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Question Number 133628 by Ñï= last updated on 23/Feb/21

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(\frac{n^2+1}{n^2}\right)=?$$$$\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(\frac{4n^2-4n+2}{4n^2-4n+1}\right)=?$$

Answered by Dwaipayan Shikari last updated on 23/Feb/21

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{n^2})=\frac{sinh\pi }{\pi }\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{x^2}{n^2})=\frac{sinh\left(\pi x\right)}{\pi x}$$

Answered by Dwaipayan Shikari last updated on 23/Feb/21

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(\frac{4n^2-4n+2}{4n^2-4n+1}\right)=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{{(2n-1)}^2})$$$$cosx=(1-\frac{2x}{\pi })(1+\frac{2x}{\pi })(1-\frac{2x}{3\pi })(1+\frac{2x}{3\pi })..$$$$cos\left(\frac{x}{2}\right)=(1-\frac{x}{\pi })(1+\frac{x}{\pi })(1-\frac{x}{3\pi })...=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{x^2}{{\pi }^2{(2n-1)}^2})$$$$cosh\left(\frac{\pi x}{2}\right)=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{x^2}{{(2n-1)}^2})$$$$cosh\left(\frac{\pi }{2}\right)=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{{(2n-1)}^2})\Rightarrow \frac{e^{\frac{\pi }{2}}+e^{-\frac{\pi }{2}}}{2}$$

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