Convergence of $\prod\limits_{n=0}^{\infty} (1+x^{2^{n}})$

Question

I never covered infinite products in my real analysis class, so I have no idea. I was wondering about the convergence of the above product. Thanks in advance.

Answer 1

By uniqueness of binary expansion, $$\prod\limits_{n=0}^{\infty} (1+x^{2^{n}})=\sum_{k=0}^\infty x^k,$$ which is $$\frac{1}{1-x}$$ if $|x|<1$.

Answer 2

$$\prod\limits_{n=0}^{\infty}( 1+x^{2^{n}}) $$ is such that every term is $1+x^{2^{n}}$ and since the exponent of $x$ is always even, you have that $x^{2^n}\ge 0$. This means that we have 3 cases:

If $x=0$ then the product is $1$,

Else, if $|x|\ge 1$ we have that the general term is $\ge 2$ so the infinite product is bigger than "$2^{+\infty}$", so it diverges.

Finally, if $0<|x|<1$, we use this criteria: $$\prod\limits_{n=0}^{\infty}( 1+x^{2^{n}}) \mbox{ converges} \iff \sum_{n=0}^{\infty}\log(1+x^{n^2}) \mbox{ converges.} $$ That series is such that $$\sum_{n=0}^{\infty}\log(1+x^{n^2}) \le \sum_{n=0}^{\infty}x^{n^2}$$ and for the values of $x$ considered it converges.

Answer 3

First of all, if $\sum_{n=0}^{\infty}\ln(1+x^{2^n})$ converges then so does the product in question.

Now $x$ must satisfy $|x|<1$ for this to converge. However, for each such $y \le 1$; $y$ positive note that $0\le \ln(1+y) \le 2y$. So $\ln(1+x^{2^n})$ is no larger than $2x^{2^n} < 2x^n$ for each $x$ satisfying $|x| < 1$.

Thus

$$0 \le \sum_{n=0}^{\infty}\ln(1+x^{2^n}) \le \sum_{n=0}^{\infty} x^n $$

and $\sum_{n=0}^{\infty} x^n$ converges.