<s>[INST] Evaluate the reciprocal of the following infinite product 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 $1-\frac{1}{2^2}$.
 [/INST] 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.
</s>