Question Number 104891 by mathmax by abdo last updated on 24/Jul/20

1) decompose the fraction F(x) =(1/(x^3 (x+1)^3 ))  2) find the sum Σ_(n=1) ^∞  (((−1)^n )/(n^3 (n+1)^3 ))

Answered by mathmax by abdo last updated on 26/Jul/20

1) F(x) =(1/(x^3 (x+1)^3 )) ⇒F(x) =((1/(x(3+1))))^3  =((1/x)−(1/(x+1)))^3   =(1/x^3 )−(3/(x^2 (x+1))) +(3/(x(x+1)^2 ))−(1/((x+1)^3 ))  =(1/x^3 ) −(3/x)((1/x)−(1/(x+1))) +(3/(x+1))((1/x)−(1/(x+1)))−(1/((x+1)^3 ))  =(1/x^3 )−(3/x^2 ) +(3/(x(x+1))) +(3/(x(x+1)))−(3/((x+1)^2 ))−(1/((x+1)^3 ))  =(1/x^3 )−(3/x^2 ) +6((1/x)−(1/(x+1)))−(3/((x+1)^2 ))−(1/((x+1)^3 ))  ⇒F(x) =(6/x)−(3/x^2 )+(1/x^3 )−(6/(x+1))−(3/((x+1)^2 ))−(1/((x+1)^3 ))

Commented bymathmax by abdo last updated on 26/Jul/20

2) let S =Σ_(n=1) ^∞  (((−1)^n )/(n^3 (n+1)^3 )) ⇒S =lim_(n→+∞) Σ_(k=1) ^n  (((−1)^k )/(k^3 (k+1)^3 ))  Σ_(k=1) ^n  (((−1)^k )/(k^3 (k+1)^3 )) =6 Σ_(k=1) ^n  (((−1)^k )/k)−3Σ_(k=1) ^n  (((−1)^k )/k^2 ) +Σ_(k=1) ^n  (((−1)^k )/k^3 )−6Σ_(k=1) ^n  (((−1)^k )/(k+1))  −3 Σ_(k=1) ^n  (((−1)^k )/((k+1)^2 ))−Σ_(k=1) ^n  (((−1)^k )/((k+1)^3 ))  but we have  Σ_(k=1) ^(n )  (((−1)^k )/k) →−ln2(n→+∞)  Σ_(k=1) ^n  (((−1)^k )/k^2 ) →Σ_(k=1) ^∞  (((−1)^k )/k^2 ) =δ(2) =(2^(1−2) −1)ξ(2) =−(1/2)×(π^2 /6) =−(π^2 /(12))  Σ_(k=1) ^n  (((−1)^k )/k^3 ) →Σ_(k=1) ^∞  (((−1)^k )/k^3 ) =δ(3) =(2^(1−3) −1)ξ(3) =−(3/4)ξ(3)  Σ_(k=1) ^n  (((−1)^k )/(k+1)) =Σ_(k=2) ^(n+1)  (((−1)^(k−1) )/k) =Σ_(k=1) ^(n+1)  (((−1)^(k−1) )/k)−1 =−Σ_(k=1) ^(n+1 ) (((−1)^k )/k)−1  →ln(2)−1  Σ_(k=1) ^n  (((−1)^k )/((k+1)^2 )) =Σ_(k=2) ^(n+1)  (((−1)^(k−1) )/k^2 ) =Σ_(k=2) ^(n+1)  (((−1)^(k−1) )/k^2 ) −1 →−δ(2)−1=(π^2 /(12))−1  Σ_(k=1) ^n  (((−1)^k )/((k+1)^3 )) =Σ_(k=2) ^(n+1)  (((−1)^(k−1) )/k^3 ) =Σ_(k=1) ^(n+1)  (((−1)^(k−1) )/k^3 )−1 →−δ(3)−1  =(3/4)ξ(3)−1 ⇒  S =−6ln(2)−3(−(π^2 /(12)))−(3/4)ξ(3)−6(ln2−1)−3((π^2 /(12))−1)+ξ(3)+1  =−12ln(2)+(π^2 /4) +(1/4)ξ(3)+6−(π^2 /4) +4  S=−12ln(2)+10 +(1/4)ξ(3)