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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 ))
1)decomposethefractionF(x)=1x3(x+1)32)findthesum∑n=1∞(−1)nn3(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 ))
1)F(x)=1x3(x+1)3⇒F(x)=(1x(3+1))3=(1x−1x+1)3=1x3−3x2(x+1)+3x(x+1)2−1(x+1)3=1x3−3x(1x−1x+1)+3x+1(1x−1x+1)−1(x+1)3=1x3−3x2+3x(x+1)+3x(x+1)−3(x+1)2−1(x+1)3=1x3−3x2+6(1x−1x+1)−3(x+1)2−1(x+1)3⇒F(x)=6x−3x2+1x3−6x+1−3(x+1)2−1(x+1)3
Commented by mathmax 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)
2)letS=∑n=1∞(−1)nn3(n+1)3⇒S=limn→+∞∑k=1n(−1)kk3(k+1)3∑k=1n(−1)kk3(k+1)3=6∑k=1n(−1)kk−3∑k=1n(−1)kk2+∑k=1n(−1)kk3−6∑k=1n(−1)kk+1−3∑k=1n(−1)k(k+1)2−∑k=1n(−1)k(k+1)3butwehave∑k=1n(−1)kk→−ln2(n→+∞)∑k=1n(−1)kk2→∑k=1∞(−1)kk2=δ(2)=(21−2−1)ξ(2)=−12×π26=−π212∑k=1n(−1)kk3→∑k=1∞(−1)kk3=δ(3)=(21−3−1)ξ(3)=−34ξ(3)∑k=1n(−1)kk+1=∑k=2n+1(−1)k−1k=∑k=1n+1(−1)k−1k−1=−∑k=1n+1(−1)kk−1→ln(2)−1∑k=1n(−1)k(k+1)2=∑k=2n+1(−1)k−1k2=∑k=2n+1(−1)k−1k2−1→−δ(2)−1=π212−1∑k=1n(−1)k(k+1)3=∑k=2n+1(−1)k−1k3=∑k=1n+1(−1)k−1k3−1→−δ(3)−1=34ξ(3)−1⇒S=−6ln(2)−3(−π212)−34ξ(3)−6(ln2−1)−3(π212−1)+ξ(3)+1=−12ln(2)+π24+14ξ(3)+6−π24+4S=−12ln(2)+10+14ξ(3)

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