Question

Am nevoie de ajutor la această limită
Mulțumesc!

Answer 1

$l = \prod\limits_{k=2}^{\infty}(2-\sqrt[k]{2}) \\ \\ f(x) = \sqrt[x]{2} = 2^{\frac{1}{x}}\\ x\geq 2\rightarrow \text{functie monoton descrescatoare}\\ \\ \\ \lim\limits_{n\to \infty}(2-\sqrt[n]{2}}) = 1^-\\ \\\\ 0< 2-\sqrt 2\approx 0,58<1 \\ 0< 2-\sqrt 2 < 2-\sqrt[3]{2} <...<2-\sqrt[n]{2} <1$

$\prod\limits_{k=2}^{\infty}0<\prod\limits_{k=2}^{\infty}(2-\sqrt[k]{2}) \leq\prod\limits_{k=2}^{\infty} x,\quad x<1\\ \\ 0 <\prod\limits_{k=2}^{\infty}(2-\sqrt[k]{2})\leq 0 \\ \\ \Rightarrow l = 0$