Prove that $\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$

Question

The question

Prove that: $$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$

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What I've tried

Knowing that: $$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$ evaluating at $z=i$ gives $$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$ so: $$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$

I'm stucked up and don't know how to continue, any help?

Answer 1

$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right),\qquad \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{1}$$ give: $$ \frac{\sin(\pi z)\sinh(\pi z)}{\pi^2 z^2(1-z^4)}=\prod_{n\geq 2}\left(1-\frac{z^4}{n^4}\right)\tag{2} $$ hence by considering $\lim_{z\to 1}LHS$ we have: $$ \prod_{n\geq 2}\left(1-\frac{1}{n^4}\right)=\frac{\sinh \pi}{4\pi} = \color{red}{\frac{e^\pi-e^{-\pi}}{8\pi}}\tag{3}$$ as wanted.

Answer 2

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

This is because $$A_{n} =\prod_{k=2}^{n}\left(1-\frac{1}{n^2}\right) = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \frac{n+1}{2n} \to \frac{1}{2}$$

We have used $\displaystyle \left(1-\frac{1}{n^4}\right) = \left(1+\frac{1}{n^2}\right) \cdot \left(1-\frac{1}{n^{2}}\right)$

Answer 3

I'll reproduce the answer @C.Dubussy have just deleted: $$ \prod_{n=2} \left( 1 - \frac{1}{n^4} \right) = \prod_{n=2}^∞ \left( 1 + \frac{1}{n^2}\right) \prod_{n=2}^∞ \left( 1 - \frac{1}{n^2}\right) = \frac{\sin{iπ}}{iπ} \prod_{n=2}^∞ \frac{n-1}{n} \prod_{n=2}^∞ \frac{n+1}{n} $$

And because the last product gives $\frac{1}{2}$, we have it!

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{N \in \mathbb{N}_{\ \geq\ 2}}$:

\begin{align} \prod_{n = 2}^{N}\pars{1 - {1 \over n^{4}}} & = \prod_{n = 2}^{N} {\pars{n - 1}\pars{n + 1}\pars{n - \ic}\pars{n + \ic} \over n^{4}} \\[5mm] & = {\pars{N - 1}!\bracks{\pars{N + 1}!/2} \over \pars{N!}^{4}}\, \verts{\pars{2 + \ic}^{\overline{N - 1}}}^{2} = {1 \over 2}\,{N + 1 \over N}\,\verts{{1 \over N!}\,{\Gamma\pars{N + 1 + \ic} \over \Gamma\pars{2 + i}}}^{\,2} \\[5mm] & = {1 \over 2}\,{N + 1 \over N}\,\verts{{1 \over 1 + \ic} \,{\pars{N + \ic}! \over N!}}^{2}\,{1 \over \verts{\Gamma\pars{1 + i}}^{2}} \end{align}

With A & S Table $\ds{\mathbf{\color{#000}{6.1.31}}}$ identity $\ds{\verts{\Gamma\pars{1 + \ic y}}^{\,2} = {\pi y \over \sinh\pars{\pi y}}}$ and the Stirling Asymptotic Expansion:

\begin{align} \prod_{n = 2}^{N}\pars{1 - {1 \over n^{4}}} & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, {1 \over 4}\verts{\root{2\pi}\pars{N + \ic}^{N + \ic + 1/2}\expo{-N - \ic} \over \root{2\pi}N^{N + 1/2}\expo{-N}}^{2}\,{1 \over \pi/\sinh\pars{\pi}} \\[5mm] & = {\expo{\pi} - \expo{-\pi} \over 8\pi} \,\verts{N^{\ic}\pars{1 + {\ic \over N}}^{N + \ic + 1/2}}^{2} = {\expo{\pi} - \expo{-\pi} \over 8\pi} \,\verts{\exp\pars{\ic\ln\pars{N}}\expo{\ic}}^{2} \\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\, \bbx{{\expo{\pi} - \expo{-\pi} \over 8\pi}} \end{align}