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Q-158528-P-n-1-n-1-3-1-n-1-3-1-P-n-1-n-1-3-1-3-n-1-3-1-3-P-n-1-n-1-1-n-2-2n-1-n-1-1-n-1-1-n-2-2n-1-n-1-1-P-n-1-

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Question Number 158751 by puissant last updated on 08/Nov/21
Q 158528 P=Π_(n=1) ^∞ ((((n+1)^3 −1)/((n+1)^3 +1))) ⇒ P = Π_(n=1) ^∞ ((((n+1)^3 −1^3 )/((n+1)^3 +1^3 ))) ⇒ P = Π_(n=1) ^∞ {(((n+1−1)(n^2 +2n+1+n+1+1))/((n+1+1)(n^2 +2n+1−n−1+1)))} ⇒ P = Π_(n=1) ^∞ {(n/(n+2))}•Π_(n=1) ^∞ {((n^2 +3n+3)/(n^2 +n+1))} = lim_(n→∞) Π_(k=1) ^n {(k/(k+2))}•lim_(n→∞) Π_(k=1) ^n {((k^2 +3k+3)/(k^2 +k+1))} = lim_(n→∞) {(1/3)•(2/4)•(3/5)•...•(n/(n+2))}×lim_(n→∞) {(7/3)•((13)/7)•...•((n^2 +3n+3)/(n^2 +n+1))} =2lim_(n→∞) {(1/((n+1)(n+2)))}×(1/3)lim_(n→∞) {n^2 +3n+3} =(2/3)lim_(n→∞) {((n^2 +3n+3)/(n^2 +3n+2))} = (2/3)lim_(n→∞) {((1+(3/n)+(3/n^2 ))/(1+(3/n)+(2/n^2 )))} = (2/3). P = Π_(n=1) ^∞ ((((n+1)^3 −1)/((n+1)^3 +1))) = (2/3).. ...............Le puissant...............
$${Q}\:\mathrm{158528} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathbb{P}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}}\right) \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} }\right) \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{\left({n}+\mathrm{1}−\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}+{n}+\mathrm{1}+\mathrm{1}\right)}{\left({n}+\mathrm{1}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}−{n}−\mathrm{1}+\mathrm{1}\right)}\right\} \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{{n}}{{n}+\mathrm{2}}\right\}\bullet\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left\{\frac{{k}}{{k}+\mathrm{2}}\right\}\bullet\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left\{\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{3}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{3}}\bullet\frac{\mathrm{2}}{\mathrm{4}}\bullet\frac{\mathrm{3}}{\mathrm{5}}\bullet…\bullet\frac{{n}}{{n}+\mathrm{2}}\right\}×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{7}}{\mathrm{3}}\bullet\frac{\mathrm{13}}{\mathrm{7}}\bullet…\bullet\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right\} \\ $$$$=\mathrm{2}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right\}×\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\right\}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}+\frac{\mathrm{3}}{{n}}+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{3}}{{n}}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }}\right\}\:=\:\frac{\mathrm{2}}{\mathrm{3}}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………\mathscr{L}{e}\:{puissant}…………… \\ $$

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