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Question Number 65674 by mathmax by abdo last updated on 01/Aug/19

find the value of Σ_(n=2) ^∞ (n/((n+1)^2 (n−1)^3 ))

$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Commented by mathmax by abdo last updated on 04/Aug/19

let S =Σ_(n=2) ^∞ (n/((n+1)^2 (n−1)^3 )) ⇒S =Σ_(n=1) ^∞ ((n+1)/((n+2)^2 n^3 )) let decompose F(x) =((x+1)/(x^3 (x+2)^2 )) ⇒ F(x) =(a/x) +(b/x^2 ) +(c/x^3 ) +(d/(x+2)) +(e/((x+2)^2 )) c =lim_(x→0) x^3 F(x) =(1/4) e =lim_(x→−2) (x+2)^2 F(x) =((−1)/(−8))=(1/8) ⇒ F(x) =(a/x) +(b/x^2 ) +(1/(4x^3 )) +(d/(x+2)) +(1/(8(x+2)^2 )) lim_(x→+∞) xF(x) =0 =a+d ⇒d =−a ⇒ F(x) =(a/x) +(b/x^2 ) +(1/(4x^3 ))−(a/(x+2)) +(1/(8(x+2)^2 )) F(1) =(2/9) =a+b+(1/4)−(a/3) +(1/(72)) =(2/3)a +b +(1/(72)) ⇒ 2 =6a+9b +(1/8) ⇒6a+9b =2−(1/8) =((15)/8) F(−1) =0 =−a+b−(1/4)−a +(1/8) =−2a+b−(1/8) ⇒ −2a+b =(1/8) ⇒b =2a+(1/8) ⇒6a+9(2a+(1/8))=((15)/8) ⇒ 24a +(9/8) =((15)/8) ⇒24a=(6/8) =(3/4) ⇒a =(3/(4.8.3)) =(1/(32)) b =(1/(16))+(1/8) =(3/(16)) ⇒F(x) =(1/(32x)) +(3/(16x^2 )) +(1/(4x^3 )) −(1/(32(x+2))) +(1/(8(x+2)^2 )) S_n =Σ_(k=1) ^n F(k) =(1/(32))Σ_(k=1) ^n (1/k) +(3/(16)) Σ_(k=1) ^n (1/k^2 ) +(1/4)Σ_(k=1) ^n (1/k^3 ) −(1/(32))Σ_(k=1) ^n (1/(k+2)) +(1/8)Σ_(k=1) ^n (1/((x+2)^2 )) Σ_(k=1) ^n (1/(k+2)) =Σ_(k=3) ^(n+2) (1/k) =Σ_(k=1) ^n (1/k)−(3/2) +(1/(n+1)) +(1/(n+2)) ⇒ (1/(32))Σ_(k=1) ^n (1/k) −(1/(32))Σ_(k=1) ^n (1/(k+2)) =−(1/(32))(−(3/2) +(1/(n+1)) +(1/(n+2)))→(3/(64)) ⇒ Σ_(k=1) ^n (1/((x+2)^2 )) =Σ_(k=3) ^(n+2) (1/k^2 ) =ξ_n (2)+(1/((n+1)^2 )) +(1/((n+2)^2 ))−1−(1/4) →ξ(2)−(5/4) ⇒ S =lim_(n→+∞) S_n =(3/(64)) +(3/(16))ξ(2) +(1/4)ξ(3) +(1/8){ξ(2)−(5/4)} =(3/(64)) +(5/(16)) (π^2 /6) +(1/4)ξ(3)−(5/(32)) =−(7/(64)) +((5π^2 )/(96)) +(1/4)ξ(3)

$${let}\:{S}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} {n}^{\mathrm{3}} } \\ $$$${let}\:{decompose}\:{F}\left({x}\right)\:=\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}^{\mathrm{3}} }\:+\frac{{d}}{{x}+\mathrm{2}}\:+\frac{{e}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${c}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{3}} \:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${e}\:={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} {F}\left({x}\right)\:=\frac{−\mathrm{1}}{−\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{3}} }\:+\frac{{d}}{{x}+\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:={a}+{d}\:\Rightarrow{d}\:=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{3}} }−\frac{{a}}{{x}+\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{2}}{\mathrm{9}}\:={a}+{b}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{{a}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{72}}\:=\frac{\mathrm{2}}{\mathrm{3}}{a}\:+{b}\:+\frac{\mathrm{1}}{\mathrm{72}}\:\Rightarrow \\ $$$$\mathrm{2}\:=\mathrm{6}{a}+\mathrm{9}{b}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow\mathrm{6}{a}+\mathrm{9}{b}\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{8}}\:=\frac{\mathrm{15}}{\mathrm{8}} \\ $$$${F}\left(−\mathrm{1}\right)\:=\mathrm{0}\:=−{a}+{b}−\frac{\mathrm{1}}{\mathrm{4}}−{a}\:+\frac{\mathrm{1}}{\mathrm{8}}\:=−\mathrm{2}{a}+{b}−\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$$−\mathrm{2}{a}+{b}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow{b}\:=\mathrm{2}{a}+\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow\mathrm{6}{a}+\mathrm{9}\left(\mathrm{2}{a}+\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\mathrm{15}}{\mathrm{8}}\:\Rightarrow \\ $$$$\mathrm{24}{a}\:+\frac{\mathrm{9}}{\mathrm{8}}\:=\frac{\mathrm{15}}{\mathrm{8}}\:\Rightarrow\mathrm{24}{a}=\frac{\mathrm{6}}{\mathrm{8}}\:=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow{a}\:=\frac{\mathrm{3}}{\mathrm{4}.\mathrm{8}.\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{32}} \\ $$$${b}\:=\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{8}}\:=\frac{\mathrm{3}}{\mathrm{16}}\:\Rightarrow{F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{32}{x}}\:+\frac{\mathrm{3}}{\mathrm{16}{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{32}\left({x}+\mathrm{2}\right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)\:=\frac{\mathrm{1}}{\mathrm{32}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:+\frac{\mathrm{3}}{\mathrm{16}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{32}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{{k}}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}−\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{32}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{1}}{\mathrm{32}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{32}}\left(−\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)\rightarrow\frac{\mathrm{3}}{\mathrm{64}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}} \left(\mathrm{2}\right)+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\rightarrow\xi\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow\:{S}\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{64}}\:+\frac{\mathrm{3}}{\mathrm{16}}\xi\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\xi\left(\mathrm{3}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\xi\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{64}}\:+\frac{\mathrm{5}}{\mathrm{16}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−\frac{\mathrm{5}}{\mathrm{32}}\:=−\frac{\mathrm{7}}{\mathrm{64}}\:+\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{96}}\:+\frac{\mathrm{1}}{\mathrm{4}}\xi\left(\mathrm{3}\right) \\ $$

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