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Question Number 70595 by mathmax by abdo last updated on 06/Oct/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 ))
calculate∑n=1∞(−1)nn2(n+1)3
Commented by mathmax by abdo last updated on 08/Oct/19
let S =Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 )) let decompose F(x)=(1/(x^2 (x+1)^3 )) F(x)=(a/x)+(b/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(e/((x+1)^3 )) b=1 and e=1 ⇒F(x)=(a/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) lim_(x→+∞) xF(x)=0=a+c ⇒c=−a ⇒ F(x)=(a/x)+(1/x^2 )−(a/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(1)=(1/8) =a+1−(a/2) +(d/4) +(1/8) ⇒(a/2) +(d/4)=−1 ⇒2a+d=−4 F(−2)=−(1/4) =−(a/2) +(1/4) +a +d −1 ⇒ −1=−2a+1 +4a+4d−4 ⇒−1=2a+4d−3 ⇒2a+4d=2 ⇒ a+2d=1 ⇒a=1−2d ⇒2(1−2d)+d=−4 ⇒2−3d=−4 ⇒ −3d=−6 ⇒d=2 ⇒a=1−4=−3 ⇒ F(x)=−(3/x) +(1/x^2 ) +(3/(x+1)) +(2/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ S=Σ_(n=1) ^∞ (−1)^n F(n)=−3Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +3 Σ_(n=1) ^∞ (((−1)^n )/(n+1)) +Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) 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+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)−1=ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^3 ) −1 =−Σ_(n=1) ^∞ (((−1)^n )/n^3 )−1 =−(2^(1−3) −1)ξ(3)−1 =−((1/4)−1)ξ(3)−1 =(3/4)ξ(3)−1 ⇒ S =3ln(2)−(π^2 /(12)) +3ln(2)−3 +(3/4)ξ(3)−1 S =6ln(2)+(3/4)ξ(3)−4−(π^2 /(12)) .
letS=∑n=1∞(−1)nn2(n+1)3letdecomposeF(x)=1x2(x+1)3F(x)=ax+bx2+cx+1+d(x+1)2+e(x+1)3b=1ande=1⇒F(x)=ax+1x2+cx+1+d(x+1)2+1(x+1)3limx→+∞xF(x)=0=a+c⇒c=−a⇒F(x)=ax+1x2−ax+1+d(x+1)2+1(x+1)3F(1)=18=a+1−a2+d4+18⇒a2+d4=−1⇒2a+d=−4F(−2)=−14=−a2+14+a+d−1⇒−1=−2a+1+4a+4d−4⇒−1=2a+4d−3⇒2a+4d=2⇒a+2d=1⇒a=1−2d⇒2(1−2d)+d=−4⇒2−3d=−4⇒−3d=−6⇒d=2⇒a=1−4=−3⇒F(x)=−3x+1x2+3x+1+2(x+1)2+1(x+1)3⇒S=∑n=1∞(−1)nF(n)=−3∑n=1∞(−1)nn+∑n=1∞(−1)nn2+3∑n=1∞(−1)nn+1+∑n=1∞(−1)n(n+1)3∑n=1∞(−1)nn=−ln(2)∑n=1∞(−1)nn2=(21−2−1)ξ(2)=−12π26=−π212∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=∑n=1∞(−1)n−1n−1=ln(2)−1∑n=1∞(−1)n(n+1)3=∑n=2∞(−1)n−1n3=∑n=1∞(−1)n−1n3−1=−∑n=1∞(−1)nn3−1=−(21−3−1)ξ(3)−1=−(14−1)ξ(3)−1=34ξ(3)−1⇒S=3ln(2)−π212+3ln(2)−3+34ξ(3)−1S=6ln(2)+34ξ(3)−4−π212.

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