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Question Number 37342 by math khazana by abdo last updated on 12/Jun/18
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1))) x^n with ∣x∣<1 2) find the value of Σ_(n=1) ^∞ (1/(n^2 (n+1)2^n )) .
calculate∑n=1∞(−1)nn2(n+1)xnwith∣x∣<12)findthevalueof∑n=1∞1n2(n+1)2n.
Commented by prof Abdo imad last updated on 15/Jun/18
let decompose F(x)=(1/(x^2 (x+1))) F(x)=(a/x) +(b/x^2 ) +(c/(x+1)) b=lim_(x→0) x^2 F(x)=1 c =lim_(x→−1) (x+1)F(x)=1 ⇒ F(x)= (a/x) +(1/x^2 ) +(1/(x+1)) F(1)=(1/2) =a +1 +(1/2) ⇒a=−1 ⇒ F(x)=−(1/x) +(1/x^2 ) +(1/(x+1)) ⇒ S(x)=Σ_(n=1) ^∞ (−1)^n {−(1/n) +(1/n^2 ) +(1/(n+1))}x^n =−Σ_(n=1) ^∞ (((−1)^n )/n) x^n +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) x^n + Σ_(n=1) ^∞ (((−1)^n )/(n+1))x^n let w(x)=Σ_(n=1) ^∞ (((−1)^n )/n)x^n w^′ (x)= Σ_(n=1) ^∞ (−1)^n x^(n−1) =(1/x){ Σ_(n=0) ^∞ (−x)^n −1} = (1/(x(x+1))) −(1/x) =(1/x){(1/(x+1)) −1}=(1/x)((−x)/(1+x)) = ((−1)/(1+x)) ⇒w(x)=−ln(1+x) +c butc=w(1)=0 ⇒w(x)= −ln(1+x) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) x^n = Σ_(n=2) ^∞ (((−1)^(n−1) )/n) x^(n−1) =−(1/x)Σ_(n=2) ^∞ (((−1)^n )/n) x^n =−(1/x){ Σ_(n=1) ^∞ (((−1)^n )/n)x^n +1} =−(1/x){−ln(1+x)} −(1/x) = (1/x)ln(1+x) −(1/x) let v(x)= Σ_(n=1) ^∞ (((−1)^n )/n^2 )x^n v^′ (x) =Σ_(n=1) ^∞ (((−1)^n )/n)x^n =−ln(1+x) ⇒ v(x)= −∫ln(1+x)dx +λ =−{ xln(1+x)−∫ (x/(1+x))dx} +λ =xln(1+x) +∫ ((1+x−1)/(1+x))dx +λ =xln(1+x) +x −ln(1+x) +λ =(x−1)ln(1+x) +x+λ but λ=v(0)=0 ⇒ v(x)=(x−1)ln(1+x) +x ⇒ S(x)=ln(1+x)+(1/x){ln(1+x)−1} +(x−1)ln(1+x) +x ⇒ S(x)=x ln(1+x) +(1/x)ln(1+x) +x−(1/x) = ((x^2 +1)/x)ln(1+x) +((x^2 −1)/x) with −1<x<1 and x≠0
letdecomposeF(x)=1x2(x+1)F(x)=ax+bx2+cx+1b=limx→0x2F(x)=1c=limx→−1(x+1)F(x)=1⇒F(x)=ax+1x2+1x+1F(1)=12=a+1+12⇒a=−1⇒F(x)=−1x+1x2+1x+1⇒S(x)=∑n=1∞(−1)n{−1n+1n2+1n+1}xn=−∑n=1∞(−1)nnxn+∑n=1∞(−1)nn2xn+∑n=1∞(−1)nn+1xnletw(x)=∑n=1∞(−1)nnxnw′(x)=∑n=1∞(−1)nxn−1=1x{∑n=0∞(−x)n−1}=1x(x+1)−1x=1x{1x+1−1}=1x−x1+x=−11+x⇒w(x)=−ln(1+x)+cbutc=w(1)=0⇒w(x)=−ln(1+x)∑n=1∞(−1)nn+1xn=∑n=2∞(−1)n−1nxn−1=−1x∑n=2∞(−1)nnxn=−1x{∑n=1∞(−1)nnxn+1}=−1x{−ln(1+x)}−1x=1xln(1+x)−1xletv(x)=∑n=1∞(−1)nn2xnv′(x)=∑n=1∞(−1)nnxn=−ln(1+x)⇒v(x)=−∫ln(1+x)dx+λ=−{xln(1+x)−∫x1+xdx}+λ=xln(1+x)+∫1+x−11+xdx+λ=xln(1+x)+x−ln(1+x)+λ=(x−1)ln(1+x)+x+λbutλ=v(0)=0⇒v(x)=(x−1)ln(1+x)+x⇒S(x)=ln(1+x)+1x{ln(1+x)−1}+(x−1)ln(1+x)+x⇒S(x)=xln(1+x)+1xln(1+x)+x−1x=x2+1xln(1+x)+x2−1xwith−1<x<1andx≠0
Commented by prof Abdo imad last updated on 15/Jun/18
Σ_(n=1) ^∞ (1/(n^2 (n+1)2^n )) =S((1/2)) =(((1/4)+1)/(1/2))ln((3/2)) +(((1/4)−1)/(1/2)) =2.(5/4)ln((3/2)) +2.(((−3)/4)) =(5/2)ln((3/2)) −(3/2)
∑n=1∞1n2(n+1)2n=S(12)=14+112ln(32)+14−112=2.54ln(32)+2.(−34)=52ln(32)−32

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