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$$\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$$

I've seen some similar questions asked. But this one is different from all these. Euler product does not apply. One cannot simply factorize $\left(1-\frac{2}{(2n+1)^2}\right)$ since the $\sqrt{2}$ on top will prevent terms from cancelling. Any help will be appreciated!

Note: we are expected to solve this in 2 mins.

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4 Answers 4

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Hint. One may recall Euler's infinite product for the cosine function

$$\cos x =\prod_{n=0}^\infty \left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right),\qquad |x|<\frac \pi2.$$

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    $\begingroup$ @Daniel Li Here is a proof of the above infinite product: math.stackexchange.com/questions/1204021/… $\endgroup$ Sep 24, 2017 at 22:48
  • $\begingroup$ I wish the 9 upvoters upvoted Jack D'Aurizio's linked answer. $\endgroup$ Sep 25, 2017 at 0:29
  • $\begingroup$ @SimplyBeautifulArt : done! Jack d'Aurizio's answers are always the best of the lot. $\endgroup$
    – Paramanand Singh
    Sep 25, 2017 at 5:26
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Maple says this is $$\sin(\pi (\sqrt{2}-1)/2)$$ and more generally $$ \prod_{n=1}^\infty \left(1 - \frac{t^2}{(2n+1)^2} \right) = \frac{\sin(\pi (1+t)/2)}{1-t^2} $$

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  • $\begingroup$ Ok Great. But where can I find result like this? Is this some well-known result? $\endgroup$
    – Daniel Li
    Sep 24, 2017 at 22:32
  • $\begingroup$ See Olivier Oloa's answer. $\endgroup$ Sep 25, 2017 at 0:44
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \prod_{n = 1}^{\infty}\bracks{1 - {2 \over \pars{2n + 1}^{2}}}}} = \prod_{n = 1}^{\infty}\bracks{1 - {1/2 \over \pars{n + 1/2}^{2}}} \\[5mm] = &\ \lim_{N \to \infty} {\bracks{\prod_{n = 1}^{N}\pars{n + 1/2 - \root{2}/2}} \bracks{\prod_{n = 1}^{N}\pars{n + 1/2 + \root{2}/2}} \over \bracks{\prod_{n = 1}^{N}\pars{n + 1/2}}^{2}} \\[5mm] = &\ \lim_{N \to \infty} {\pars{3/2 - \root{2}/2}^{\overline{N}}\pars{3/2 + \root{2}/2}^{\overline{N}} \over \bracks{\pars{3/2}^{\overline{N}}}^{2}} \\[5mm] = &\ {\Gamma^{2}\pars{3/2} \over \Gamma\pars{3/2 + \root{2}/2}\Gamma\pars{3/2 - \root{2}/2}}\, \lim_{N \to \infty} {\pars{N + 1/2 - \root{2}/2}!\pars{N + 1/2 + \root{2}/2}! \over \bracks{\pars{N + 1/2}!}^{2}} \\[5mm] = &\ {\pi/4 \over \pars{-1/4}\Gamma\pars{1/2 + \root{2}/2}\Gamma\pars{1/2 - \root{2}/2}}\ \times \\[2mm] &\ \lim_{N \to \infty} {\bracks{\pars{N + 1/2 - \root{2}/2}^{N + 1 - \root{2}/2}} \bracks{\pars{N + 1/2 + \root{2}/2}^{N + 1 + \root{2}/2}} \expo{-2N - 1} \over \pars{N + 1/2}^{2N + 2}\expo{-2N - 1}} \\[5mm] = &\ -\sin\pars{\pi\bracks{{1 \over 2} + {\root{2} \over 2}}} \times \\[2mm] &\ \lim_{N \to \infty} {\bracks{1 + \pars{1/2 - \root{2}/2}/N}^{N} \bracks{1 + \pars{1/2 + \root{2}/2}/N}^{N} \over \braces{\bracks{1 + \pars{1/2}/N}^{N}}^{2}} \\[5mm] = & -\cos\pars{\pi\,{\root{2} \over 2}}\, {\exp\pars{1/2 - \root{2}/2}\exp\pars{1/2 \root{2}/2} \over \bracks{\exp\pars{1/2}}^{2}} = \bbx{-\,\cos\pars{{\root{2} \over 2}\,\pi}} \approx 0.6057 \end{align}

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Considering the partial products

$$A_p=\prod_{n=1}^{p} \left(1-\frac{2}{(2n+1)^2}\right)$$ a CAS produced $$A_p=-\cos \left(\frac{\pi }{\sqrt{2}}\right) \frac{\Gamma \left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma \left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma \left(p+\frac{3}{2}\right)^2}$$ and using Stirling approximetion for large $p$, this leads to $$\log\left(\frac{\Gamma \left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma \left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma \left(p+\frac{3}{2}\right)^2} \right)=\frac{1}{2p}+O\left(\frac{1}{p^2}\right)$$ then $$A_p =-\cos \left(\frac{\pi }{\sqrt{2}}\right)\left(1+\frac{1}{2p}+O\left(\frac{1}{p^2}\right)\right)$$

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