How can one prove that
\begin{equation} \lim\limits_{n \rightarrow \infty} \prod\limits_{k=1}^n \frac{(2 k)^2 \cdot (4k-3)}{(2k-1)^2 \cdot (4k+1)} = \frac{\pi}{4}? \label{eq:limit} \end{equation}
It is very likely that this limit is closely related to the Wallis product. I also validated using Maple that this limit is actually correct:
> limit(product((2*k)*(2*k)*(4*k-3)/((2*k-1)*(2*k-1)*(4*k+1)), k = 1 .. n), n = infinity);
Pi/4
Thanks in advance.
$\begin{aligned}\prod_{k=1}^\infty \frac{(2 k)^2 \cdot (4k-3)}{(2k-1)^2 (4k+1)} & =\prod_{k=1}^\infty \frac{(2 k)^2 (4k-3)(4k^2-1)}{(2k-1)^2 \cdot (4k+1)(4k^2-1)} \\& =\prod_{k=1}^\infty \frac{(2 k)^2 (4k-3)(2k+1)}{(2k-1) (4k+1)(4k^2-1)} \\& =\prod_{k=1}^\infty \frac{(2 k)^2 }{(4k^2-1)} \cdot \prod_{k=1}^\infty \frac{(4k-3)(2k+1)}{(2k-1)(4k+1)} \\& = \frac{\pi}{2} \cdot \lim_{ n \to \infty} \prod_{k=1}^{n} \frac{(4k-3)}{(2k-1) } \cdot \prod_{k=1}^{n} \frac{(2k+1)}{(4k+1) } \\& = \frac{\pi}{2} \cdot \lim_{ n \to \infty} \prod_{k=1}^{n} \frac{(4k-3)}{(2k-1) } \cdot \prod_{k=0}^{n-1} \frac{(2k-1)}{(4k-3) } \\& = \frac{\pi}{2} \cdot \lim_{ n \to \infty} \prod_{k=1}^{n} \frac{(4k-3)}{(2k-1) } \cdot \prod_{k=1}^{n} \frac{(2k-1)}{(4k-3) } \cdot \frac{2n+1}{4n+1} \\& = \frac{\pi}{2} \cdot \lim_{ n \to \infty} \frac{2n+1}{4n+1} = \frac{\pi}{2} \cdot \lim_{n \to \infty} \frac{2+\frac{1}{n}}{4+\frac{1}{n}} = \frac{\pi}{2} \cdot \frac{1}{2} \\& = \frac{\pi}{4}. \end{aligned}$
Where the fourth equality is given by the Wallis product.