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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) 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))

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

Answered by SEKRET last updated on 19/Jan/23

b) lim_(n→∞) ( ((1/n))^(2022) +((2/n))^(2022) +....+1 −(1/(2023))) = 0+0+0...+0+1−(n/(2023))= −∞

$$\left.\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\:\left(\:\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +\left(\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +....+\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2023}}\right) \\ $$$$=\:\mathrm{0}+\mathrm{0}+\mathrm{0}...+\mathrm{0}+\mathrm{1}−\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2023}}=\:−\infty \\ $$

Commented by aba last updated on 19/Jan/23

b) lim_(n→∞) (n/n) ( ((1/n))^(2022) +((2/n))^(2022) +....+1 )−(n/(2023)) = 0+0+0...+0+1−(n/(2023))= −∞

$$ \\ $$$$\left.\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\frac{\mathrm{n}}{\mathrm{n}}\:\left(\:\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +\left(\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2022}} +....+\mathrm{1}\:\right)−\frac{\mathrm{n}}{\mathrm{2023}} \\ $$$$=\:\mathrm{0}+\mathrm{0}+\mathrm{0}...+\mathrm{0}+\mathrm{1}−\frac{\mathrm{n}}{\mathrm{2023}}=\:−\infty \\ $$

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