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Question Number 197740 by cortano12 last updated on 27/Sep/23

Answered by mr W last updated on 27/Sep/23

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{k\sqrt{k}}{n}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{n\sqrt{n}}{n}\underset{k=1}{\sum ^n}\frac{k\sqrt{k}}{n\sqrt{n}}$$$$=\left(\underset{n\to \mathrm{\infty }}{\lim }n\sqrt{n}\right)\left[\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}\left(\frac{k}{n}\sqrt{\frac{k}{n}}\right)\right]$$$$=\left(\underset{n\to \mathrm{\infty }}{\lim }n\sqrt{n}\right)\left[{\int }_0^{1}x\sqrt{x}dx\right]$$$$=\left(\underset{n\to \mathrm{\infty }}{\lim }n\sqrt{n}\right)\times \frac{2}{5}$$$$=\mathrm{\infty }$$

Answered by witcher3 last updated on 27/Sep/23

$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\mathrm{k}\sqrt{\mathrm{k}}⩾\mathrm{n}\sqrt{\mathrm{n}}$$$$\frac{\mathrm{\Sigma }\mathrm{k}\sqrt{\mathrm{k}}}{\mathrm{n}}⩾\sqrt{\mathrm{n}}\to \mathrm{\infty }$$

Answered by Frix last updated on 27/Sep/23

$$\frac{1+\stackrel{?}{2}\sqrt{2}+3\sqrt{3}+...+n\sqrt{n}}{n}=s+\sqrt{n}$$$$s>0\wedge \underset{n\to \mathrm{\infty }}{\lim }\sqrt{n}=\mathrm{\infty }$$

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