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Question Number 63665 by mathmax by abdo last updated on 07/Jul/19
find the value of Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 ))
findthevalueof∑n=1∞(−1)nn2(n+1)3
Commented by mathmax by abdo last updated on 07/Jul/19
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=lim_(x→0) x^2 F(x)=1 e =lim_(x→−1) (x+1)^3 F(x)=1 ⇒F(x)=(a/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(x)−(1/x^2 ) =(1/x^2 ){(1/((x+1)^3 )) −1} =(1/x^2 ){((1−x^3 −3x^2 −3x−1)/((x+1)^3 ))} =−((x^3 +3x^2 +3x)/(x^2 (x+1)^3 )) =−((x^2 +3x +3)/(x(x+1)^3 )) =(a/x) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ −((x^2 +3x +3)/((x+1)^3 )) =a +((cx)/(x+1)) +(dx/((x+1)^2 )) +(x/((x+1)^3 )) (x→0) ⇒ a =−3 ⇒F(x) =−(3/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) lim_(x→+∞) xF(x)=0 =−3 +c ⇒c=3 ⇒ F(x)=−(3/x) +(1/x^2 ) +(3/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(−2) =−(1/4) =(3/2) +(1/4) −3 +d −1 =(7/4)−4 +d ⇒ d =−(1/4)−(7/4) +4 =−2+4 =2 ⇒ 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) =−3 Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +2Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) +Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) we have Σ_(n=1) ^∞ (x^n /n) =−ln(1−x) if ∣x∣<1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) if δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) we have proved that δ(x)=(2^(1−x) −1)ξ(x) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(−1) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−{−(π^2 /(12)) +1}=(π^2 /(12)) −1 also Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^3 ) =−{ δ(3) +1} =−(2^(−2) −1)ξ(3)−1 =−(−(3/4))ξ(3)−1 =(3/4)ξ(3)−1 ⇒ S = 3ln(2)−(π^2 /(12)) +((2π^2 )/(12)) −2 +(3/4)ξ(3)−1 =3ln(2) +(π^2 /(12)) +(3/4)ξ(3)−3 .
letdecomposeF(x)=1x2(x+1)3⇒F(x)=ax+bx2+cx+1+d(x+1)2+e(x+1)3b=limx→0x2F(x)=1e=limx→−1(x+1)3F(x)=1⇒F(x)=ax+1x2+cx+1+d(x+1)2+1(x+1)3F(x)−1x2=1x2{1(x+1)3−1}=1x2{1−x3−3x2−3x−1(x+1)3}=−x3+3x2+3xx2(x+1)3=−x2+3x+3x(x+1)3=ax+cx+1+d(x+1)2+1(x+1)3⇒−x2+3x+3(x+1)3=a+cxx+1+dx(x+1)2+x(x+1)3(x→0)⇒a=−3⇒F(x)=−3x+1x2+cx+1+d(x+1)2+1(x+1)3limx→+∞xF(x)=0=−3+c⇒c=3⇒F(x)=−3x+1x2+3x+1+d(x+1)2+1(x+1)3F(−2)=−14=32+14−3+d−1=74−4+d⇒d=−14−74+4=−2+4=2⇒F(x)=−3x+1x2+3x+1+2(x+1)2+1(x+1)3S=∑n=1∞(−1)nF(n)=−3∑n=1∞(−1)nn+∑n=1∞(−1)nn2+2∑n=1∞(−1)n(n+1)2+∑n=1∞(−1)n(n+1)3wehave∑n=1∞xnn=−ln(1−x)if∣x∣<1⇒∑n=1∞(−1)nn=−ln(2)ifδ(x)=∑n=1∞(−1)nnxwehaveprovedthatδ(x)=(21−x−1)ξ(x)⇒∑n=1∞(−1)nn2=(2−1−1)ξ(2)=−12π26=−π212∑n=1∞(−1)n(n+1)2=∑n=2∞(−1)n−1n2=−∑n=2∞(−1)nn2=−{−π212+1}=π212−1also∑n=1∞(−1)n(n+1)3=∑n=2∞(−1)n−1n3=−∑n=2∞(−1)nn3=−{δ(3)+1}=−(2−2−1)ξ(3)−1=−(−34)ξ(3)−1=34ξ(3)−1⇒S=3ln(2)−π212+2π212−2+34ξ(3)−1=3ln(2)+π212+34ξ(3)−3.

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