Question

Să se calculeze limita :

Answer 1

$\displaystyle\bf\\propun~sa~ne~ocupam~de~suma~prima~data.\\\sum_{k=1}^n \frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} =\sum_{k=1}^n \frac{1}{k\sqrt{k}+\sqrt{k}+k\sqrt{k+1}} = \sum_{k=1}^n\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k} )} =\\\\\sum_{k=1}^n \frac{\sqrt{k+1}-\sqrt{k} }{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k}) } = \sum_{k=1}^n\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}\sqrt{k}} =\\\\\sum_{k=1}^n \frac{\sqrt{k+1}}{\sqrt{k+1}\sqrt{k}}-\frac{\sqrt{k}}{\sqrt{k+1}\sqrt{k}}.$$\displaystyle\bf\\simplifiam \implies vom~avea~:~\sum_{k=1}^n\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}}=\\\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=1-\frac{1}{\sqrt{n+1}} .\\acum~putem~calcula~limita!\\\lim_{n \to +\infty} 1-\frac{1}{\sqrt{n+1}}=\lim_{n\to +\infty}1-\lim_{n\to +\infty}\frac{1}{\sqrt{n+1}}=1-0=1.$