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Question Number 32041 by abdo imad last updated on 18/Mar/18

$$findthenatureof\mathrm{\Sigma }{u}_n/$$$$u_n=\frac{\sqrt{1}+\sqrt{2}+....+\sqrt{n}}{n^3}.$$

Commented by abdo imad last updated on 20/Mar/18

$$wehave{u}_n=\frac{1}{n^2}\left(\frac{1}{\sqrt{n}}\sum _{k=1}^n\sqrt{\frac{k}{n}}\right)$$$$=\frac{1}{n\sqrt{n}}\left(\frac{1}{n}\sum _{k=1}^n\sqrt{\frac{k}{n}}\right)but{lim}_{n\to \mathrm{\infty }}\frac{1}{n}\sum _{k=1}^n\sqrt{\frac{k}{n}}$$$$={\int }_0^1\sqrt{x}dx={\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^1=\frac{2}{3}\Rightarrow u_n\sim \frac{2}{3n\sqrt{n}}and$$$$theserie\sum _{n⩾1}\frac{2}{3n\sqrt{n}}isconvergentso\mathrm{\Sigma }{u}_{n}converges.$$

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