# How to find this infinite product

How to find this infinite product ?

$$\prod_{n=0}^\infty \left(1-\dfrac{2}{4(2n+1)^2+1}\right)$$

I try to use infinite product of $\cos{x}$ but it doesn't work.

Thank you.

• See Basel problem. As an exercise, can you also evaluate $\displaystyle\prod_{n=0}^\infty \left(1+\dfrac2{4(2n+1)^2+1}\right)$ ? Commented Jul 12, 2015 at 17:15

$$\begin{eqnarray*}\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) &=& \prod_{n\geq 0}\left(\frac{4(2n+1)^2-1}{4(2n+1)^2+1}\right)\\&=&\color{purple}{\prod_{n\geq 0}\left(1-\frac{1}{4(2n+1)^2}\right)}\color{blue}{\prod_{n\geq 0}\left(1+\frac{1}{4(2n+1)^2}\right)^{-1}}\end{eqnarray*}$$

hence by exploiting the Weierstrass products for the $\color{purple}{\cos}$ and $\color{blue}{\cosh}$ function we get:

$$\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) = \frac{1}{\sqrt{2}\cosh\frac{\pi}{4}}=\color{red}{\frac{\sqrt{2}}{e^{\pi/4}+e^{-\pi/4}}}.$$

• $\dfrac{\cos\dfrac\pi4}{\cosh\dfrac\pi4}$ has a certain ring to it... :-$)$ Commented Jul 12, 2015 at 17:20

\begin{align} &\prod_{n=0}^\infty\left(1-\frac2{4(2n+1)^2+1}\right)\\ &=\prod_{n=0}^\infty\frac{(4n+1)(4n+3)}{(4n+2-i)(4n+2+i)}\tag{1}\\ &=\prod_{n=0}^\infty\frac{(n+\frac14)(n+\frac34)}{(n+\frac{2-i}4)(n+\frac{2+i}4)}\tag{2}\\ &=\lim_{n\to\infty}\left.\frac{\Gamma(n+\frac14)}{\Gamma(\frac14)}\frac{\Gamma(n+\frac34)}{\Gamma(\frac34)}\middle/\frac{\Gamma(n+\frac{2-i}4)}{\Gamma(\frac{2-i}4)}\frac{\Gamma(n+\frac{2+i}4)}{\Gamma(\frac{2+i}4)}\right.\tag{3}\\ &=\lim_{n\to\infty}\frac{\Gamma(\frac{2-i}4)\Gamma(\frac{2+i}4)}{\Gamma(\frac14)\Gamma(\frac34)}\cdot\frac{\Gamma(n+\frac14)\Gamma(n+\frac34)}{\Gamma(n+\frac{2-i}4)\Gamma(n+\frac{2+i}4)}\tag{4}\\ &=\frac{\pi\csc\left(\pi\frac{2-i}4\right)}{\pi\csc\left(\frac\pi4\right)}\tag{5}\\ &=\frac1{\sqrt2\cosh(\frac\pi4)}\tag{6} \end{align} Explanation:
$(1)$: subtract and factor the numerator and denominator
$(2)$: factor out the $4$ so that it is easier to see the relation to $\Gamma$
$(3)$: $\prod\limits_{k=0}^{n-1}(k+x)=\frac{\Gamma(n+x)}{\Gamma(x)}$ applied four times
$(4)$: rearrange terms for easier cancelling
$(5)$: Euler's Reflection Formula and Gautschi's Inequality
$(6)$: $\csc\left(\frac\pi2-i\frac\pi4\right)=\sec\left(i\frac\pi4\right)=\mathrm{sech}(\frac\pi4)$

This is a method of directly evaluating the product, without any prior knowledge, which I have been applying to a number of products. I gave this method of proof here. My answer to the current question involves some different features to my answer to the other question.

Set

\begin{align*} f(x) = \prod_{n=1}^\infty \left( 1 - \frac{2x}{4 (2n-1)^2 + 1} \right) \end{align*}

Taking logs and differentiating with respect to $$x$$ gives:

\begin{align*} \frac{d}{dx} \ln f(x) = \sum_{n=1}^\infty \dfrac{\left( - \frac{2}{4 (2n-1)^2 + 1} \right)}{\left( 1 - \frac{2x}{4 (2n-1)^2 + 1} \right)} = - \sum_{n=1}^\infty \frac{2}{4 (2n-1)^2 + 1 - 2x} \end{align*}

