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Question Number 159681 by mnjuly1970 last updated on 20/Nov/21

$$$$$$provethat:$$$$\mathrm{P}=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{1}{n(n+2)})\stackrel{?}{=}\frac{-\sqrt{2}sin\left(\pi \sqrt{2}\right)}{\pi }$$$$m.n$$

Answered by mindispower last updated on 21/Nov/21

$$\underset{n=1}{\prod ^m}(1-\frac{1}{n(n+2)})=\underset{n=1}{\prod ^m}\frac{(n+1-\sqrt{2})(n+1+\sqrt{2})}{n(n+2)}$$$$=\frac{2.\mathrm{\Gamma }(m+2-\sqrt{2})\mathrm{\Gamma }(m+2+\sqrt{2})}{\mathrm{\Gamma }(m+1)\mathrm{\Gamma }(m+3)\mathrm{\Gamma }(2+\sqrt{2})\mathrm{\Gamma }(2-\sqrt{2})}$$$$\mathrm{\Gamma }(1+\sqrt{2})=\sqrt{2}\mathrm{\Gamma }\left(\sqrt{2}\right)$$$$\iff \frac{2\mathrm{\Gamma }(m+2-\sqrt{2})\mathrm{\Gamma }(m+2+\sqrt{2})}{m!.(m+2)!\mathrm{\Gamma }\left(\sqrt{2}\right)\mathrm{\Gamma }(1-\sqrt{2}).\sqrt{2}.(1-\sqrt{2})(1+\sqrt{2})}$$$$=-\frac{\sqrt{2}}{\frac{\pi }{sin\left(\pi \sqrt{2}\right)}}.\frac{\mathrm{\Gamma }(m+2-\sqrt{2})\mathrm{\Gamma }(m+2+\sqrt{2})}{m!.(m+2)!}$$$$\mathrm{\Gamma }(z+a)\sim \sqrt{2\pi .}{z}^{a-\frac{1}{2}+z}{e}^{-z}$$$$z=m+2$$$$\mathrm{\Gamma }(m+2-\sqrt{2})\mathrm{\Gamma }(m+2+\sqrt{2})\sim 2\pi .{(m+2)}^{2m+3}{e}^{-2(m+2)}$$$$\mathrm{\Gamma }(m+3)\mathrm{\Gamma }(m+1)\sim 2\pi {(m+3)}^{m+\frac{5}{2}}{(m+1)}^{m+\frac{1}{2}}{e}^{-2m-4}$$$$\underset{m\to \mathrm{\infty }}{\lim }\frac{\mathrm{\Gamma }(m+2-\sqrt{2})\mathrm{\Gamma }(m+2+\sqrt{2})}{m!.(m+2)!}$$$$=\underset{m\to \mathrm{\infty }}{\lim }\frac{{(m+2)}^{2m+3}}{{(m+1)}^{m+\frac{1}{2}}{(m+3)}^{m+\frac{5}{2}}}$$$$=\underset{m\to \mathrm{\infty }}{\lim }{\left(\frac{{(m+2)}^2}{m^2+4m+3}\right)}^m=\underset{m\to \mathrm{\infty }}{\lim }.{(1+\frac{1}{m^2+4m+3})}^m=1$$$$\Rightarrow$$$$\prod _{n⩾1}(1-\frac{1}{n(n+2)})=\underset{m\to \mathrm{\infty }}{\lim }\underset{n=1}{\prod ^m}(1-\frac{1}{n(n+2)})=-\sqrt{2}.\frac{sin\left(\pi \sqrt{2}\right)}{\pi }.\underset{m\to \mathrm{\infty }}{\lim }.\frac{\mathrm{\Gamma }(m+2-\sqrt{2})\mathrm{\Gamma }(m+2+\sqrt{2})}{\mathrm{\Gamma }(m+1)\mathrm{\Gamma }(m+3)}=-\sqrt{2}.\frac{sin\left(\pi \sqrt{2}\right)}{\pi }$$$$$$$$$$

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