Compute $\lim_{n\to\infty}(1+a)(1+a^2)(1+a^4)...(1+a^{2^n})$?

Question

Compute

$$\lim_{n\to\infty}(1+a)(1+a^2)(1+a^4)...(1+a^{2^n})$$

Where $|a|<1$.

Answer 1

$$(1-a)f=(1-a)(1+a)(1+a^2)(1+a^4)(1+a^8)...(1+a^{2^n})=(1-a^{2^{n+1}})$$ Use $(x-y)(x+y)=x^2-y^2$ sucsessively to get $$f_n=\frac{1-a^{2^{n +1}}}{1-a}.$$ As $|a|<1$ $$\lim_{n \rightarrow \infty} f_n= \frac{1}{1-a}.$$

Answer 2

Perhaps interesting even though a valid answer already exists:

Since every positive integer has a unique representation as the sum of distinct powers of $2$ (that is, a unique binary expansion), you see that $$(1+a)(1+a^2)\dots(1+a^{2^n})=\sum_{k=0}^{2^{n+1}-1}a^k.$$(Give a formal proof of that by induction.) So the limit is $\sum a^k=1/(1-a)$.