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Question Number 185329 by liuxinnan last updated on 20/Jan/23

$${lim}_{k\to +\mathrm{\infty }}\underset{i=1}{\sum ^k}\frac{1}{\sqrt{i-1}+\sqrt{i}}=$$

Answered by aleks041103 last updated on 20/Jan/23

$$\sqrt{i-1}<\sqrt{i}\Rightarrow \sqrt{i-1}+\sqrt{i}<2\sqrt{i}⩽2i$$$$\Rightarrow \underset{i=1}{\sum ^k}\frac{1}{\sqrt{i-1}+\sqrt{i}}>\frac{1}{2}\underset{i=1}{\sum ^k}\frac{1}{\sqrt{i}}>\frac{1}{2}\underset{i=1}{\sum ^k}\frac{1}{i}$$$$since\underset{i=1}{\sum ^k}\frac{1}{i}diverges,then$$$$\underset{k\to \mathrm{\infty }}{lim}\underset{i=1}{\sum ^k}\frac{1}{\sqrt{i-1}+\sqrt{i}}\to \mathrm{\infty }$$

Answered by mr W last updated on 20/Jan/23

$$\frac{1}{\sqrt{i-1}+\sqrt{i}}=\sqrt{i}-\sqrt{i-1}$$$$\underset{i=1}{\sum ^k}\frac{1}{\sqrt{i-1}+\sqrt{i}}=\underset{i=1}{\sum ^k}(\sqrt{i}-\sqrt{i-1})=\sqrt{k}$$$$\underset{k\to \mathrm{\infty }}{\lim }\underset{i=1}{\sum ^k}\frac{1}{\sqrt{i-1}+\sqrt{i}}=\underset{k\to \mathrm{\infty }}{\lim }\sqrt{k}=\mathrm{\infty }$$

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