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Question Number 130753 by Algoritm last updated on 28/Jan/21

Answered by Dwaipayan Shikari last updated on 28/Jan/21

a)Σ_(n=1) ^∞ ((2n−1)/2^(n−1) )=4Σ_(n=1) ^∞ (n/2^n )−Σ^∞ (1/2^(n−1) )=8−2=6 b) S=x+3x^2 +7x^3 +15x^4 +... S(1−x)=x+2x^2 +4x^3 +8x^4 +.. S(1−x)=(1/2)(2x+4x^2 +8x^3 +16x^4 +...) S(1−x)=(x/(1−2x))⇒S=(x/((1−x)(1−2x))) x=(1/2) (1/2)+(3/4)+(7/8)+((15)/(16))+...→Diverges

$$\left.{a}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }=\mathrm{4}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{2}^{{n}} }−\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }=\mathrm{8}−\mathrm{2}=\mathrm{6} \\ $$$$\left.\:\:\:\:\:\:{b}\right)\:\:\:\:\:\:\:{S}={x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{4}} +... \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{4}} +.. \\ $$$${S}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{4}} +...\right) \\ $$$${S}\left(\mathrm{1}−{x}\right)=\frac{{x}}{\mathrm{1}−\mathrm{2}{x}}\Rightarrow{S}=\frac{{x}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−\mathrm{2}{x}\right)} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{15}}{\mathrm{16}}+...\rightarrow{Diverges} \\ $$

Answered by JDamian last updated on 28/Jan/21

(ii) Σ_(k=1) ^(∞) (((2^k −1)/2^k ))=Σ_(k=1) ^(∞) (1−(1/2^k ))=Σ_(k=1) ^(∞) 1 − Σ_(k=1) ^(∞) ((1/2^k ))= = ∞−1 = ∞

$$\left({ii}\right)\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\left(\frac{\mathrm{2}^{{k}} −\mathrm{1}}{\mathrm{2}^{{k}} }\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\mathrm{1}\:−\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\left(\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\right)= \\ $$$$=\:\infty−\mathrm{1}\:=\:\infty \\ $$

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