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Question Number 72018 by mathmax by abdo last updated on 23/Oct/19
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 02/Nov/19
Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^2 )) =lim_(n→+∞) Σ_(k=1) ^n (((−1)^k )/(k^3 (k+1)^2 )) 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=lim_(x→0) x^3 F(x)=1 e=lim_(x→−1) (x+1)^2 F(x)=−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+4−2a−1 ⇒ 2a+4b+2=0 ⇒a+2b+1=0 ⇒a+2b=−1 F(−2)=(1/(−8)) =−(a/2) +(b/4)−(1/8)+a−1 ⇒(1/8)=−(a/2)−(b/4)+(1/8)+1 ⇒ 1=−4a−2b +1 +8 ⇒−4a−2b +8=0 ⇒2a+b=4 ⇒ 2(−2b−1)+b=4 ⇒−4b−2+b =4 ⇒−3b =6 ⇒b=−2 ⇒a=3 ⇒ F(x)=(3/x)−(2/x^2 )+(1/x^3 )−(3/(x+1))−(1/((x+1)^2 )) ⇒ S =3Σ_(n=1) ^∞ (((−1)^n )/n)−2Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +Σ_(n=1) ^∞ (((−1)^n )/n^3 )−3Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) we have Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(1−2) −1)ξ(2) (formula Σ_(n=1) ^∞ (((−1)^n )/n^x )=(2^(1−x) −1)ξ(x)) =−(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=1) ^∞ (((−1)^(n−1) )/n)−1 =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )−1 =(π^2 /(12))−1 ⇒ S =−3ln(2)−2(−(π^2 /(12)))−(3/4)ξ(3)−3(ln(2)−1)−((π^2 /(12))−1) =−6ln(2)+(π^2 /6)−(3/4)ξ(3) +4 −(π^2 /(12)) =−6ln(2)+(π^2 /(12)) +4−(3/4)ξ(3) ⇒ S =(π^2 /(12)) +4−(3/4)ξ(3)−6ln(2)
∑n=1∞(−1)nn3(n+1)2=limn→+∞∑k=1n(−1)kk3(k+1)2letdecomposeF(x)=1x3(x+1)2⇒F(x)=ax+bx2+cx3+dx+1+e(x+1)2c=limx→0x3F(x)=1e=limx→−1(x+1)2F(x)=−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+4−2a−1⇒2a+4b+2=0⇒a+2b+1=0⇒a+2b=−1F(−2)=1−8=−a2+b4−18+a−1⇒18=−a2−b4+18+1⇒1=−4a−2b+1+8⇒−4a−2b+8=0⇒2a+b=4⇒2(−2b−1)+b=4⇒−4b−2+b=4⇒−3b=6⇒b=−2⇒a=3⇒F(x)=3x−2x2+1x3−3x+1−1(x+1)2⇒S=3∑n=1∞(−1)nn−2∑n=1∞(−1)nn2+∑n=1∞(−1)nn3−3∑n=1∞(−1)nn+1−∑n=1∞(−1)n(n+1)2wehave∑n=1∞(−1)nn=−ln(2)∑n=1∞(−1)nn2=(21−2−1)ξ(2)(formula∑n=1∞(−1)nnx=(21−x−1)ξ(x))=−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=1∞(−1)n−1n−1=ln(2)−1∑n=1∞(−1)n(n+1)2=∑n=2∞(−1)n−1n2=∑n=1∞(−1)n−1n2−1=π212−1⇒S=−3ln(2)−2(−π212)−34ξ(3)−3(ln(2)−1)−(π212−1)=−6ln(2)+π26−34ξ(3)+4−π212=−6ln(2)+π212+4−34ξ(3)⇒S=π212+4−34ξ(3)−6ln(2)

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