Find the value of the Product $$P=\prod_{k=1}^{\infty} \frac{2k(2k+2)}{(2k+1)^2}$$
we have
$$P=\prod_{k=1}^{\infty}\left(1-\frac{1}{(2k+1)^2}\right)$$ Taking $ln$ on both sides we get
$$\ln(P)=\sum_{k=1}^{\infty}\ln\left(1-\frac{1}{2k+1}\right)+\sum_{k=1}^{\infty}\ln\left(1+\frac{1}{2k+1}\right)$$
Is there any way to continue further?
By the Weierstrass product for the cosine function $$ \cos(z) = \prod_{m\geq 0}\left(1-\frac{4z^2}{(2m+1)^2\pi^2}\right)\tag{1} $$ hence $$ \prod_{k\geq 1}\left(1-\frac{1}{(2k+1)^2}\right) = \lim_{z\to 1}\frac{\cos(\pi z/2)}{1-z^2}\stackrel{\text{de l'Hospital}}{=}\lim_{z\to 1}\frac{-\frac{\pi}{2}\sin(\pi z/2)}{-2z} = \color{red}{\frac{\pi}{4}}\tag{2}$$ and the same conclusion also follows from Wallis' product.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\prod_{k = 1}^{N}{2k\pars{2k + 2} \over \pars{2k + 1}^{2}} = \prod_{k = 1}^{N}{k\pars{k + 1} \over \pars{k + 1/2}\pars{k + 1/2}} = {N! \over \pars{3/2}^{\overline{N}}}\,{\pars{N + 1}! \over \pars{3/2}^{\overline{N}}} \\[5mm] = &\ \pars{N + 1}\,\bracks{N!\,{\Gamma\pars{3/2} \over \pars{N + 1/2}!}}^{2}\qquad\qquad\qquad \pars{~\bbox[#ffd,15px]{\ds{\Gamma\pars{3 \over 2} = {1 \over 2}\,\Gamma\pars{1 \over 2} = {\root{\pi} \over 2}}}~} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&\ {\pi \over 4}\,\pars{N + 1}\, \bracks{\root{2\pi}N^{N + 1/2}\expo{-N} \over \root{2\pi}\pars{N + 1/2}^{N + 1}\expo{-\pars{N + 1/2}}}^{2} \\[5mm] = &\ {\pi \over 4}\,\pars{N + 1}\, \braces{{1 \over \root{N}}\,{\expo{1/2} \over \bracks{1 + \pars{1/2}/N}^{\,N + 1}}}^{2} \\[5mm] = & {\pi \over 4}\,\pars{1 + {1 \over N}}\,\braces{{\expo{1/2} \over \bracks{1 + \pars{1/2}/N}^{\,\,N}}}^{2} \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\, \bbx{\pi \over 4} \end{align}
From the formula $$\prod_{k\geq0}\frac{\left(k+a\right)\left(k+b\right)}{\left(k+c\right)\left(k+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\,a+b=c+d$$ we have $$\prod_{k\geq1}\frac{2k\left(2k+2\right)}{\left(2k+1\right)^{2}}=\prod_{k\geq0}\frac{\left(k+1\right)\left(k+2\right)}{\left(k+3/2\right)^{2}}=\frac{\Gamma^{2}\left(3/2\right)}{\Gamma\left(1\right)\Gamma\left(2\right)}=\color{red}{\frac{\pi}{4}}.$$
