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Question Number 134067 by bramlexs22 last updated on 27/Feb/21

 Calculate lim_(n→∞)  Π_(k=2) ^n  ((k^3 −1)/(k^3 +1)) = ?

$$\:\mathrm{Calculate}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{n}} {\prod}}\:\frac{\mathrm{k}^{\mathrm{3}} −\mathrm{1}}{\mathrm{k}^{\mathrm{3}} +\mathrm{1}}\:=\:? \\ $$

Answered by john_santu last updated on 27/Feb/21

 since ((k^3 −1)/(k^3 +1)) = (((k−1)(k^2 +k+1))/((k+1)(k^2 −k+1)))=(((k−1)((k+1)^2 −(k+1)+1))/((k+1)(k^2 −k+1)))  we get Π_(k=2) ^n  ((k^3 −1)/(k^3 +1)) = (2/3).((n^2 +n+1)/(n^2 +n)) →_(n→∞)  (2/3).

$$\:{since}\:\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}} +\mathrm{1}}\:=\:\frac{\left({k}−\mathrm{1}\right)\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)}=\frac{\left({k}−\mathrm{1}\right)\left(\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\left({k}+\mathrm{1}\right)+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)} \\ $$$${we}\:{get}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\:\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}} +\mathrm{1}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}.\frac{{n}^{\mathrm{2}} +{n}+\mathrm{1}}{{n}^{\mathrm{2}} +{n}}\:\underset{{n}\rightarrow\infty} {\rightarrow}\:\frac{\mathrm{2}}{\mathrm{3}}. \\ $$$$ \\ $$

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