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Question Number 86958 by abdomathmax last updated on 01/Apr/20
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^2 ))
calculate∑n=1∞(−1)nn3(n+1)2
Commented by mathmax by abdo last updated on 04/Apr/20
let decompose F(x)=(1/(x^3 (x+1)^2 )) ⇒ F(x)=(a/x)+(b/x^2 ) +(c/x^3 ) +(d/(x+1)) +(e/((x+1)^2 )) c=1 and e =−1 ⇒ F(x)=(a/x) +(b/x^2 )+(1/x^3 )+(d/(x+1))−(1/((x+1)^2 )) lim_(x→+∞) xF(x)=0 =a+d ⇒d=−a ⇒ F(x)=(a/x) +(b/x^2 ) +(1/x^3 )−(a/(x+1))−(1/((x+1)^2 )) F(1) =(1/4) =a+b+1−(a/2)−(1/4) ⇒1 =4a+4b−2a−1 ⇒ 2a+4b =2 ⇒a +2b =1 F(−2) =−(1/8) =−(a/2)+(b/4)−(1/8) +a−1 ⇒ −1 =−4a +2b−1+8a−8 ⇒0 =4a+2b −8 ⇒ 2a+b−4 =0 ⇒2(1−2b)+b−4 =0 ⇒2−4b+b−4 =0 ⇒ −2b−2 =0 ⇒b=−1 ⇒a =1−2(−1) =1 ⇒ F(x)=(1/x)−(1/x^2 )+(1/x^3 )−(1/(x+1))−(1/((x+1)^2 )) ⇒ S =Σ_(n=1) ^∞ (−1)^n F(n) =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)) −Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) we have Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) let δ(x)=Σ_(n=1) ^∞ (((−1)^n )/n^x ) we have δ(x) =(2^(1−x) −1)ξ(x) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =δ(2)=(2^(−1) −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=2) ^∞ (((−1)^n )/n) =−(Σ_(n=1) ^∞ (((−1)^n )/n) +1) =−(−ln(2)+1) =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−( Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +1) =−(−(π^2 /(12))+1) =(π^2 /(12))−1 ⇒ S =−ln(2) +(π^2 /(12))−(3/4)ξ(3)−ln(2)+1 −(π^2 /(12)) +1 ⇒ S =2−2ln(2)−(3/4)ξ(3)
letdecomposeF(x)=1x3(x+1)2⇒F(x)=ax+bx2+cx3+dx+1+e(x+1)2c=1ande=−1⇒F(x)=ax+bx2+1x3+dx+1−1(x+1)2limx→+∞xF(x)=0=a+d⇒d=−a⇒F(x)=ax+bx2+1x3−ax+1−1(x+1)2F(1)=14=a+b+1−a2−14⇒1=4a+4b−2a−1⇒2a+4b=2⇒a+2b=1F(−2)=−18=−a2+b4−18+a−1⇒−1=−4a+2b−1+8a−8⇒0=4a+2b−8⇒2a+b−4=0⇒2(1−2b)+b−4=0⇒2−4b+b−4=0⇒−2b−2=0⇒b=−1⇒a=1−2(−1)=1⇒F(x)=1x−1x2+1x3−1x+1−1(x+1)2⇒S=∑n=1∞(−1)nF(n)=∑n=1∞(−1)nn−∑n=1∞(−1)nn2+∑n=1∞(−1)nn3−∑n=1∞(−1)nn+1−∑n=1∞(−1)n(n+1)2wehave∑n=1∞(−1)nn=−ln(2)letδ(x)=∑n=1∞(−1)nnxwehaveδ(x)=(21−x−1)ξ(x)∑n=1∞(−1)nn2=δ(2)=(2−1−1)ξ(2)=−12×π26=−π212∑n=1∞(−1)nn3=(21−3−1)ξ(3)=(14−1)ξ(3)=−34ξ(3)∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=−∑n=2∞(−1)nn=−(∑n=1∞(−1)nn+1)=−(−ln(2)+1)=ln(2)−1∑n=1∞(−1)n(n+1)2=∑n=2∞(−1)n−1n2=−∑n=2∞(−1)nn2=−(∑n=1∞(−1)nn2+1)=−(−π212+1)=π212−1⇒S=−ln(2)+π212−34ξ(3)−ln(2)+1−π212+1⇒S=2−2ln(2)−34ξ(3)

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