Closed form for an infinite product: $\displaystyle \prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)$

Question

The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9): $$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$ Unfortunately, it's given without any reference or explanation. Could anyone give one?

Answer 1

This formula is an application of the theorem $1.1$ for $n=2$ from your Chamberland & Straub paper (when $a+b=c+d$) : $$\tag{1}\prod_{k\ge 0}\frac {(k+a)(k+b)}{(k+c)(k+d)}=\frac {\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}$$ Observe first that : $\quad\displaystyle n^2-n-1=\left(n-\frac{\sqrt{5}+1}2\right)\left(n+\frac{\sqrt{5}-1}2\right)\;$
An idea is to use $\quad a:=-\dfrac{\sqrt{5}+1}2,\;b:=\dfrac{\sqrt{5}-1}2$.

note that $\;a+b=-1\;$ and take $c=0,\;d=-1$ but we would have $\;0\cdot (-1)\;$ at the denominator so let's shift all these values with an offset $2$ (i.e. set $n:=k+2$) then : $$a:=2-\frac{\sqrt{5}+1}2,\;b:=2+\frac{\sqrt{5}-1}2,\;c:=2,\;d:=1$$ we then get : \begin{align} \prod_{k\ge 0}\frac {(k+2)^2-(k+2)-1}{(k+2)(k+1)}&=\frac {\Gamma(2)\Gamma(1)}{\Gamma\left(2-\frac{\sqrt{5}+1}2\right)\Gamma\left(2+\frac{\sqrt{5}-1}2\right)}\\ &=\frac 1{\left(1-\frac{\sqrt{5}+1}2\right)\left(1+\frac{\sqrt{5}-1}2\right)\Gamma\left(1-\frac{\sqrt{5}+1}2\right)\Gamma\left(1+\frac{\sqrt{5}-1}2\right)}\\ \tag{2}&=-\frac 1{\Gamma\left(1-\frac{\sqrt{5}+1}2\right)\Gamma\left(\frac{\sqrt{5}+1}2\right)} \end{align}

Using the Euler reflexion formula $\displaystyle \Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin(\pi z)}$ for $z=\dfrac {\sqrt{5}+1}2$ allows to get : $$\prod_{n\ge 1}\frac {(n+1)^2-(n+1)-1}{n(n+1)}=-\frac {\sin\left(\pi(\sqrt{5}+1)/2\right)}{\pi}$$ and conclude that indeed : $$\tag{3}\prod_{n\ge 1}\left(1-\frac 1{n(n+1)}\right)=-\frac {\sin\left(\pi(\sqrt{5}+1)/2\right)}{\pi}=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right)$$

Answer 2

Indeed noticing $n^2-n-1= (n-\phi_+)(n-\phi_-)$ where $\phi_+=\frac{1+\sqrt{5}}{2}$ and $\phi_-=\frac{1-\sqrt{5}}{2}$, We have $$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=\prod_{n=2}^{\infty}\frac{(n-\phi_+)(n-\phi_-)}{(n-0)(n-1)}=\\ \prod_{n=0}^{\infty}\frac{(n+2-\phi_+)(n+2-\phi_-)}{(n+2)(n+1)}=\frac{\Gamma(2)\Gamma(1)}{\Gamma(2-\phi_+)\Gamma(2-\phi_-)}$$

Where I used the formula $$\prod_{n=0}^{\infty} \frac{(n+a_1)(n+a_2)\cdots(n+a_i)}{(n+b_1)(n+b_2)\cdots(n+b_i)}=\frac{\Gamma(b_1)\Gamma(b_2)\cdots\Gamma(b_i)}{\Gamma(a_1)\Gamma(a_2)\cdots\Gamma(a_i)}$$ which holds whenever $a_1+a_2+\dots+a_i=b_1+b_2+\dots+b_i$.

This follows from Euler's product form of the Gamma function. (and is also the topic of that paper)

The rest follows from the reflection formula $$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$$, and some adjustments.

