Proving a result for $\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)$

Question

$$\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)=\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)}$$

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.

Answer 1

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.)

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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} $\ds{\color{red}{\mbox{This result is equivalent to the proposed one}}}$.