Question

Aceste exerciții. Vă mulțumesc pentru ajutor!​

Answer 1

Explicație pas cu pas:

$a)\displaystyle \lim_{ x \to \infty } a_n=\frac{1}{\infty} =0\\\\b)\ a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}} =\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n} =\sqrt{n+1}-\sqrt{n}\\a_1+a_2+...+a_n=\sqrt2-\sqrt1+\sqrt3-\sqrt2+\sqrt4-\sqrt3+...+\sqrt{n-1}+\sqrt{n-2}+\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1, \forall n\in\mathbb{N^&*}$

$c) \ \displaystyle \lim_{ x \to \infty } \frac{ a_1+a_2+...+a_n }{ \sqrt{n} }=\displaystyle \lim_{ x \to \infty } \frac{ \sqrt{n+1}-1 }{ \sqrt{n} }=\displaystyle \lim_{ x \to \infty } \bigg(\sqrt{\frac{n+1}{n} } - \frac{1}{\sqrt{n}} \bigg)=\displaystyle \lim_{ x \to \infty } \bigg(\sqrt{1+\frac{1}{n}}- \frac{1}{\sqrt{n}} \bigg)=1-0=1$