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Question Number 185328 by liuxinnan last updated on 20/Jan/23

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

Answered by aleks041103 last updated on 20/Jan/23

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

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