[tex]\lim\limits_{n\to\infty} \Big(1+\dfrac{1}{2}\Big)\Big(1+\dfrac{1}{2^2}\Big)\Big(1+\dfrac{1}{2^4}\Big)...\Big(1+\dfrac{1}{2^{2^{n-1}}}\Big)\\ \\ \text{Observam ca secventa e convergenta.}\\ \\ x = \dfrac{1}{2} \\ \\ (1+x)(1+x^2)(1+x^4)...(1+x^{2^{n-1}})= \\ \\= 1+x+x^2+x^3+x^4+.... \Big|_{n\to \infty}\\ \\ \text{Deoarece:} \\ \\ (1+x)(1+x^2) = 1+x+x^2+x^3\\ (1+x)(1+x^2)(1+x^4) = 1+x+x^2+x^3+x^4+x^5+x^6+x^7\\ .....\\ (1+x)(1+x^2)(1+x^4)(1+x^8)... = 1+x+x^2+x^3+x^4+...[/tex]
[tex]1+x+x^2+x^3+x^4+... = \dfrac{1}{1-x},\quad |x|<1\quad (\text{identitate)}\\ \\ \Rightarrow \dfrac{1}{1-x} = \dfrac{1}{1-\dfrac{1}{2}} = \dfrac{1}{\dfrac{1}{2}} = 2 \\ \\ \Rightarrow \lim\limits_{n\to\infty} \Big(1+\dfrac{1}{2}\Big)\Big(1+\dfrac{1}{2^2}\Big)\Big(1+\dfrac{1}{2^4}\Big)...\Big(1+\dfrac{1}{2^{n-1}}\Big) = \boxed{2}[/tex]