Answer 3

We will be only dealing with the following identities:

$$ \Gamma (z) = \frac{ e^{ -\gamma z }}{ z } \, \prod_{ k=1 }^{ \infty } \left\{ \left( 1+\frac{ z }{ k } \right)^{ -1 } \, e^{ \frac { z }{ k } } \right\} \hspace{10mm} \Gamma (-z)\Gamma (z)\quad =\quad \frac { -\pi }{ z \sin { (\pi z) } } $$

Let's start then ^^:

\begin{align} \prod_{ n=2 }^{ \infty }{ \left( 1-\frac { 1 }{ n(n-1) } \right) } &= \prod_{ n=2 }^{ \infty }{ \left( \frac { { n }^{ 2 }-n-1 }{ n(n-1) } \right) } \\ &= \prod_{ n=2 }^{ \infty }{ \left( \frac { (n-{ \phi }_{ + }) }{ n } \right) } \prod _{ n=2 }^{ \infty }{ \left( \frac { (n-{ \phi }_{ - }) }{ (n-1) } \right) } \\ &= \prod_{ n=2 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ n } \right) } \prod _{ k=1 }^{ \infty }{ \left( \frac { k+1-{ \phi }_{ - } }{ k } \right) } \\ &= \frac { \prod _{ k=1 }^{ \infty }{ \left( 1-\frac { { \emptyset }_{ + } }{ k } \right) } }{ 1-{ \phi }_{ + } } \prod _{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) } \end{align}

Now let's calculate each product using our identity: \begin{align} \prod_{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) { e }^{ -\frac { 1-{ \phi }_{ - } }{ k } } } &= \frac { { e }^{ -\gamma (1-{ \phi }_{ - }) } }{ (1-{ \phi }_{ - })\Gamma (1-{ \phi }_{ - }) } \\ \prod_{ k=1 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ k } \right) } { e }^{ \frac { { \phi }_{ + } }{ k } } &= \frac { { e }^{ \gamma { \phi }_{ + } } }{ -{ \phi }_{ + }\Gamma (-{ \phi }_{ + }) } \end{align}

Now we will multiply both the products(observe that ${ \phi }_{ + }+{ \phi }_{ - } = 1$): \begin{align} \prod_{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) } { e }^{ \frac { { \phi }_{ - }-1 }{ k } }\prod _{ k=1 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ k } \right) } { e }^{ \frac { { \phi }_{ + } }{ k } } &= \prod _{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) \left( 1-\frac { {\phi }_{ + } }{ k } \right) } \\ &= \frac { 1 }{ -\Gamma (-{ \phi }_{ + })\Gamma (1-{ \phi }_{ - }){ \phi }_{ + }(1-{ \phi }_{ - }) } \end{align}

Observe that:

$$ \Gamma (-{ \phi }_{ + })\Gamma (1-{ \phi }_{ - }) = \Gamma (-{ \phi }_{ + })\Gamma ({ \phi }_{ + }) = \frac { -\pi \csc { (\pi { \phi }_{ + }) } }{ { \phi }_{ + } } $$

so: $$ -\Gamma (-{ \phi }_{ + })\Gamma (1-{ \phi }_{ - }){ { \left( { \phi }_{ + } \right) }^{ 2 } } = \pi \csc { (\pi { \phi }_{ + }){ \phi }_{ + } } $$

We get: \begin{align} \prod_{ n=1 }^{ \infty }{ \left( 1-\frac { 1 }{ n(n-1) } \right) } &= \frac { \prod _{ k=1 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ k } \right) } \prod _{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) } }{ 1-{ \phi }_{ + } } \\ &= \frac { \sin { (\pi { \phi }_{ + }) } }{ \pi { \phi }_{ + }(1-{ \phi }_{ + }) } \\ &= -\frac { \sin { (\frac { \pi }{ 2 } +\frac { \sqrt { 5 } }{ 2 } \pi ) } }{ \pi } \\ &= -\frac { \cos { \left( \frac { \sqrt { 5 } \pi }{ 2 } \right) } }{ \pi } \end{align}

