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Question Number 29461 by prof Abdo imad last updated on 09/Feb/18

$$letgive{u}_n=\frac{1}{\sqrt{n}}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}})$$$$find{lim}_{n\to +\mathrm{\infty }}{u}_n.$$

Commented by prof Abdo imad last updated on 13/Feb/18

$$wehave{u}_n=\frac{\sqrt{n}}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}})$$$$=\frac{1}{n}(\frac{1}{\sqrt{\frac{1}{n}}}+\frac{1}{\sqrt{\frac{2}{n}}}+....+\frac{1}{\sqrt{\frac{n}{n}}})$$$$=\frac{1}{n}\sum _{k=1}^n\frac{1}{\sqrt{\frac{k}{n}}}so{u}_{n}isaRiemansumand$$$${lim}_{n\to \mathrm{\infty }}{u}_n={\int }_0^1\frac{dx}{\sqrt{x}}={\left[2\sqrt{x}\right]}_0^1=2.$$

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