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Question Number 185293 by SANOGO last updated on 19/Jan/23

$$\underset{k=1}{\sum ^n}\frac{1}{\sqrt{k+n}}$$

Commented by aleks041103 last updated on 20/Jan/23

$$thatiswrong.$$$$\sqrt{k+n}⩾\sqrt{1+n}\Rightarrow \frac{1}{\sqrt{k+n}}⩽\frac{1}{\sqrt{1+n}}$$$$youthought\frac{1}{\sqrt{k+n}}⩾\frac{1}{\sqrt{1+n}}$$

Commented by Frix last updated on 20/Jan/23

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{\underset{k=1}{\sum ^n}\frac{1}{\sqrt{k+n}}}{\sqrt{n}}=2(\sqrt{2}-1)\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{1}{\sqrt{k+n}}=\mathrm{\infty }$$

Answered by 123564 last updated on 21/Jan/23

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