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Question Number 67013 by mathmax by abdo last updated on 21/Aug/19
find the value of Σ_(n=1) ^∞ (((−1)^n )/((n+1)n^3 ))
findthevalueof∑n=1∞(−1)n(n+1)n3
Commented by mathmax by abdo last updated on 24/Aug/19
let S =Σ_(n=1) ^∞ (((−1)^n )/((n+1)n^3 )) we have S =lim_(n→+∞) S_n with S_n =Σ_(k=1) ^n (((−1)^k )/((k+1)k^3 )) let first decompose F(x) =(1/((x+1)x^3 )) F(x)=(a/(x+1))+(b/x) +(c/x^2 ) +(d/x^3 ) we have a =−1 d =1 ⇒F(x) =−(1/(x+1)) +(b/x)+(c/x^2 ) +(1/x^3 ) lim_(x→+∞) xF(x) =0 =−1+b ⇒b=1 ⇒F(x)=−(1/(x+1))+(1/x) +(c/x^2 ) +(1/x^3 ) F(1)=(1/2) =−(1/2) +2+c ⇒1 =2+c ⇒c =−1 ⇒ F(x)=−(1/(x+1))+(1/x)−(1/x^2 )+(1/x^3 ) ⇒S_n =Σ_(k=1) ^n (−1)^k F(k) =−Σ_(k=1) ^n (((−1)^k )/(k+1)) +Σ_(k=1) ^n (((−1)^k )/k)−Σ_(k=1) ^n (((−1)^k )/k^2 ) +Σ_(k=1) ^n (((−1)^k )/k^3 ) but Σ_(k=1) ^n (((−1)^k )/(k+1)) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k) =−Σ_(k=2) ^(n+1) (((−1)^k )/k) =−(Σ_(k=1) ^(n+1) (((−1)^k )/k)+1)=1−Σ_(k=1) ^(n+1) (((−1)^k )/k) ⇒ S_n =−1 +Σ_(k=1) ^(n+1) (((−1)^k )/k) +Σ_(k=1) ^∞ (((−1)^k )/k)−Σ_(k=1) ^n (((−1)^k )/k^2 )+Σ_(k=1) ^n (((−1)^k )/k^3 ) → −1+2Σ_(k=1) ^∞ (((−1)^k )/k) −Σ_(k=1) ^∞ (((−1)^k )/k^2 ) +Σ_(k=1) ^∞ (((−1)^k )/k^3 ) we have Σ_(k=1) ^∞ (((−1)^k )/k) =−ln(2) Σ_(k=1) ^∞ (((−1)^k )/k^x ) =δ(x)=(2^(1−x) −1)ξ(x) ⇒ Σ_(k=1) ^∞ (((−1)^k )/k^2 ) =(2^(−1) −1)ξ(2) =−(1/2)×(π^2 /6) =−(π^2 /(12)) Σ_(k=1) ^∞ (((−1)^k )/k^3 ) =(2^(1−3) −1)ξ(3) =−(3/4)ξ(3) ⇒ S =lim_(n→+∞) S_n =−1+2(−ln(2))+(π^2 /(12))−(3/4)ξ(3) =(π^2 /(12))−1−2ln(2)−(3/4)ξ(3) .
letS=∑n=1∞(−1)n(n+1)n3wehaveS=limn→+∞SnwithSn=∑k=1n(−1)k(k+1)k3letfirstdecomposeF(x)=1(x+1)x3F(x)=ax+1+bx+cx2+dx3wehavea=−1d=1⇒F(x)=−1x+1+bx+cx2+1x3limx→+∞xF(x)=0=−1+b⇒b=1⇒F(x)=−1x+1+1x+cx2+1x3F(1)=12=−12+2+c⇒1=2+c⇒c=−1⇒F(x)=−1x+1+1x−1x2+1x3⇒Sn=∑k=1n(−1)kF(k)=−∑k=1n(−1)kk+1+∑k=1n(−1)kk−∑k=1n(−1)kk2+∑k=1n(−1)kk3but∑k=1n(−1)kk+1=∑k=2n+1(−1)k−1k=−∑k=2n+1(−1)kk=−(∑k=1n+1(−1)kk+1)=1−∑k=1n+1(−1)kk⇒Sn=−1+∑k=1n+1(−1)kk+∑k=1∞(−1)kk−∑k=1n(−1)kk2+∑k=1n(−1)kk3→−1+2∑k=1∞(−1)kk−∑k=1∞(−1)kk2+∑k=1∞(−1)kk3wehave∑k=1∞(−1)kk=−ln(2)∑k=1∞(−1)kkx=δ(x)=(21−x−1)ξ(x)⇒∑k=1∞(−1)kk2=(2−1−1)ξ(2)=−12×π26=−π212∑k=1∞(−1)kk3=(21−3−1)ξ(3)=−34ξ(3)⇒S=limn→+∞Sn=−1+2(−ln(2))+π212−34ξ(3)=π212−1−2ln(2)−34ξ(3).

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