<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] 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 $$P=\frac{\Gamma\left(1/4\right)\Gamma\left(3/4\right)}{\Gamma\left(1/2\right)^{2}}=\color{red}{\sqrt{2}}.$$
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