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Question Number 28981 by abdo imad last updated on 02/Feb/18

find the values of Π_(n=2) ^∞ (1−(2/(n(n+1)))) .

$${find}\:{the}\:{values}\:{of}\:\prod_{{n}=\mathrm{2}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}\right)\:. \\ $$

Commented by abdo imad last updated on 03/Feb/18

Π_(n=2) ^∞ (1−(2/(n(n+1))))=lim_(n→+∞) S_n with S_n = Π_(k=2) ^n (1−(2/(k(k+1))))=Π_(k=2) ^n ( ((k^2 +k−2)/(k(k+1)))) but k^2 +k−2=k^2 −k+ 2k−2=k(k−1) +2(k−1) =(k−1)(k+2) S_n = Π_(k=2) ^n ((k−1)/k) .((k+2)/(k+1))=Π_(k=1) ^(n−1) (k/(k+1)) .Π_(k=3) ^(n+1) ((k+1)/k) =(1/2).(2/3)Π_(k=3) ^(n−1) (k/(k+1)).Π_(k=3) ^(n−1) ((k+1)/k).((n+1)/n).((n+2)/(n+1)) S_n = (1/3).((n+2)/n) ⇒lim_(n→+ ∞) S_n =(1/3) so Π_(n=2) ^∞ (1−(2/(n(n+1))))= (1/3) .

$$\prod_{{n}=\mathrm{2}} ^{\infty} \:\left(\mathrm{1}−\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}\right)={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:{with}\: \\ $$$${S}_{{n}} =\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\mathrm{1}−\frac{\mathrm{2}}{{k}\left({k}+\mathrm{1}\right)}\right)=\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\:\frac{{k}^{\mathrm{2}} \:+{k}−\mathrm{2}}{{k}\left({k}+\mathrm{1}\right)}\right)\:{but} \\ $$$${k}^{\mathrm{2}} \:+{k}−\mathrm{2}={k}^{\mathrm{2}} \:−{k}+\:\mathrm{2}{k}−\mathrm{2}={k}\left({k}−\mathrm{1}\right)\:+\mathrm{2}\left({k}−\mathrm{1}\right) \\ $$$$=\left({k}−\mathrm{1}\right)\left({k}+\mathrm{2}\right) \\ $$$${S}_{{n}} =\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{{k}−\mathrm{1}}{{k}}\:.\frac{{k}+\mathrm{2}}{{k}+\mathrm{1}}=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{{k}}{{k}+\mathrm{1}}\:.\prod_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{{k}+\mathrm{1}}{{k}}\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}\prod_{{k}=\mathrm{3}} ^{{n}−\mathrm{1}} \:\frac{{k}}{{k}+\mathrm{1}}.\prod_{{k}=\mathrm{3}} ^{{n}−\mathrm{1}} \:\frac{{k}+\mathrm{1}}{{k}}.\frac{{n}+\mathrm{1}}{{n}}.\frac{{n}+\mathrm{2}}{{n}+\mathrm{1}} \\ $$$${S}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{3}}.\frac{{n}+\mathrm{2}}{{n}}\:\Rightarrow{lim}_{{n}\rightarrow+\:\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\:{so} \\ $$$$\prod_{{n}=\mathrm{2}} ^{\infty} \:\left(\mathrm{1}−\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}\right)=\:\frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$$$ \\ $$

Answered by iv@0uja last updated on 03/Feb/18

1−(2/(n(n+1)))=((n^2 +n−2)/(n(n+1))) =(((n−1)(n+2))/(n(n+1))) Π_(n=2) ^m (((n−1)(n+2))/(n(n+1))) =((1∙4)/(2∙3))×((2∙5)/(3∙4))×((3∙6)/(4∙5))×...×(((m−1)(m+2))/(m(m+1))) =(1/3)×((m+2)/m) lim_(m→∞) ((m+2)/m)=1 Π_(n=2) ^∞ (((n−1)(n+2))/(n(n+1)))=(1/3)×1=(1/3)

$$\mathrm{1}−\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}=\frac{{n}^{\mathrm{2}} +{n}−\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{{m}} {\prod}}\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}\centerdot\mathrm{4}}{\mathrm{2}\centerdot\mathrm{3}}×\frac{\mathrm{2}\centerdot\mathrm{5}}{\mathrm{3}\centerdot\mathrm{4}}×\frac{\mathrm{3}\centerdot\mathrm{6}}{\mathrm{4}\centerdot\mathrm{5}}×...×\frac{\left({m}−\mathrm{1}\right)\left({m}+\mathrm{2}\right)}{{m}\left({m}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{m}+\mathrm{2}}{{m}} \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\frac{{m}+\mathrm{2}}{{m}}=\mathrm{1} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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