Question Number 185209 by Rupesh123 last updated on 18/Jan/23
Answered by som(math1967) last updated on 18/Jan/23
![a) lim_(n→∞) (1/n)[ ((1/n))^(2022) +((2/n))^(2022) +...((n/n))^(2022) ] lim_(n→∞) (1/n)Σ_(r=1) ^n ((r/n))^(2022) lim_(h→0) hΣ_(r=1) ^n (rh)^(2022) [h=(1/n)] =∫_0 ^1 x^(2022) dx =[(x^(2023) /(2023))]_0 ^1 =(1/(2023))](https://www.tinkutara.com/question/Q185216.png)
$$\left.{a}\right)\:\underset{{n}\rightarrow\infty} {{lim}}\:\frac{\mathrm{1}}{{n}}\left[\:\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2022}} +\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2022}} +…\left(\frac{{n}}{{n}}\right)^{\mathrm{2022}} \right] \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{r}}{{n}}\right)^{\mathrm{2022}} \\ $$$$\:\underset{{h}\rightarrow\mathrm{0}} {{lim}}\:{h}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left({rh}\right)^{\mathrm{2022}} \:\:\:\left[{h}=\frac{\mathrm{1}}{{n}}\right] \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2022}} {dx} \\ $$$$=\left[\frac{{x}^{\mathrm{2023}} }{\mathrm{2023}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2023}} \\ $$
Answered by SEKRET last updated on 19/Jan/23
$$\left.\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\:\left(\:\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +\left(\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +….+\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2023}}\right) \\ $$$$=\:\mathrm{0}+\mathrm{0}+\mathrm{0}…+\mathrm{0}+\mathrm{1}−\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2023}}=\:−\infty \\ $$
Commented by aba last updated on 19/Jan/23
$$ \\ $$$$\left.\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\frac{\mathrm{n}}{\mathrm{n}}\:\left(\:\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +\left(\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +….+\mathrm{1}\:\right)−\frac{\mathrm{n}}{\mathrm{2023}} \\ $$$$=\:\mathrm{0}+\mathrm{0}+\mathrm{0}…+\mathrm{0}+\mathrm{1}−\frac{\mathrm{n}}{\mathrm{2023}}=\:−\infty \\ $$