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Question Number 38520 by math khazana by abdo last updated on 26/Jun/18
find the value of Σ_(n=2) ^∞ ((3n^2 +1)/((n−1)^3 (n+1)^3 ))
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by abdo mathsup 649 cc last updated on 28/Jun/18
let S_n =Σ_(k=2) ^n ((3k^2 +1)/((k−1)^3 )(k+1)^3 )) we have (k+1)^3 −(k−1)^3 =(k+1−k+1)( (k+1)^2 +(k−1)(k+1) +(k−1)^2 ) =2{ k^2 +2k+1 +k^2 −1 +k^(2 ) −2k+1} =2{3k^2 +1}⇒ S_n =(1/2) Σ_(k=2) ^n (((n+1)^3 −(n−1)^3 )/((n+1)^3 (n−1)^3 )) =(1/2) Σ_(k=2) ^n (1/((n−1)^3 )) −(1/2) Σ_(k=2) ^n (1/((n+1)^3 )) =(1/2)Σ_(k=1) ^(n−1) (1/k^3 ) −(1/2) Σ_(k=3) ^(n+1) (1/k^3 ) =(1/2) ξ_(n−1) (3) −(1/2){ ξ_(n+1) (3)−1 −(1/8)} =(1/2){ ξ_(n−1) (3)−ξ_(n+1) (3)} +(9/(16)) but lim_(n→∞) ξ_(n−1) (3)=ξ(3) lim_(n→∞) ξ_(n+1) (3)=ξ(3) ⇒lim_(n→+∞) S_n =(9/(16)) .
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{3}{k}^{\mathrm{2}} \:+\mathrm{1}}{\left.\left({k}−\mathrm{1}\right)^{\mathrm{3}} \right)\left({k}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${we}\:{have}\:\left({k}+\mathrm{1}\right)^{\mathrm{3}} −\left({k}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$=\left({k}+\mathrm{1}−{k}+\mathrm{1}\right)\left(\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:+\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)\:+\left({k}−\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$$=\mathrm{2}\left\{\:\:{k}^{\mathrm{2}} \:+\mathrm{2}{k}+\mathrm{1}\:+{k}^{\mathrm{2}} −\mathrm{1}\:+{k}^{\mathrm{2}\:} −\mathrm{2}{k}+\mathrm{1}\right\} \\ $$$$=\mathrm{2}\left\{\mathrm{3}{k}^{\mathrm{2}} \:+\mathrm{1}\right\}\Rightarrow \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} \:−\left({n}−\mathrm{1}\right)^{\mathrm{3}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} \left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\xi_{{n}−\mathrm{1}} \left(\mathrm{3}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{8}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\xi_{{n}−\mathrm{1}} \left(\mathrm{3}\right)−\xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)\right\}\:+\frac{\mathrm{9}}{\mathrm{16}}\:{but} \\ $$$${lim}_{{n}\rightarrow\infty} \xi_{{n}−\mathrm{1}} \left(\mathrm{3}\right)=\xi\left(\mathrm{3}\right) \\ $$$${lim}_{{n}\rightarrow\infty} \xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)=\xi\left(\mathrm{3}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {S}_{{n}} =\frac{\mathrm{9}}{\mathrm{16}}\:. \\ $$$$ \\ $$$$ \\ $$

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