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

$$findthevaluesof\prod _{n=2}^{\mathrm{\infty }}(1-\frac{2}{n(n+1)}).$$

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

$$\prod _{n=2}^{\mathrm{\infty }}(1-\frac{2}{n(n+1)})={lim}_{n\to +\mathrm{\infty }}{S}_{n}with$$$$S_n=\prod _{k=2}^n(1-\frac{2}{k(k+1)})=\prod _{k=2}^n\left(\frac{k^2+k-2}{k(k+1)}\right)but$$$$k^2+k-2=k^2-k+2k-2=k(k-1)+2(k-1)$$$$=(k-1)(k+2)$$$$S_n=\prod _{k=2}^n\frac{k-1}{k}.\frac{k+2}{k+1}=\prod _{k=1}^{n-1}\frac{k}{k+1}.\prod _{k=3}^{n+1}\frac{k+1}{k}$$$$=\frac{1}{2}.\frac{2}{3}\prod _{k=3}^{n-1}\frac{k}{k+1}.\prod _{k=3}^{n-1}\frac{k+1}{k}.\frac{n+1}{n}.\frac{n+2}{n+1}$$$$S_n=\frac{1}{3}.\frac{n+2}{n}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{1}{3}so$$$$\prod _{n=2}^{\mathrm{\infty }}(1-\frac{2}{n(n+1)})=\frac{1}{3}.$$$$$$

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

$$1-\frac{2}{n(n+1)}=\frac{n^2+n-2}{n(n+1)}$$$$=\frac{(n-1)(n+2)}{n(n+1)}$$$$\underset{n=2}{\prod ^m}\frac{(n-1)(n+2)}{n(n+1)}$$$$=\frac{1\cdot 4}{2\cdot 3}\times \frac{2\cdot 5}{3\cdot 4}\times \frac{3\cdot 6}{4\cdot 5}\times ...\times \frac{(m-1)(m+2)}{m(m+1)}$$$$=\frac{1}{3}\times \frac{m+2}{m}$$$$\underset{m\to \mathrm{\infty }}{\lim }\frac{m+2}{m}=1$$$$\underset{n=2}{\prod ^{\mathrm{\infty }}}\frac{(n-1)(n+2)}{n(n+1)}=\frac{1}{3}\times 1=\frac{1}{3}$$

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