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Question Number 40897 by abdo.msup.com last updated on 28/Jul/18
calculate Σ_(n=1) ^∞ (n/((n+1)^2 (n+2)))
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)} \\ $$
Commented by math khazana by abdo last updated on 03/Aug/18
let decompose F(x)=(x/((x+2)(x+1)^2 )) F(x)=(a/(x+2)) +(b/(x+1)) +(c/((x+1)^2 )) a=lim_(x→−2) (x+2)F(x)= −2 c=lim_(x→−1) (x+1)^2 F(x)=−1 ⇒ F(x)=((−2)/(x+2)) +(b/(x+1)) −(1/((x+1)^2 )) F(0)=0=−1 +b −1=b−2⇒b=2 ⇒ F(x)=((−2)/(x+2)) +(2/(x+1)) −(1/((x+1)^2 )) S_n =Σ_(k=1) ^n (k/((k+2)(k+1)^2 )) =Σ_(k=1) ^n F(k) =−2Σ_(k=1) ^n (1/(k+2)) +2Σ_(k=1) ^n (1/(k+1)) −Σ_(k=1) ^n (1/((k+1)^2 )) but Σ_(k=1) ^n (1/(k+2)) = Σ_(k=3) ^(n+2) (1/k)=H_(n+2) −(3/2) Σ_(k=1) ^n (1/(k+1)) =Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 Σ_(k=1) ^n (1/((k+1)^2 )) =Σ_(k=2) ^(n+1) (1/k^2 ) =ξ_(n+1) (2)−1 ⇒ S_n =−2{H_(n+2) −(3/2)} +2{H_(n+1) −1} −ξ_(n+1) (2) +1 =−2(H_(n+2) −H_(n+1) ) +2 −ξ_(n+1) (2) but lim_(n→+∞) H_(n+2) −H_(n+1) =0 and lim_(n→+∞) ξ_(n+1) (2)=(π^2 /6) ⇒ S=lim_(n→+∞) S_n =2−(π^2 /6) .
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{{x}}{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{2}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)=\:−\mathrm{2} \\ $$$${c}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−\mathrm{2}}{{x}+\mathrm{2}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}=−\mathrm{1}\:+{b}\:−\mathrm{1}={b}−\mathrm{2}\Rightarrow{b}=\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−\mathrm{2}}{{x}+\mathrm{2}}\:+\frac{\mathrm{2}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{{k}}{\left({k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right) \\ $$$$=−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\:\frac{\mathrm{1}}{{k}}={H}_{{n}+\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} −\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1}\:\Rightarrow \\ $$$${S}_{{n}} =−\mathrm{2}\left\{{H}_{{n}+\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\right\}\:+\mathrm{2}\left\{{H}_{{n}+\mathrm{1}} −\mathrm{1}\right\}\:−\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:+\mathrm{1} \\ $$$$=−\mathrm{2}\left({H}_{{n}+\mathrm{2}} −{H}_{{n}+\mathrm{1}} \right)\:+\mathrm{2}\:−\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:{but} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{H}_{{n}+\mathrm{2}} −{H}_{{n}+\mathrm{1}} =\mathrm{0}\:{and}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow \\ $$$${S}={lim}_{{n}\rightarrow+\infty} {S}_{{n}} =\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$

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