The sum can be evaluated by employing the fact that the function

\begin{align*} \dfrac{\sin^2 \dfrac{\pi z}{2}}{\sin \pi z} \end{align*}

has simple poles at all odd integers $$k$$ with residues $$(-1)^k/ \pi$$. This allows us to write

\begin{align*} - \sum_{n=1}^\infty \frac{2}{4 (2n-1)^2 + 1 - 2x} & = \sum_{n=1}^\infty (-1)^n \frac{2 \sin^2 \dfrac{\pi n}{2}}{4 n^2 + 1 - 2x} \\ & = \frac{1}{2 i} \oint_C \dfrac{2 \sin^2 \dfrac{\pi z}{2}}{(4z^2+1-2x) \sin \pi z} dz \\ & = \frac{1}{2 i} \oint_C \dfrac{\tan \dfrac{\pi z}{2}}{(4z^2+1-2x)} dz \end{align*}

where the contour $$C$$ is defined in fig (a). The dots are poles at odd integer values of $$z$$, whereas the crosses mark the location of the other poles of $$\dfrac{\tan \dfrac{\pi z}{2}}{(4z^2+1-2x)}$$. Since we will be putting $$x=1$$ at the end of the calculation and we want to avoid the poles $$z = \pm \sqrt{2x-1}/2$$ coinciding with odd integer values, we will restrict ourselves to $$x < 5/2$$. For $$0 \leq x < 1/2$$ the poles take imaginary values: $$\pm i \sqrt{1-2x}$$.

We have

\begin{align*} \sum_{n=1}^\infty (-1)^n \frac{2 \sin^2 \dfrac{\pi n}{2}}{4 n^2 + 1 - 2x} & = \frac{1}{2} \sum_{n=1}^\infty (-1)^n \frac{2 \sin^2 \dfrac{\pi n}{2}}{4 n^2 + 1 - 2x} + \frac{1}{2} \sum_{n=-1}^{-\infty} (-1)^n \frac{2 \sin^2 \dfrac{\pi n}{2}}{4 n^2 + 1 - 2x} \\ & = \frac{1}{4 i} \oint_{C+C'} \dfrac{\tan \dfrac{\pi z}{2}}{(4z^2+1-2x)} dz \end{align*}

where the contour $$C'$$ is defined in fig (a). We complete the path of integration along semicircles at infinity (see fig (b)) since the integration along them vanishes. Since the resulting enclosed area contains a finite number of singularities marked by crosses, we can shrink this contour down to a contour, $$C_s$$, surrounding these finite number of poles (see fig (c)). So that

\begin{align*} \frac{d}{dx} \ln f(x) = \frac{1}{4 i} \oint_{C_s} \dfrac{\tan \dfrac{\pi z}{2}}{(4z^2+1-2x)} dz \end{align*}

The are poles at $$z = \pm \sqrt{2x-1}/2$$, so

\begin{align*} \frac{d}{dx} \ln f(x) & = \frac{1}{4 i} (- 2 \pi i) \left[ \dfrac{\tan \dfrac{\pi \sqrt{2x-1}}{4}}{4 \sqrt{2x-1}} + \dfrac{\tan -\dfrac{\pi \sqrt{2x-1}}{4}}{-4 \sqrt{2x-1}} \right] \\ & = - \frac{\pi}{4} \dfrac{\tan \dfrac{\pi \sqrt{2x-1}}{4}}{\sqrt{2x-1}} \\ & = \frac{d}{dx} \ln \cos (\dfrac{\pi \sqrt{2x-1}}{4}) \end{align*}

Implying

\begin{align*} f(x) = C \cos (\dfrac{\pi \sqrt{2x-1}}{4}) = \prod_{n=1}^\infty \left( 1 - \frac{2x}{4 (2n-1)^2 + 1} \right) \end{align*}

Putting $$x=0$$ we obtain $$C = \dfrac{1}{\cosh (\dfrac{\pi}{4})}$$, so that

\begin{align*} \prod_{n=1}^\infty \left( 1 - \frac{2x}{4 (2n-1)^2 + 1} \right) = \dfrac{\cos (\dfrac{\pi \sqrt{2x-1}}{4})}{\cosh (\dfrac{\pi}{4})} \end{align*}

Finally, putting $$x=1$$ we obtain

\begin{align*} \prod_{n=1}^\infty \left( 1 - \frac{2}{4 (2n-1)^2 + 1} \right) = \dfrac{\cos (\dfrac{\pi}{4})}{\cosh (\dfrac{\pi}{4})} = \frac{1}{\sqrt{2} \cosh (\dfrac{\pi}{4})} \end{align*}

NOTE: If we replace $$2x$$ with $$x^2+1$$, then we would have:

\begin{align*} \prod_{n=1}^\infty \left( 1 - \frac{x^2+1}{4 (2n-1)^2 + 1} \right) = \dfrac{\cos (\dfrac{\pi x}{4})}{\cosh (\dfrac{\pi}{4})} \end{align*}

which checks out as

\begin{align*} \prod_{n=1}^\infty \left( 1 - \frac{x^2+1}{4 (2n-1)^2 + 1} \right) = \dfrac{\prod_{n=1}^\infty \left( 1 - \frac{x^2}{4 (2n-1)^2} \right)}{\prod_{n=1}^\infty \left( 1 + \frac{1}{4 (2n-1)^2} \right)} \end{align*}

and by use of the Weierstrass products for the cos and cosh.