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Question Number 36837 by Rio Mike last updated on 06/Jun/18

Σ_(r=0) ^∞ 2^(4r−2)

$$\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{\mathrm{4}{r}−\mathrm{2}} \: \\ $$

Commented by prof Abdo imad last updated on 06/Jun/18

let S_n =Σ_(k=0) ^n 2^(4k −2) ⇒S_n =(1/4) Σ_(k=0) ^n (2^4 )^k =(1/4) (((2^4 )^(n+1) −1)/(2^4 −1)) = (1/(4(2^4 −1))){ 2^(4n+4) −1} but 2>1 ⇒ lim_(n→+∞) 2^(4n+4) =+∞ ⇒lim_(n→+∞) S_n =+∞

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\mathrm{2}^{\mathrm{4}{k}\:−\mathrm{2}} \Rightarrow{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(\mathrm{2}^{\mathrm{4}} \right)^{{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\left(\mathrm{2}^{\mathrm{4}} \right)^{{n}+\mathrm{1}} \:−\mathrm{1}}{\mathrm{2}^{\mathrm{4}} \:−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}^{\mathrm{4}} −\mathrm{1}\right)}\left\{\:\mathrm{2}^{\mathrm{4}{n}+\mathrm{4}} \:−\mathrm{1}\right\}\:{but} \\ $$$$\mathrm{2}>\mathrm{1}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \mathrm{2}^{\mathrm{4}{n}+\mathrm{4}} \:=+\infty\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {S}_{{n}} =+\infty \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

s=2^(4.0−2) +2^(4.1−2) +2^(4.2−2) +2^(4.3−2) +... =2^(−2) +2^2 +2^6 +2^(10) +.. =((2^(−2) {(2^2 )^n −1})/(2^2 −1)) =((2^(−2) (2^(2n) −1))/3)=((2^(2n) −1)/(12)) since common ratio is 2^2 i,e 4>1 so s→∞

$${s}=\mathrm{2}^{\mathrm{4}.\mathrm{0}−\mathrm{2}} +\mathrm{2}^{\mathrm{4}.\mathrm{1}−\mathrm{2}} +\mathrm{2}^{\mathrm{4}.\mathrm{2}−\mathrm{2}} +\mathrm{2}^{\mathrm{4}.\mathrm{3}−\mathrm{2}} +... \\ $$$$=\mathrm{2}^{−\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{6}} +\mathrm{2}^{\mathrm{10}} +.. \\ $$$$=\frac{\mathrm{2}^{−\mathrm{2}} \left\{\left(\mathrm{2}^{\mathrm{2}} \right)^{{n}} −\mathrm{1}\right\}}{\mathrm{2}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{2}^{−\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\mathrm{3}}=\frac{\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{12}} \\ $$$${since}\:{common}\:{ratio}\:{is}\:\mathrm{2}^{\mathrm{2}} \:\:{i},{e}\:\mathrm{4}>\mathrm{1} \\ $$$${so}\:{s}\rightarrow\infty \\ $$

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