Let $x$ be a real number such that $|x|<1$. Determine $\lim_{n \to \infty} \prod_{i=1}^{n} \left(1+x^{2^{i}}\right)$

Question

I've written out the first few terms of the series up to n = 3 and I've figured out that they always take the form $1+x^2+x^4+...+x^{2k}$. How can I figure out what the limit will be as $n$ goes to infinity?

Answer 1

Well if the form you found is the right one and x<1 then I would say that limit of $x^k=0$(it's a geometric progession with a common ratio inferior to one) when n goes to infinity. then it means you'll get $\frac{1}{1-x^2}$

I hope it'll help

Answer 2

Hint: Multiply by $ ( 1 - x^2)$.

There's a telescoping series going on.

Answer 3

If you have not figured it out, you can use a telescoping product by multiplying the product by $\frac{1-x^2}{1-x^2}$ and repeatedly using difference of squares. Then you get $$\prod_{k=1}^{n}{(1+x^{2^k})} = \frac{1-x^{2^{n+1}}}{1-x^2}=\frac{1}{1-x^2}-\frac{x^{2^{n+1}}}{1-x^2}.$$ As $n\to\infty,$ the first term stays constant, the denominator of the second term stay constant and the numerator of the second term goes to $0$ because $|x|<1.$ Thus, the answer is $\frac{1}{1-x^2}.$