Let $(a_n)_{n\geq 0}$ be a sequence of positive real numbers given by $a_n=\prod\limits_{i=0}^n\bigg(1+\frac{1}{2^{2^i}}\bigg)$. Study the monotonicity and the boundedness of $(a_n)_{n\geq 0}$.
It is easy to show that the sequence is strictly monotonic increasing and a lower bound would be $0$. What can we say about the upper bound? One can show that $a_n\leq (1+\frac{1}{2})^{n+1}$, but the right hand side of the inequality tends to infinity when $n$ tends to infinity.
Using $\log(1+x)\le x$ along with $n\le 2^{n}$ we assert that
$$\prod_{i=0}^N \left(1+\frac1{2^{2^i}}\right)=e^{\sum_{i=0}^N\log\left(1+\frac1{2^{2^i}}\right)}\le e^{\sum_{i=0}^N\frac1{2^{2^i}}}\le e^{\sum_{i=0}^N \frac1{2^i}}<e^{2}$$