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Question Number 32997 by abdo imad last updated on 09/Apr/18
calculate Σ_(n=0) ^∞ ((2n+3)/((n+1)^2 (n+2)^2 ))
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{n}+\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$
Commented by abdo imad last updated on 10/Apr/18
let find a and b / F(n)= ((2n+3)/((n+1)^2 (n+2^ )^2 )) = (a/((n+1)^2 )) +(b/((n+2)^2 )) we have lim_(n→+∞) (n+1)^2 F(n)=0 =a +b ⇒b=−a F(0) =(3/4)=a +(b/4) ⇒ 3 =4a+b =a+b +3a=3a ⇒a=1 and b=−1 ⇒((2n+3)/((n+1)^2 (n+2)^2 )) = (1/((n+1)^2 )) −(1/((n+2)^2 )) let put S_N =Σ_(n=0) ^N ((2n+3)/((n+1)^2 (n+2)^2 )) and S =Σ_(n=0) ^∞ ((2n+3)/((n+1)^2 (n+2)^2 )) we have S=lim_(N→∞) S_N but S_N =Σ_(n=0) ^N ( (1/((n+1)^2 )) −(1/((n+2)^2 )))=Σ_(n=0) ^N (u_n − u_(n+1) ) with u_n = (1/((n+1)^2 )) ⇒ S_N =u_0 −u_1 +u_1 −u_2 +...u_n −u_(n+1) =u_0 −u_(n+1) = 1− (1/((n+2)^2 )) ⇒ lim_(n→+∞) S_N =1 ⇒ S =1 .
$${let}\:{find}\:{a}\:{and}\:{b}\:/\:\:{F}\left({n}\right)=\:\:\frac{\mathrm{2}{n}+\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}^{} \right)^{\mathrm{2}} }\:=\:\frac{{a}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{b}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \:{F}\left({n}\right)=\mathrm{0}\:={a}\:+{b}\:\Rightarrow{b}=−{a} \\ $$$${F}\left(\mathrm{0}\right)\:=\frac{\mathrm{3}}{\mathrm{4}}={a}\:+\frac{{b}}{\mathrm{4}}\:\Rightarrow\:\mathrm{3}\:=\mathrm{4}{a}+{b}\:={a}+{b}\:+\mathrm{3}{a}=\mathrm{3}{a}\:\Rightarrow{a}=\mathrm{1} \\ $$$${and}\:{b}=−\mathrm{1}\:\Rightarrow\frac{\mathrm{2}{n}+\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${let}\:{put}\:{S}_{{N}} \:=\sum_{{n}=\mathrm{0}} ^{{N}} \:\:\frac{\mathrm{2}{n}+\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }\:\:{and} \\ $$$${S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{n}+\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }\:{we}\:{have}\:{S}={lim}_{{N}\rightarrow\infty} {S}_{{N}} \\ $$$${but}\:{S}_{{N}} =\sum_{{n}=\mathrm{0}} ^{{N}} \:\left(\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }\right)=\sum_{{n}=\mathrm{0}} ^{{N}} \left({u}_{{n}} −\:{u}_{{n}+\mathrm{1}} \right) \\ $$$${with}\:{u}_{{n}} =\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{S}_{{N}} \:={u}_{\mathrm{0}} \:−{u}_{\mathrm{1}} \:+{u}_{\mathrm{1}} −{u}_{\mathrm{2}} \:+…{u}_{{n}} −{u}_{{n}+\mathrm{1}} \\ $$$$={u}_{\mathrm{0}} \:−{u}_{{n}+\mathrm{1}} =\:\mathrm{1}−\:\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{N}} =\mathrm{1} \\ $$$$\Rightarrow\:{S}\:=\mathrm{1}\:. \\ $$

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