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Question Number 64819 by mathmax by abdo last updated on 22/Jul/19
calculate Σ_(n=1) ^∞ (((2n+1)(−1)^n )/(n^3 (n+1)^2 ))
calculate∑n=1∞(2n+1)(−1)nn3(n+1)2
Commented by mathmax by abdo last updated on 24/Jul/19
let S =Σ_(n=1) ^∞ (−1)^n ((2n+1)/(n^3 (n+1)^2 )) letdecompose F(x) =((2x+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)/(−1)) =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) =(3/4) =a+b +1−(a/2) +(1/4) ⇒3 =4a+4b+4−2a +1 ⇒ 3=2a +4b+5 ⇒2a+4b =−2 ⇒a+2b =−1 F(−2) =((−3)/(−8)) =(3/8) =−(a/2) +(b/4) −(1/8) +a+1 ⇒ 3 =−4a +2b−1 +8a+8 ⇒3 =4a +2b+7 ⇒4a+2b =−4 ⇒ 2a+b =−2 ⇒b =−2−2a ⇒a +2(−2−2a) =−1 ⇒ a−4−4a =−1 ⇒−3a =3 ⇒a =−1 ⇒b =−2−2(−1) =0 ⇒ F(x) =−(1/x) +(1/x^3 ) +(1/(x+1)) +(1/((x+1)^2 )) ⇒ S =−Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^3 ) +Σ_(n=1) ^∞ (((−1)^n )/(n+1)) +Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) we know Σ_(n=0) ^∞ (x^n /n) =−ln(1−x) if ∣x∣<1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) also δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) =(2^(1−x) −1)ξ(x) ⇒ 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=2) ^∞ (((−1)^n )/n^2 ) =−(Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +1) =−(2^(1−2) −1)ξ(2)−1 =(1/2)ξ(2)−1 =(1/2)(π^2 /6) −1 =(π^2 /(12)) −1⇒ S =ln(2)−(3/4)ξ(3) +ln(2)−1+(π^2 /(12)) −1 S=2ln(2)−(3/4)ξ(3)+(π^2 /2) −2 .
letS=∑n=1∞(−1)n2n+1n3(n+1)2letdecomposeF(x)=2x+1x3(x+1)2F(x)=ax+bx2+cx3+dx+1+e(x+1)2c=limx→0x3F(x)=1e=limx→−1(x+1)2F(x)=−1−1=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)=34=a+b+1−a2+14⇒3=4a+4b+4−2a+1⇒3=2a+4b+5⇒2a+4b=−2⇒a+2b=−1F(−2)=−3−8=38=−a2+b4−18+a+1⇒3=−4a+2b−1+8a+8⇒3=4a+2b+7⇒4a+2b=−4⇒2a+b=−2⇒b=−2−2a⇒a+2(−2−2a)=−1⇒a−4−4a=−1⇒−3a=3⇒a=−1⇒b=−2−2(−1)=0⇒F(x)=−1x+1x3+1x+1+1(x+1)2⇒S=−∑n=1∞(−1)nn+∑n=1∞(−1)nn3+∑n=1∞(−1)nn+1+∑n=1∞(−1)n(n+1)2weknow∑n=0∞xnn=−ln(1−x)if∣x∣<1⇒∑n=1∞(−1)nn=−ln(2)alsoδ(x)=∑n=1∞(−1)nnx=(21−x−1)ξ(x)⇒∑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=2∞(−1)nn2=−(∑n=1∞(−1)nn2+1)=−(21−2−1)ξ(2)−1=12ξ(2)−1=12π26−1=π212−1⇒S=ln(2)−34ξ(3)+ln(2)−1+π212−1S=2ln(2)−34ξ(3)+π22−2.
Commented by mathmax by abdo last updated on 24/Jul/19
S =2ln(2)−(3/4)ξ(3)+(π^2 /(12)) −2 .
S=2ln(2)−34ξ(3)+π212−2.

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