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Question Number 140636 by Mathspace last updated on 10/May/21

calculate Σ_(n=1) ^∞   (((−1)^n )/(n^2 (n+1)(2n+1)^2 ))

calculaten=1(1)nn2(n+1)(2n+1)2

Answered by Dwaipayan Shikari last updated on 10/May/21

Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)(2n+1)))=Σ_(n=1) ^∞ (((−1)^n )/(n^2 (2n+1)))−(((−1)^n )/(n(n+1)(2n+1)))  =Σ_(n=1) ^∞ (((−1)^n )/n)((1/n)−(2/(2n+1)))−Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)))+Σ_(n=1) ^∞ (((−1)^n )/((n+1)(2n+1)))  =−(π^2 /(12))−3Σ(((−1)^n )/(n(2n+1)))+Σ_(n=1) ^∞ ((2(−1)^n )/(2n+1))−(((−1)^n )/(n+1))  =−(π^2 /(12))−3Σ_(n=1) ^∞ (((−1)^n )/n)−(((−1)^n 2)/(2n+1))−2((1/3)−(1/5)+..)+((1/2)−(1/3)+...)  =−(π^2 /(12))+3log(2)+8((π/4)−1)+(1−log(2))  =((−π^2 )/(12))+2log(2)+2π−7

n=1(1)nn2(n+1)(2n+1)=n=1(1)nn2(2n+1)(1)nn(n+1)(2n+1)=n=1(1)nn(1n22n+1)n=1(1)nn(2n+1)+n=1(1)n(n+1)(2n+1)=π2123Σ(1)nn(2n+1)+n=12(1)n2n+1(1)nn+1=π2123n=1(1)nn(1)n22n+12(1315+..)+(1213+...)=π212+3log(2)+8(π41)+(1log(2))=π212+2log(2)+2π7

Answered by mathmax by abdo last updated on 11/May/21

let  S=Σ_(n=1) ^∞  (((−1)^n )/(n^2 (n+1)(2n+1)^2 )) let decompose  F(x)=(1/(x^2 (x+1)(2x+1))) ⇒F(x)=(a/x)+(b/x^2 ) +(c/(x+1)) +(d/(2x+1))  b=1  ,c=−1 ,d =(1/(((1/4))((1/2))))=8 ⇒F(x)=(a/x)+(1/x^2 )−(1/(x+1)) +(8/(2x+1))  lim_(x→+∞) xF(x)=0 =a−1+4 ⇒a=−3 ⇒  F(x)=−(3/x)+(1/x^2 )−(1/(x+1)) +(8/(2x+1)) ⇒  S=−3Σ_(n=1) ^∞  (((−1)^n )/n) +Σ_(n=1) ^∞  (((−1)^n )/n^2 )−Σ_(n=1) ^∞  (((−1)^n )/(n+1)) +8Σ_(n=1) ^∞  (((−1)^n )/(2n+1))  we have Σ_(n=1) ^∞  (((−1)^n )/n) =−log2  Σ_(n=1) ^∞  (((−1)^n )/n^2 ) =(2^(1−2) −1)ξ(2) =−(1/2).(π^2 /6)=−(π^2 /(12))  Σ_(n=1) ^∞  (((−1)^n )/(n+1)) =Σ_(n=2) ^∞  (((−1)^(n−1) )/n) =−Σ_(n=2) ^∞  (((−1)^n )/n)  =−(−log2 +1) =−1+log2  Σ_(n=1) ^∞  (((−1)^n )/(2n+1)) =(π/4)−1 ⇒  S=−3(−log2)−(π^2 /(12)) +1−log2 +2π−8  =2log2 −(π^2 /(12)) +2π−7

letS=n=1(1)nn2(n+1)(2n+1)2letdecomposeF(x)=1x2(x+1)(2x+1)F(x)=ax+bx2+cx+1+d2x+1b=1,c=1,d=1(14)(12)=8F(x)=ax+1x21x+1+82x+1limx+xF(x)=0=a1+4a=3F(x)=3x+1x21x+1+82x+1S=3n=1(1)nn+n=1(1)nn2n=1(1)nn+1+8n=1(1)n2n+1wehaven=1(1)nn=log2n=1(1)nn2=(2121)ξ(2)=12.π26=π212n=1(1)nn+1=n=2(1)n1n=n=2(1)nn=(log2+1)=1+log2n=1(1)n2n+1=π41S=3(log2)π212+1log2+2π8=2log2π212+2π7

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