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Question Number 185209 by Rupesh123 last updated on 18/Jan/23

Answered by som(math1967) last updated on 18/Jan/23

$$a)\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}[{\left(\frac{1}{n}\right)}^{2022}+{\left(\frac{2}{n}\right)}^{2022}+...{\left(\frac{n}{n}\right)}^{2022}]$$$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\underset{r=1}{\sum ^n}{\left(\frac{r}{n}\right)}^{2022}$$$$\underset{h\to 0}{lim}h\underset{r=1}{\sum ^n}{\left(rh\right)}^{2022}[h=\frac{1}{n}]$$$$={\int }_0^1{x}^{2022}dx$$$$={\left[\frac{x^{2023}}{2023}\right]}_0^1=\frac{1}{2023}$$

Answered by SEKRET last updated on 19/Jan/23

$$\mathbf{b})$$$$\underset{\mathbf{n}\to \mathrm{\infty }}{\mathbf{lim}}({\left(\frac{1}{\mathbf{n}}\right)}^{2022}+{\left(\frac{2}{\mathbf{n}}\right)}^{2022}+....+1-\frac{1}{2023})$$$$=0+0+0...+0+1-\frac{\mathbf{n}}{2023}=-\mathrm{\infty }$$

Commented by aba last updated on 19/Jan/23

$$$$$$\mathbf{b})$$$$\underset{\mathbf{n}\to \mathrm{\infty }}{\mathbf{lim}}\frac{\mathrm{n}}{\mathrm{n}}({\left(\frac{1}{\mathbf{n}}\right)}^{2022}+{\left(\frac{2}{\mathbf{n}}\right)}^{2022}+....+1)-\frac{\mathrm{n}}{2023}$$$$=0+0+0...+0+1-\frac{\mathrm{n}}{2023}=-\mathrm{\infty }$$

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