Infinite Product of One Plus Reciprocals of Squares

 $$\prod _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n^2})=(1+\frac{1}{1})\times (1+\frac{1}{4})\times (1+\frac{1}{9})\times \cdots$$ $$=\frac{\sinh \pi }{\pi }=\frac{e^{\pi }-e^{-\pi }}{2\pi }$$

$$\prod _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n^2})=\prod _{n=1}^{\mathrm{\infty }}\frac{n^2+1}{n^2}$$ $$=\prod _{n=1}^{\mathrm{\infty }}\frac{(n-i)(n+i)}{(n-0)(n-0)}\text{(Difference of Two Squares)}$$ $$=\frac{\mathrm{\Gamma }(1)\mathrm{\Gamma }(1)}{\mathrm{\Gamma }(1+i)\mathrm{\Gamma }(1-i)}\text{(Infinite Product Formula)}$$

$$=\frac{\mathrm{\Gamma }(1)\mathrm{\Gamma }(1)}{i\mathrm{\Gamma }(i)\mathrm{\Gamma }(1-i)}\text{(Gamma Difference Equation)}$$ $$=\frac{0!\times 0!\sin (i\pi )}{i\pi }\text{(Gamma Function Extends Factorial, Euler's Reflection Formula)}$$ $$=\frac{i\sinh \pi }{i\pi }\text{(Hyperbolic Sine Definition)}$$ $$=\frac{\sinh \pi }{\pi }=\frac{e^{\pi }-e^{-\pi }}{2\pi }$$Από wikipedia: