Convergence and the limit of $x_n=(1+\frac{1}{2})(1+\frac{1}{2^2})...(1+\frac{1}{2^{2^n}})=\prod_{i=0}^n (1+\frac{1}{2^{2^i}})$ for $n \ge 0$ It is easy to see that the sequence is bounded below by $1$ and increasing as we multiply the terms by a number larger than $1$ each time.
I was also able to show that $x_n \le a_n$ for all $n \ge 0$ where $a_n=\sum_{i=0}^n \frac{1}{2^i}$. Since we know $(a_n)_n$ is bounded above by $2$, so do $(x_n)_n$ and now we know it also converges.
However, I couldn't find its limit. Can you help me on finding its limit?
Thanks for your effort and time in advance.

$$x_n=\left(1+\frac 1 2\right)\left(1+\frac 1 {2^2}\right)\cdots\left(1+\frac 1 {2^{2^n}}\right)\\=\frac{\left(1-\frac 1 2\right)\left[\left(1+\frac 1 2\right)\left(1+\frac 1 {2^2}\right)\cdots\left(1+\frac 1 {2^{2^n}}\right)\right]}{\left(1-\frac 1 2\right)}\\=\frac{\left(1-\frac 1 {2^{2^{n+1}}}\right)}{\left(1-\frac 1 2\right)}.$$
Thus, $$\lim x_n=\frac 1{1/2}=2.$$
See $x_n=1+\frac 1 2+\frac 1 {2^2}+\frac 1 {2^3}+\frac 1 {2^4}+\cdots+\frac 1 {2^{1+2+\cdots +2^{2^n}}}>\sum_{i=0}^n \frac 1 {2^i}.$
