Proving a result for $\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)$ $$\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)=\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)}$$
According to WA. I attempted using
$$\prod_{k=0}^{\infty}\Bigl(1-\frac{x^2}{\pi^2k^2}\Bigr)=\frac{\sin x}{x}$$
But I couldn’t reindex the product appropriately to use it the way I wanted to (factoring out a 4 and then continuing from there). I’d like to have at least a direction to go in or an idea on how to do the product.
 A: It isn't pretty, but we can prove this by working backwards using Euler's product definition of the gamma function (seen here):
$$
\Gamma(x) = \lim_{n\to\infty} n!(n+1)^x \prod_{k=0}^n (x+k)^{-1}.
$$
When we substitute that in for the right hand side, we get
$$
\begin{align}
\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)} &= \frac{a^{2}-4}{a^{2}}\frac{\left(n+1\right)^{\frac{a+4}{2}}n!^{2}\prod_{k=0}^{n}\left(\frac{a+4}{4}+k\right)^{-2}}{\left(n+1\right)^{\frac{a+4}{2}}n!^{2}\left(\prod_{k=0}^{n}\left(\frac{a+2}{4}+k\right)^{-1}\right)\left(\prod_{k=0}^{n}\left(\frac{a+6}{4}+k\right)^{-1}\right)}\\
&= \frac{a^{2}-4}{a^{2}}\frac{\prod_{k=0}^{n}\left(\frac{a+4}{4}+k\right)^{-2}}{\prod_{k=0}^{n}\left(\frac{a+2}{4}+k\right)^{-1}\left(\frac{a+6}{4}+k\right)^{-1}}\\
&= \frac{a^{2}-4}{a^{2}}\prod_{k=0}^{\infty}\frac{\left(\frac{a+2}{4}+k\right)\left(\frac{a+6}{4}+k\right)}{\left(\frac{a+4}{4}+k\right)^{2}}\\
&= \frac{a^{2}-4}{a^{2}}\prod_{k=1}^{\infty}\frac{\left(\frac{a}{4}+k+\frac{1}{2}\right)\left(\frac{a}{4}+k-\frac{1}{2}\right)}{\left(\frac{a}{4}+k\right)^{2}}\\
&=\frac{a^{2}-4}{a^{2}}\prod_{k=1}^{\infty}\frac{\left(\frac{a}{4}+k\right)^{2}-\frac{1}{4}}{\left(\frac{a}{4}+k\right)^{2}}\\
&= \left(1-\frac{4}{a^{2}}\right)\prod_{k=1}^{\infty}\left(1-\frac{4}{\left(a+4k\right)^{2}}\right)\\
&= \prod_{k=0}^{\infty}\left(1-\frac{4}{\left(4k+a\right)^{2}}\right).
\end{align}
$$
(I omitted the limit of $n$ in the equations because it already takes up so much space.)
A: Use (proved by induction)
$$
\prod_{k=0}^{n-1} (a+4k) = \frac{4^n\Gamma(n+\frac{a}{4})}{\Gamma(\frac{a}{4})}
$$
together with
$$
1-\frac{4}{(4\,k+a)^2}={\frac { \left( a+4\,k+2 \right) 
 \left( a+4\,k-2 \right) }{ \left( 4\,k+a \right) ^{2}}}
$$
to get
$$
\prod _{k=0}^{n-1}{\frac { \left( a+4\,k+2 \right)  \left( a+4\,k-2
 \right) }{ \left( 4\,k+a \right) ^{2}}}
={\frac { \left( a-2
 \right)  \left( a+4\,n-2 \right)   \Gamma \left(\frac{a-2}{4}+n
 \right) ^{2} \;\Gamma \left( \frac{a-2}{4} \right) ^{2}
}{16 \; \Gamma \left( \frac{a}{4} +n\right) ^{2} \; \Gamma
 \left( \frac{a+2}{4} \right)  ^{2}}}
$$
Than take limit as $n \to \infty$.  My result is
$$
\frac{\displaystyle\Gamma\left(\frac{a}{4}\right)^2}{\displaystyle \Gamma\left(\frac{a-2}{4}\right)\Gamma\left(\frac{a+2}{4}\right)}
$$
which agrees with Polygon.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{N \in \mathbb{N}_{\ \ge\ 0}}$:
\begin{align}
&\bbox[10px,#ffd]{\prod_{k = 0}^{N}
\bracks{1 - {4 \over \pars{4k + a}^{2}}}} =
\prod_{k = 0}^{N}{\pars{4k + a + 2}\pars{4k + a - 2} \over
\pars{4k + a}^{2}}
\\[5mm] = &\
\prod_{k = 0}^{N}{\pars{k + a/4 + 1/2}\pars{k + a/4 - 1/2} \over
\pars{k + a/4}\pars{k + a/4}}
\\[5mm] = &\
\bracks{\prod_{k = 1}^{N + 1}\pars{k + {a \over 4} - {1 \over 2}}}
\prod_{k = 0}^{N}{k + a/4 - 1/2 \over \pars{k + a/4}\pars{k + a/4}}
\\[5mm] = &\
{N + a/4 + 1/2 \over a/4 - 1/2}
\pars{\prod_{k = 0}^{N}{k + a/4 - 1/2 \over k + a/4}}^{2}
\\[5mm] = &\
{4N + a + 2 \over a - 2}
\bracks{\pars{a/4 - 1/2}^{\overline{N + 1}} \over \pars{a/4}^{\overline{N + 1}}}^{2}
\\[5mm] = &\
{4N + a + 2 \over a - 2}
\bracks{\Gamma\pars{N + a/4 + 1/2}/\Gamma\pars{a/4 - 1/2} \over \Gamma\pars{N + a/4 + 1}/\Gamma\pars{a/4}}^{2}
\\[5mm] = &\
{4N + a + 2 \over a - 2}
{\Gamma^{2}\pars{a/2} \over \Gamma^{2}\pars{a/4 - 1/2}}
\bracks{\pars{N + a/4 - 1/2}! \over \pars{N + a/4}!}^{2}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, &
{4N + a + 2 \over a - 2}
{\Gamma^{2}\pars{a/4} \over \Gamma^{2}\pars{a/4 - 1/2}}
\\[2mm] \times &\
\bracks{\root{2\pi}\pars{N + a/4 - 1/2}^{N + a/4}
\expo{-N - a/4 + 1/2} \over
\root{2\pi}\pars{N + a/4}^{N + a/4 +1/2}\expo{-N - a/4}}^{2}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, &
{4N + a + 2 \over a - 2}
{\Gamma^{2}\pars{a/4} \over \Gamma^{2}\pars{a/4 - 1/2}}
\\[2mm] \times &\
\bracks{{N^{N + a/4}\,\bracks{1 + \pars{a/4 - 1/2}/N}^{\, N}
 \over N^{N + a/4 + 1/2}
\bracks{1 + \pars{a/4}/N}^{\, N}}\,\root{\expo{}}}^{2}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, &
{4N + a + 2 \over a - 2}
{\Gamma^{2}\pars{a/4} \over \Gamma^{2}\pars{a/4 - 1/2}}\,{1 \over N}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\,&
\bbx{{4 \over a - 2}\,\,
{\Gamma^{2}\pars{a/4} \over \Gamma^{2}\pars{a/4 - 1/2}}}
\\ &\
\end{align}
$\ds{\color{red}{\mbox{This result is equivalent to the proposed one}}}$.
