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Question Number 74887 by abdomathmax last updated on 03/Dec/19
calculate Σ_(n=1) ^∞   (((−1)^n )/((n+1)n^3 ))
calculaten=1(1)n(n+1)n3
Commented by mathmax by abdo last updated on 21/Dec/19
let decompose F(x)=(1/((x+1)x^3 )) ⇒F(x)=(a/x)+(b/x^2 ) +(c/x^3 ) +(d/(x+1))  c =x^3 F(x)∣_(x=0)   =1  d=(x+1)F(x)∣_(x=−1)    =−1 ⇒F(x)=(a/x)+(b/x^2 ) +(1/x^3 )−(1/(x+1))  lim_(x→+∞) xF(x)=0 =a−1 ⇒a=1 ⇒F(x)=(1/x) +(b/x^2 ) +(1/x^3 )−(1/(x+1))  F(1)=(1/2) =1+b+1−(1/2) ⇒b+2 =1 ⇒b=−1 ⇒  F(x)=(1/x)−(1/x^2 ) +(1/x^3 )−(1/(x+1)) ⇒  Σ_(n=1) ^∞  (((−1)^n )/((n+1)n^3 ))  =Σ_(n=1) ^∞  (((−1)^n )/n)−Σ_(n=1) ^∞  (((−1)^n )/n^2 ) +Σ_(n=1) ^∞  (((−1)^n )/n^3 )  −Σ_(n=1) ^∞  (((−1)^n )/(n+1))  we have  Σ_(n=1) ^∞  (((−1)^n )/n) =−ln(2)  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^3 ) =(2^(1−3) −1)ξ(3) =((1/4)−1)ξ(3) =−(3/4)ξ(3)  Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞  (((−1)^(n−1) )/n) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) −1 =ln(2)−1 ⇒  S =−ln(2)+(π^2 /(12))−(3/4)ξ(3)−ln(2)+1  =1−2ln(2)+(π^2 /(12))−(3/4)ξ(3)
letdecomposeF(x)=1(x+1)x3F(x)=ax+bx2+cx3+dx+1c=x3F(x)x=0=1d=(x+1)F(x)x=1=1F(x)=ax+bx2+1x31x+1limx+xF(x)=0=a1a=1F(x)=1x+bx2+1x31x+1F(1)=12=1+b+112b+2=1b=1F(x)=1x1x2+1x31x+1n=1(1)n(n+1)n3=n=1(1)nnn=1(1)nn2+n=1(1)nn3n=1(1)nn+1wehaven=1(1)nn=ln(2)n=1(1)nn2=(2121)ξ(2)=12×π26=π212n=1(1)nn3=(2131)ξ(3)=(141)ξ(3)=34ξ(3)n=1(1)nn+1=n=2(1)n1n=n=1(1)n1n1=ln(2)1S=ln(2)+π21234ξ(3)ln(2)+1=12ln(2)+π21234ξ(3)

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