Answer 4

$$ \begin{align} \prod_{n=2}^\infty\left(1-\frac1{n(n-1)}\right) &=\prod_{n=2}^\infty\frac{n^2-n-1}{n(n-1)}\tag{1}\\ &=\lim_{N\to\infty}\prod_{n=2}^N\frac{(n-\phi)(n+\phi-1)}{n(n-1)}\tag{2}\\ &=\lim_{N\to\infty}\overbrace{\frac{\Gamma(N+1-\phi)}{\Gamma(2-\phi)}}^{\large\prod(n-\phi)} \overbrace{\frac{\Gamma(N+\phi)}{\Gamma(1+\phi)}}^{\large\prod(n+\phi-1)} \overbrace{\frac{\Gamma(2)}{\Gamma(N+1)}}^{\large\prod\frac1n} \overbrace{\frac{\Gamma(1)}{\Gamma(N)}}^{\large\prod\frac1{n-1}}\tag{3}\\ &=\frac{\Gamma(2)\Gamma(1)}{\Gamma(2-\phi)\Gamma(1+\phi)}\tag{4}\\ &=\frac1{\Gamma(2-\phi)\Gamma(\phi-1)}\tag{5}\\ &=\frac{\sin(\pi/\phi)}\pi\tag{6}\\ &=-\frac1\pi\cos\left(\frac{\sqrt5}2\pi\right)\tag{7} \end{align} $$ Explanation:
$(1)$: make difference a single quotient
$(2)$: factor numerator
$(3)$: use $\Gamma(x+1)=x\Gamma(x)$
$(4)$: use Gautschi's Inequality and the Squeeze Theorem
$(5)$: $\Gamma(2)=\Gamma(1)=1$ and $\Gamma(\phi+1)=\phi(\phi-1)\Gamma(\phi-1)=\Gamma(\phi-1)$
$(6)$: Euler Reflection Formula
$(7)$: $\sin(x-\pi/2)=-\cos(x)$

Answer 5

We want to compute: $$S=\sum_{n\geq 1}\log\left(1-\frac{1}{n(n+1)}\right)=\sum_{n\geq 1}\int_{0}^{+\infty}\frac{e^{-(n^2+n-1)x}-e^{-(n^2+n)x}}{x}\,dx$$ where the identity follows from Frullani's theorem. Since $\int_{0}^{+\infty}\frac{f(x)}{x}\,dx=\int_{0}^{+\infty}\mathcal{L}(f)(s)\,ds$, $$ S = \int_{0}^{+\infty}\sum_{n\geq 1}\left(\frac{1}{n^2+n-1+s}-\frac{1}{n^2+n+s}\right)\,ds $$ that is just: $$ S = 4\int_{0}^{1}\sum_{n\geq 1}\frac{1}{(2n+1)^2+(s-5)}\,ds=\int_{0}^{1}\left(\frac{1}{1-s}+\frac{\pi\tan\left(\frac{\pi}{2}\sqrt{5-4s}\right)}{\sqrt{5-4s}}\right)\,ds$$ or: $$ S = \int_{0}^{4}\left(\frac{1}{s}+\frac{\pi\tan\left(\frac{\pi}{2}\sqrt{1+s}\right)}{4\sqrt{1+s}}\right)\,ds=\int_{1}^{\sqrt{5}}\left(\frac{2s}{s^2-1}+\frac{\pi}{2}\,\tan\left(\frac{\pi s}{2}\right)\right)\,ds $$ so: $$ S = \left. \log(s^2-1)-\log\cos\frac{\pi s}{2}\right|_{1}^{\sqrt{5}} = \log\left(-\pi\sec\frac{\pi\sqrt{5}}{2}\right)$$ and by exponentiating: $$ \prod_{n\geq 1}\left(1-\frac{1}{n(n+1)}\right) = -\frac{1}{\pi}\,\cos\frac{\pi\sqrt{5}}{2}$$ as wanted.