Question

Решить уравнение с лимитом​

Answer 1

$\dfrac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=\dfrac{(\sqrt{2n+1}-\sqrt{2n-1})}{(\sqrt{2n-1}+\sqrt{2n+1})(\sqrt{2n+1}-\sqrt{2n-1})}=\\ \dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{2}= \dfrac{\sqrt{2n+1}}{2}- \dfrac{\sqrt{2n-1}}{2}$

$\lim\limits_{n\to\infty}\dfrac{1}{\sqrt{n}} (\dfrac{1}{\sqrt 1 + \sqrt 3}+...+\dfrac{1}{\sqrt{2n-1}+\sqrt{2n+1}})=\lim\limits_{n\to\infty}\dfrac{1}{\sqrt{n}} (\dfrac{\sqrt{3}}{2}- \dfrac{\sqrt{1}}{2}+...+\dfrac{\sqrt{2n+1}}{2}- \dfrac{\sqrt{2n-1}}{2})=\lim\limits_{n\to\infty}\dfrac{1}{\sqrt{n}} (-\dfrac{\sqrt{1}}{2}+\dfrac{\sqrt{2n+1}}{2})\\ =-\lim\limits_{n\to\infty}\dfrac{1}{\sqrt{n}} \dfrac{\sqrt{1}}{2}+\lim\limits_{n\to\infty}\dfrac{\sqrt{2n+1}}{2\sqrt n}=0+\dfrac{\sqrt 2}{2}=\dfrac{\sqrt 2}{2}$