Question Number 102784 by bobhans last updated on 11/Jul/20

Σ_(n=1) ^∞  (n/4^n ) =?

Answered by Dwaipayan Shikari last updated on 11/Jul/20

Σ_(n=1) ^∞ (n/4^n )=(1/4)+(2/(16))+(3/(64))+(4/(256))+(5/(1024))+....=S_n         (S_n /4)=          (1/(16))+(2/(64))+(3/(256))+(4/(1024))+......  subtracting  ((3S_n )/4)=(1/4)+(1/(16))+(1/(64))+(1/(256))+(1/(1024))+......  ((3S_n )/4)=(1/3)⇒S_n =(4/9)

Answered by bramlex last updated on 11/Jul/20

let S = Σ_(n = 1) ^∞  ((n+1)/4^n ) & X=Σ_(n = 1) ^∞  (n/4^n )  ⇔S = X+Σ_(n = 1) ^∞  (1/4^n )   S = X + ((1/4)/(1−(1/4))) ;(geometric P )  S = X+(1/3) . in the other hand   we have Σ_(n = 1) ^∞ ((n+1)/4^n ) = 4 Σ_(n = 1) ^∞ (n/4^n ) −1  S = 4X−1 = X+(1/3)⇒3X=(4/3); X = (4/9)  Σ_(n = 1 ) ^∞  (n/4^n ) = (4/9) •

Answered by mathmax by abdo last updated on 11/Jul/20

S =Σ_(n=1) ^∞  (n/4^n )     let w(x) =Σ_(n=0) ^∞  x^n    with ∣x∣<1 ⇒  w^′ (x) =Σ_(n=1) ^∞  nx^(n−1)  ⇒xw^′ (x) =Σ_(n=1) ^∞  nx^n  ⇒Σ_(n=1) ^∞  n((1/4))^n  =(1/4)w^′ ((1/4))  we have w(x)=(1/(1−x)) ⇒ w^′ (x) =(1/((1−x)^2 )) ⇒xw^′ (x) =(x/((1−x)^2 )) ⇒  Σ_(n=1) ^∞  (n/4^n ) =(1/(4(1−(1/4))^2 )) =(1/4)×((4/3))^2  =(4/9)