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Question Number 46860 by 786786AM last updated on 01/Nov/18
Find the sum to infinity whose n^(th) term is (n/2^(n−1) ) .
$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{whose}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{is}\:\frac{\mathrm{n}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\:. \\ $$
Commented by maxmathsup by imad last updated on 01/Nov/18
let s(x)=Σ_(n=1) ^∞ nx^(n−1) with ∣x∣<1 we have Σ_(n=1) ^∞ (n/2^(n−1) ) =s((1/2)) but Σ_(n=0) ^∞ x^n = (1/(1−x)) ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ s(x)=(1/((1−x)^2 )) ⇒ s((1/2)) = (1/((1−(1/2))^2 )) =4 ⇒ Σ_(n=1) ^∞ nx^(n−1) =4 .
$${let}\:{s}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:{we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{\mathrm{2}^{{n}−\mathrm{1}} }\:={s}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\:{s}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${s}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\mathrm{4}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\mathrm{4}\:. \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
anther method let S_n (x)=Σ_(k=1) ^n k x^(k−1) with ∣x∣<1 we have S_∞ ((1/2))=Σ_(k=1) ^∞ k x^(k−1) let determine S_n (x) we have Σ_(k=0) ^n x^k =((x^(n+1) −1)/(x−1)) ⇒ Σ_(k=1) ^n k x^(k−1) =((nx^(n+1) −(n+1)x^n +1)/((1−x)^2 ))=S_n (x) ⇒ S_n ((1/2)) = 4{(n/2^(n+1) ) −((n+1)/2^n ) +1} →4 (n→+∞) ⇒ S_(∞ ) ((1/2))=4 =Σ_(n=1) ^∞ (n/2^(n−1) )
$${anther}\:{method}\:\:{let}\:\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}−\mathrm{1}} \:\:{with}\:\mid{x}\mid<\mathrm{1}\:{we}\:{have} \\ $$$${S}_{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}\:{x}^{{k}−\mathrm{1}} \:\:{let}\:{determine}\:{S}_{{n}} \left({x}\right)\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \:=\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}−\mathrm{1}} \:=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} +\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }={S}_{{n}} \left({x}\right)\:\Rightarrow \\ $$$${S}_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{4}\left\{\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} }\:−\frac{{n}+\mathrm{1}}{\mathrm{2}^{{n}} }\:+\mathrm{1}\right\}\:\rightarrow\mathrm{4}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow\:{S}_{\infty\:} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{4}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
S_n =(1/2^0 )+(2/2^1 )+(3/2^2 )+(4/2^3 )+...+(n/2^(n−1) ) (S_n /2)= +(1/2^1 )+(2/2^2 )+(3/2^3 )+...+((n−1)/2^(n−1) )+(n/2^n ) S_n −(S_n /2)=((1/2^0 )+(1/2^1 )+(1/2^2 )+...+(1/2^(n−1) ))−(n/2^n ) (S_n /2)=(((1/2^0 )(1−(1/2^n )))/(1−(1/2)))−(n/2^n ) S_n = 2×(((1−(1/2^n )))/(1/2))−(n/2^n ) when n→∞ S_∞ =4
$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{0}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{3}} }+…+\frac{{n}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$\frac{{S}_{{n}} }{\mathrm{2}}=\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+…+\frac{{n}−\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }+\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$${S}_{{n}} −\frac{{S}_{{n}} }{\mathrm{2}}=\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)−\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$$\frac{{S}_{{n}} }{\mathrm{2}}=\frac{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{0}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}−\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$${S}_{{n}} =\:\mathrm{2}×\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)}{\frac{\mathrm{1}}{\mathrm{2}}}−\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$${when}\:{n}\rightarrow\infty\:\:{S}_{\infty} \:=\mathrm{4} \\ $$

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