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Question Number 130006 by Adel last updated on 21/Jan/21

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

S=x+2x^2 +3x^3 +.... S(1−x)=x+x^2 +x^3 +... S=(x/((1−x)^2 )) x=(1/3) (1/3)+(2/3^2 )+(3/3^3 )+(4/3^4 )+...=(3/4)

$${S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +.... \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +... \\ $$$${S}=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \:\:}\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{4}} }+...=\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by Adel last updated on 21/Jan/21

s(1−x)=x+x^2 +x^3 +∙∙∙∙ is way?

$$\mathrm{s}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\centerdot\centerdot\centerdot\centerdot \\ $$$$\mathrm{is}\:\:\:\:\mathrm{way}? \\ $$

Commented by Dwaipayan Shikari last updated on 21/Jan/21

S=x+2x^2 +3x^3 +... −xS= −x^2 −2x^3 −... S(1−x)=x+x^2 +x^3 +... (x+x^2 +x^3 +..=(x/(1−x)))

$$\:\:\:\:\:\:\:\:{S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +... \\ $$$$−{xS}=\:\:\:\:\:\:\:−{x}^{\mathrm{2}} \:\:−\mathrm{2}{x}^{\mathrm{3}} −... \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\:\:\:\:\:\:\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +..=\frac{{x}}{\mathrm{1}−{x}}\right) \\ $$

Commented by Adel last updated on 21/Jan/21

thanks

$$\mathrm{thanks} \\ $$

Answered by mathmax by abdo last updated on 21/Jan/21

S=Σ_(n=1) ^∞ (n/3^n ) =s((1/3))with s(x)=Σ_(n=1) ^∞ nx^n we take ∣x∣<1 we have Σ_(n=0) ^∞ x^n =(1/(1−x)) by derivation we get Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 ))=s(x) ⇒s((1/3))=(1/(3(1−(1/3))^2 )) =(1/(3×(4/9))) =(3/4)

$$\mathrm{S}=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{n}}{\mathrm{3}^{\mathrm{n}} }\:\:\:=\mathrm{s}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\mathrm{with}\:\mathrm{s}\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \mathrm{nx}^{\mathrm{n}} \:\:\:\mathrm{we}\:\mathrm{take}\:\mid\mathrm{x}\mid<\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:\mathrm{by}\:\mathrm{derivation}\:\mathrm{we}\:\mathrm{get}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}} \:=\frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }=\mathrm{s}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{s}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}×\frac{\mathrm{4}}{\mathrm{9}}}\:=\frac{\mathrm{3}}{\mathrm{4}} \\ $$

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