Question-194128

Question Number 194128 by tri26112004 last updated on 28/Jun/23

Answered by MM42 last updated on 28/Jun/23

a) Σ_(n=2023) ^∞ (−(1/2))^n =Σ_(n=0) ^∞ (−(1/2))^n −Σ_(n=0) ^(2022) (−(1/2))^n =(2/3)−((1−((1/2))^(2022) )/(3/2)) = (1/3)×((1/2))^(2021) b) Σ_(n=1) ^∞ ((22)/(n(n+22))) = Σ_(n=1) ^∞ ((1/n)−(1/(n+22)) ) S_k =1+(1/2)+(1/3)+...+(1/(22))+...−(1/(k+21))−(1/(k+22)) Σ_(n=1) ^∞ ((22)/(n(n+22))) =lim_(k→∞) S_k =1+(1/2)+(1/3)+...+(1/(22))

$$\left.\:\:{a}\right)\:\:\:\:\underset{{n}=\mathrm{2023}} {\overset{\infty} {\sum}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} −\underset{{n}=\mathrm{0}} {\overset{\mathrm{2022}} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2022}} }{\frac{\mathrm{3}}{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2021}} \\ $$$$\left.{b}\right)\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{22}}{{n}\left({n}+\mathrm{22}\right)}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{22}}\:\right) \\ $$$${S}_{{k}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{22}}+…−\frac{\mathrm{1}}{{k}+\mathrm{21}}−\frac{\mathrm{1}}{{k}+\mathrm{22}} \\ $$$$\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{22}}{{n}\left({n}+\mathrm{22}\right)}\:={lim}_{{k}\rightarrow\infty} \:{S}_{{k}} \:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{22}} \\ $$$$ \\ $$

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