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Question Number 41051 by turbo msup by abdo last updated on 01/Aug/18
find the value of Σ_(n=1) ^∞ ((2n+3)/(n^2 (n+1)^2 ))
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{n}+\mathrm{3}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by math khazana by abdo last updated on 01/Aug/18
let S =Σ_(n=1) ^∞ ((2n+3)/(n^2 (n+1)^2 )) and S_n =Σ_(k=1) ^n ((2k+3)/(k^2 (k+1)^2 )) we have lim_(n→+∞) S_n =S let decompose F(x) = ((2x+3)/(x^2 (x+1)^2 )) F(x)= (a/x) +(b/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) b=lim_(x→0) x^2 F(x) =3 d =lim_(x→−1) (x+1)^2 F(x)=1 ⇒ F(x)=(a/x) +(3/x^2 ) +(c/(x+1)) +(1/((x+1)^2 )) lim_(x→+∞) xF(x) =0=a+c ⇒c=−a ⇒ F(x)= (a/x) +(3/x^2 ) −(a/(x+1)) +(1/((x+1)^2 )) F(1) = (5/4) =a+3−(a/2) +(1/4) =(a/2) +((13)/4) ⇒5 =2a +13 ⇒2a =−8 ⇒a =−4 ⇒ F(x) =−(4/x) +(3/x^2 ) +(4/(x+1)) +(1/((x+1)^2 )) ⇒ S_n =Σ_(k=1) ^n F(k) =−4Σ_(k=1) ^n (1/k) +3Σ_(k=1) ^n (1/k^2 ) + 4 Σ_(k=1) ^n (1/(k+1)) +Σ_(k=1) ^n (1/((k+1)^2 )) =−4 H_n + 3 ξ_n (2) +4( H_(n+1) −1) +ξ_(n+1) (2) −1 =4(H_(n+1) −H_n ) +3ξ_n (2) +ξ_(n+1) (2)−5 lim_(n→+∞) S_n =4×0 +3(π^2 /6) +(π^2 /6) −5 =((2π^2 )/3) −5 ⇒ S =((2π^2 )/3) −5 .
$${let}\:{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{n}+\mathrm{3}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{and}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}+\mathrm{3}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} ={S}\:\:{let}\:{decompose} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)\:=\mathrm{3} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:−\frac{{a}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{5}}{\mathrm{4}}\:={a}+\mathrm{3}−\frac{{a}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{{a}}{\mathrm{2}}\:+\frac{\mathrm{13}}{\mathrm{4}}\: \\ $$$$\Rightarrow\mathrm{5}\:=\mathrm{2}{a}\:+\mathrm{13}\:\Rightarrow\mathrm{2}{a}\:=−\mathrm{8}\:\Rightarrow{a}\:=−\mathrm{4}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{4}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)\:=−\mathrm{4}\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}\:+\mathrm{3}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$+\:\mathrm{4}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\mathrm{4}\:{H}_{{n}} \:+\:\mathrm{3}\:\xi_{{n}} \left(\mathrm{2}\right)\:+\mathrm{4}\left(\:{H}_{{n}+\mathrm{1}} −\mathrm{1}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:−\mathrm{1} \\ $$$$=\mathrm{4}\left({H}_{{n}+\mathrm{1}} −{H}_{{n}} \right)\:+\mathrm{3}\xi_{{n}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{5} \\ $$$${lim}_{{n}\rightarrow+\infty} {S}_{{n}} =\mathrm{4}×\mathrm{0}\:+\mathrm{3}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\mathrm{5} \\ $$$$=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{5}\:\Rightarrow\:{S}\:=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{5}\:. \\ $$

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