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Question Number 156924 by cortano last updated on 17/Oct/21

Π_(k=2) ^∞ (((k^3 −1)/(k^3 +1))) =?

$$\:\underset{{k}=\mathrm{2}} {\overset{\infty} {\prod}}\:\left(\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}} +\mathrm{1}}\right)\:=? \\ $$

Answered by puissant last updated on 17/Oct/21

P=Π_(k=2) ^∞ (((k^3 −1)/(k^3 +1))) = lim_(n→∞) Π_(k=2) ^n (((k^3 −1)/(k^3 +1))) = lim_(n→∞) Π_(k=2) ^n ((((k−1)(k^2 +k+1))/((k+1)(k^2 −k+1)))) =lim Π_(k=2) ^n (((k−1)/(k+1)))×lim Π_(k=2) ^n (((k^2 +k+1)/(k^2 −k+1)))_(n→∞) =lim_(n→∞) {(1/3)•(2/4)•(3/5)•...•((n−2)/n)•((n−1)/(n+1))}×lim_(n→∞) {(7/3)•((13)/7)•((21)/(13))•...•((n^2 +n+1)/(n^2 −n+1))} =lim_(n→∞) {(2/(n(n+1)))}×lim_(n→∞) {((n^2 +n+1)/3)}= (2/3)lim_(n→∞) {((n^2 (1+(1/n)+(1/n^2 )))/(n^2 (1+(1/n))))}=(2/3).. ∴∵ P = Π_(k=2) ^∞ (((k^3 −1)/(k^3 +1)))=(2/3)... ......................Le puissant..................

$${P}=\underset{{k}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}} +\mathrm{1}}\right)\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}} +\mathrm{1}}\right) \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\frac{\left({k}−\mathrm{1}\right)\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)}\right)\:\underset{{n}\rightarrow\infty} {=\mathrm{lim}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)×\mathrm{lim}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\frac{{k}^{\mathrm{2}} +{k}+\mathrm{1}}{{k}^{\mathrm{2}} −{k}+\mathrm{1}}\right)} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{3}}\bullet\frac{\mathrm{2}}{\mathrm{4}}\bullet\frac{\mathrm{3}}{\mathrm{5}}\bullet...\bullet\frac{{n}−\mathrm{2}}{{n}}\bullet\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}}\right\}×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{7}}{\mathrm{3}}\bullet\frac{\mathrm{13}}{\mathrm{7}}\bullet\frac{\mathrm{21}}{\mathrm{13}}\bullet...\bullet\frac{{n}^{\mathrm{2}} +{n}+\mathrm{1}}{{n}^{\mathrm{2}} −{n}+\mathrm{1}}\right\} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}\right\}×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{{n}^{\mathrm{2}} +{n}+\mathrm{1}}{\mathrm{3}}\right\}=\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}{{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\right\}=\frac{\mathrm{2}}{\mathrm{3}}.. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:{P}\:=\:\underset{{k}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}} +\mathrm{1}}\right)=\frac{\mathrm{2}}{\mathrm{3}}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......................\mathscr{L}{e}\:{puissant}.................. \\ $$

Commented by mnjuly1970 last updated on 17/Oct/21

very nice sir puissant..

$${very}\:{nice}\:{sir}\:{puissant}.. \\ $$

Commented by puissant last updated on 17/Oct/21

thanks sir Mnjuly...

$${thanks}\:{sir}\:{Mnjuly}... \\ $$

Commented by Tawa11 last updated on 17/Oct/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Commented by cortano last updated on 18/Oct/21

yes gave kudos

$${yes}\:\:{gave}\:{kudos} \\ $$

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