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Question Number 80792 by john santu last updated on 06/Feb/20

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}[\frac{2n}{2n-1}.\frac{2n}{2n+1}]=?$$

Answered by mind is power last updated on 06/Feb/20

$$ln\left(\prod _{n⩾1}\frac{2n.2n}{(2n-1)(2n+1)}\right)=\sum _{n⩾1_{}}ln\left(\frac{1}{(1-\frac{1}{2n})(1+\frac{1}{2n})}\right)$$$$=-\sum _{n⩾1}ln(1-\frac{1}{4n^2})\sim \frac{1}{4n^2}\Rightarrow oursumexist\Rightarrow productexist$$$$\underset{n=1}{\prod ^{+\mathrm{\infty }}}\frac{4n^2}{(2n-1)(2n+1)}=\frac{1}{\mathrm{\Pi }\frac{(2n-1)(2n+1)}{4n^2}}$$$$=\frac{1}{\prod _{n⩾1}(1-\frac{1}{4n^2})}$$$$wehaveeulerformula\Rightarrow sin\left(x\right)=x\prod _{k⩾1}(1-\frac{x^2}{k^2.{\pi }^2})$$$$x=\frac{\pi }{2}\Rightarrow 1=\frac{\pi }{2}\prod _{k⩾1}(1-\frac{1}{4k^2})\Rightarrow \frac{1}{\prod _{n⩾1}(1-\frac{1}{4n^2})}=\frac{\pi }{2}$$$$\Rightarrow \underset{n=1}{\prod ^{+\mathrm{\infty }}}\frac{2n}{2n-1}.\frac{2n}{2n+1}=\frac{\pi }{2}$$$$$$

Commented by jagoll last updated on 07/Feb/20

$$thankyousir$$

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