I hae to evaluate the reciprocal of the following product to infinity
$$\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19}{2 \cdot 2 \cdot 6 \cdot 6 \cdot 10 \cdot 10 \cdot 14 \cdot 14 \cdot 18 \cdot 18}\cdot\ldots $$
I am guessing you express each number as a sum. Example, the first $\frac{1}{2}$ can be expressed as $\frac{2-1}{2}$ and the next $\frac{3}{2}$ as $\frac{2+1}{2}$, hence their product equals $1-\frac{1}{2^2}$.
On
Note that $$1-\frac{1}{\left(4n-2\right)^{2}}=\frac{\left(4n-3\right)\left(4n-1\right)}{\left(4n-2\right)\left(4n-2\right)} $$ and $$P=\prod_{n\geq1}\left(1-\frac{1}{\left(4n-2\right)^{2}}\right)^{-1}=\prod_{n\geq1}\frac{\left(4n-2\right)\left(4n-2\right)}{\left(4n-3\right)\left(4n-1\right)} $$ $$=\prod_{n\geq0}\frac{\left(4n+2\right)\left(4n+2\right)}{\left(4n+1\right)\left(4n+3\right)}=\prod_{n\geq0}\frac{\left(n+1/2\right)\left(n+1/2\right)}{\left(n+1/4\right)\left(n+3/4\right)}$$ and now we can use the well known identity $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d $$ which follows from the representation of the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}. $$ Hence $$P=\frac{\Gamma\left(1/4\right)\Gamma\left(3/4\right)}{\Gamma\left(1/2\right)^{2}}=\color{red}{\sqrt{2}}.$$
Since $\frac{1\cdot 3}{2\cdot 2}=1-\frac{1}{2^2}$, $\frac{5\cdot 7}{6\cdot 6}=1-\frac{1}{6^2}$ and so on, we want to compute: $$ P = \prod_{n\geq 1}\left(1-\frac{1}{(4n-2)^2}\right)^{-1} $$ while the Weierstrass product for the sine function $\frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)$ gives $$ \prod_{n\geq 1}\left(1-\frac{1}{(4n)^2}\right)^{-1}=\frac{\pi}{2\sqrt{2}},\qquad \prod_{n\geq 1}\left(1-\frac{1}{(2n)^2}\right)^2=\frac{\pi}{2}$$ hence $$\boxed{\, P = \frac{\frac{\pi}{2}}{\frac{\pi}{2\sqrt{2}}} = \color{red}{\sqrt{2}}\;}$$ since every number of the form $(4n-2)$ is an even number that is not a multiple of $4$.
An alternative approach, following Did's comment, is to notice that $$ \prod_{n=1}^{N}\left(1-\frac{1}{(4n-2)^2}\right)^{-1} =\frac{4^N\,((2N)!)^3}{(N!)^2\,(4N)!}, $$ hence $P=\sqrt{2}$ also follows from Stirling's approximation. That is no wonder since we already know that Wallis product and Stirling's approximation are closely related.