How is the infinite product 2?

Question

I found this problem and according to Wolfram Alpha, the answer is 2.

$$\prod_{n=0}^{\infty} \left (1+\frac{1}{2^{2^n}}\right)$$

Please do the favor of explaining me how the product is 2.

Answer 1

Multiply it by $(1-\frac12):$

$$\begin{align}(1-\frac12)P&\vphantom{\cfrac11}=(1-\frac12)(1+\frac12)(1+\frac1{2^2})(1+\frac1{2^4})\dots\\\vphantom{\cfrac11}&=(1-\frac1{2^2})(1+\frac1{2^2})(1+\frac1{2^4})\dots\\\vphantom{\cfrac11}&=(1-\frac1{2^4})(1+\frac1{2^4})\dots\\&=(1-\frac1{2^8})\dots\\&=\vdots\\\vphantom{\cfrac11}&=1\end{align}$$

This uses the fact that $(1-a)(1+a)=(1-a^2)$.

Thus,

$$(1-\frac12)P=1\implies P=2$$

Answer 2

(migrated from comment) In general, for $|x| < 1$ we have

$$ \prod_{n=0}^{\infty} (1 + x^{2^n}) = 1 + x + x^2 + x^3 + \cdots = \frac{1}{1-x}. \tag{*}$$

This is essentially the statement that there is a unique binary expansion for each non-negative integer. For the proof, notice that

$$ (1-x)\prod_{n=0}^{N-1} (1 + x^{2^n}) = 1 - x^{2^N}. $$

This can be obtained by applying the formula $(1 - X)(1+X) = 1-X^2$ repeatedly. Now taking $N \to \infty$ you get $\text{(*)}$.

Answer 3

Multiplying a sum of $n$ terms by $1+2^{-n}$ will replicate all terms and scale down the replicas by $2^n$.

The partial products are

$$1$$ $$\color{green}{1+\frac12}$$ $$\left(\color{green}{1+\frac12}\right)+\frac14\left(\color{green}{1+\frac12}\right)=\color{blue}{1+\frac12+\frac14+\frac18}$$ $$\left(\color{blue}{1+\frac12+\frac14+\frac18}\right)+\frac1{16}\left(\color{blue}{1+\frac12+\frac14+\frac18}\right)=1+\frac12+\frac14+\frac18+\frac1{16}+\frac1{32}+\frac1{64}+\frac1{128}$$ $$\cdots$$