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Question Number 64820 by mathmax by abdo last updated on 22/Jul/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)(n+2)))
calculate∑n=1∞(−1)nn2(n+1)(n+2)
Commented by mathmax by abdo last updated on 22/Jul/19
let decompose F(x) =(1/(x^2 (x+1)(x+2))) ⇒ F(x) =(a/x) +(b/x^2 ) +(c/(x+1)) +(d/(x+2)) b=lim_(x→0) x^2 F(x) =(1/2) c =lim_(x→−1) (x+1)F(x) =1 d =lim_(x→−2) (x+2)F(x) =−(1/4) ⇒F(x)=(a/x) +(1/(2x^2 )) +(1/(x+1)) −(1/(4(x+2))) lim_(x→+∞) xF(x) =0 =a+1 ⇒a =−1 ⇒ F(x) =−(1/x) +(1/(2x^2 )) +(1/(x+1)) −(1/(4(x+2))) ⇒ S =−Σ_(n=1) ^∞ (((−1)^n )/n) +(1/2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −(1/4)Σ_(n=1) ^∞ (((−1)^n )/(n+2)) we know Σ_(n=1) ^∞ (x^n /n) =−ln(1−x) if ∣x∣<1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln2 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) =−(−ln2 +1) =ln2−1 Σ_(n=1) ^∞ (((−1)^n )/(n+2)) =Σ_(n=3) ^(+∞) (((−1)^(n−2) )/n) =Σ_(n=3) ^∞ (((−1)^n )/n) =Σ_(n=1) ^∞ (((−1)^n )/n) −(−1+(1/2)) =−ln2 +(1/2) let δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) with x>1 we have proved that 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)) ⇒ S =ln(2)−(π^2 /(24)) +ln2−1 −(1/4)(−ln2 +(1/2)) =2ln2 −(π^2 /(24)) −1 +(1/4)ln2 −(1/8) =(9/4)ln2 −(9/8) −(π^2 /(24))
letdecomposeF(x)=1x2(x+1)(x+2)⇒F(x)=ax+bx2+cx+1+dx+2b=limx→0x2F(x)=12c=limx→−1(x+1)F(x)=1d=limx→−2(x+2)F(x)=−14⇒F(x)=ax+12x2+1x+1−14(x+2)limx→+∞xF(x)=0=a+1⇒a=−1⇒F(x)=−1x+12x2+1x+1−14(x+2)⇒S=−∑n=1∞(−1)nn+12∑n=1∞(−1)nn2+∑n=1∞(−1)nn+1−14∑n=1∞(−1)nn+2weknow∑n=1∞xnn=−ln(1−x)if∣x∣<1⇒∑n=1∞(−1)nn=−ln2∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=−∑n=2∞(−1)nn=−(∑n=1∞(−1)nn+1)=−(−ln2+1)=ln2−1∑n=1∞(−1)nn+2=∑n=3+∞(−1)n−2n=∑n=3∞(−1)nn=∑n=1∞(−1)nn−(−1+12)=−ln2+12letδ(x)=∑n=1∞(−1)nnxwithx>1wehaveprovedthatδ(x)=(21−x−1)ξ(x)⇒∑n=1∞(−1)nn2=δ(2)=(2−1−1)ξ(2)=−12π26=−π212⇒S=ln(2)−π224+ln2−1−14(−ln2+12)=2ln2−π224−1+14ln2−18=94ln2−98−π224

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