Infinite product

$\begin{aligned} P & = \lim_{n \to \infty} \frac {2^2}{1\times 3} \times \frac {4^2}{3\times 5} \times \frac {6^2}{5\times 7} \times \cdots \times \frac {(2n)^2}{(2n-1)(2n+1)} \\ & = \lim_{n \to \infty} \frac {2^{2n} (n!)^2}{(2n-1)!!(2n+1)!!} & \small \color{#3D99F6} \text{By Stirling's formula: }n! \sim \sqrt{2\pi} n^{n+\frac 12}e^{-n} \\ & = \lim_{n \to \infty} \frac {2^{2n} (n!)^2}{\frac {(2n)!}{2^nn!} \times \frac {(2n+2)!}{2^{n+1}(n+1)!}} \\ & = \lim_{n \to \infty} \frac {2^{4n+1} (n!)^3(n+1)!}{(2n)!(2n+2)!} \\ & = \lim_{n \to \infty} \frac {2^{4n+1} \left(\sqrt {2\pi} n^{n+\frac 12}e^{-n}\right)^3 \left(\sqrt {2\pi} (n+1)^{n+\frac 32}e^{-n-1}\right)}{\left(\sqrt {2\pi} (2n)^{2n+\frac 12}e^{-2n}\right)\left(\sqrt {2\pi} (2n+2)^{2n+\frac 52}e^{-2n-2}\right)} \\ & = \lim_{n \to \infty} \frac {2^{4n+3}\pi^2{\color{#3D99F6}n^{3n+\frac 32}(n+1)^{n+\frac 32}}e^{-4n-1}}{2^{4n+4}\pi{\color{#3D99F6}n^{2n+\frac 12}(n+1)^{2n+\frac 52}}e^{-4n-2}} \\ & = \lim_{n \to \infty} \frac {2^{4n+3}\pi^2{\color{#3D99F6}n^{4n+3}}e^{-4n-1}}{2^{4n+4}\pi{\color{#3D99F6}n^{4n+3}}e^{-4n-2}} \\ & = \frac \pi 2 \approx \boxed{1.5708} \end{aligned}$

The given infinite product $\begin{aligned} P&=\Big(\frac{2^2}{1\cdot 3}\Big)\cdot\Big(\frac{4^2}{3\cdot 5}\Big)\cdot\Big(\frac{6^2}{5\cdot 7}\Big)\cdot .....& \\ &=\prod_{n=1}^\infty\bigg[\frac{(2n)^2}{(2n-1)(2n+1)}\bigg]=\prod_{n=1}^\infty\bigg[\frac{(2n)^2}{(2n)^2-1}\bigg]& \\ &=\prod_{n=1}^\infty\bigg[\frac{1}{1-\frac{1}{(2n)^2}}\bigg]\end{aligned}$

Product expansion of $\sin z$ is $\quad\color{#3D99F6}\displaystyle\sin z=z\prod_{n=1}^\infty\Big(1-\frac{z^2}{n^2\pi^2}\Big)$

So, $\displaystyle P_1=\prod_{n=2}^\infty\Big(1-\frac{1}{n^2}\Big)=\lim_{z\rightarrow\pi} \frac{\sin z}{z(1-\frac{z^2}{n^2\pi^2})}=\lim_{z\rightarrow\pi} \frac{\cos z}{(1-\frac{3z^2}{n^2\pi^2})}=\frac{1}{2}\quad\small \color{#3D99F6} [\text{using L'Hospital's Rule}]$

Again, product expansion of $\cos z$ is $\quad\color{#3D99F6}\displaystyle\cos z=\prod_{n=1}^\infty\Big[1-\frac{z^2}{(n-\frac{1}{2})^2\pi^2}\Big]$

So, $\displaystyle P_2 =\prod_{n=2}^\infty\Big[1-\frac{1}{(2n-1)^2}\Big]=\lim_{z\rightarrow\frac{\pi}{2}} \frac{\cos z}{1-\frac{z^2}{(\pi/2)^2}}=\lim_{z\rightarrow\frac{\pi}{2}} \frac{\sin z}{\frac{2z}{(\pi/2)^2}}=\frac{\pi}{4} \quad\small \color{#3D99F6} [\text{using L'Hospital's Rule}]$

Then, $\displaystyle P=\frac{1}{P_1/P_2}=\frac{P_2}{P_1}=\frac{\pi}{2}= \boxed{1.5708}$