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Question Number 33709 by math khazana by abdo last updated on 22/Apr/18
find Σ_(n=0) ^∞ (n+1)x^(3n) 2) calculate Σ_(n=0) ^∞ ((n+1)/8^n ) .
$${find}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({n}+\mathrm{1}\right){x}^{\mathrm{3}{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{n}+\mathrm{1}}{\mathrm{8}^{{n}} }\:. \\ $$
Commented by prof Abdo imad last updated on 27/Apr/18
let put for ∣x∣<1 w(x)= Σ_(n=0) ^∞ (n+1)x^(3n) we have x^3 w(x)=Σ_(n=0) ^∞ (n+1)(x^3 )^(n+1) =ϕ(x^3 ) with ϕ(t)=Σ_(n=0) ^∞ (n+1)t^(n+1) but we have Σ_(n=0) ^∞ t^n = (1/(1−t)) ⇒Σ_(n=1) ^∞ n t^(n−1) = (1/((1−t)^2 )) ⇒ Σ_(n=1) ^∞ n t^n = (t/((1−t)^2 )) but ϕ(t)=Σ_(n=1) ^∞ n t^n ⇒ ϕ(t) = (t/((1−t)^2 )) with ∣t∣<1 ⇒x^3 w(x)= (x^3 /((1−x^3 )^2 )) ⇒ w(x)= (1/((1−x^3 )^2 )) 2)we have Σ_(n=0) ^∞ ((n+1)/8^n ) = Σ_(n=0) ^∞ (n+1)((1/2))^(3n) =w((1/2)) = (1/((1−(1/8))^2 )) = (1/(((7/8))^2 )) = ((64)/(49)) .
$${let}\:{put}\:{for}\:\mid{x}\mid<\mathrm{1}\:{w}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({n}+\mathrm{1}\right){x}^{\mathrm{3}{n}} \:\:{we}\:{have} \\ $$$${x}^{\mathrm{3}} {w}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{1}\right)\left({x}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} =\varphi\left({x}^{\mathrm{3}} \right)\:{with} \\ $$$$\varphi\left({t}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({n}+\mathrm{1}\right){t}^{{n}+\mathrm{1}} \:\:{but}\:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{t}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:{t}^{{n}−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:{t}^{{n}} \:\:\:=\:\:\frac{{t}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:{but}\:\varphi\left({t}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:{t}^{{n}} \Rightarrow \\ $$$$\varphi\left({t}\right)\:=\:\frac{{t}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:\:{with}\:\mid{t}\mid<\mathrm{1}\:\:\Rightarrow{x}^{\mathrm{3}} {w}\left({x}\right)=\:\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{w}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{n}+\mathrm{1}}{\mathrm{8}^{{n}} }\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{n}} \\ $$$$={w}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{64}}{\mathrm{49}}\:\:. \\ $$

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