15:13

Any $n\in\mathbb{N}\setminus\{0\}$ can be written in a unique way as a sum of different powers of $2$, hence: $$\prod_{i\geq 0}(1+x^{2^i}) = \sum_{n\geq 0}x^n = \frac{1}{1-x}$$ and the radius of convergence is obviously $1$. The previous identity can also be proved by induction: $$\frac{1}{1-x} = \frac{1+x}{1-x^2} = \frac{(1+x)(1+x^2)}{1-x^4} = \frac{(1+x)(1+x^2)(1+x^4)}{1-x^8}=\ldots$$
Because you can write all numbers in base of $2$ so you can achieve $x^n$ for all $n\in \mathbb N$ by multiplication of $(1+x^{2^i})$ when $2^i$s appear in representation of $n$ in base $2$.
For example $13=2^3+2^2+1$ and so $x^{13}$ is in the expansion of $(1+x^8)(1+x^4)(1+x)$.
And all $x^n$ appears exactly once in $\prod_{i=0}^{\infty}(1+x^{2^i})$ because every number has unique representation in base $2$.