Finding $\lim_{n\to\infty}\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)$ I have a trouble with this limit of the infinite product:
$$\lim _{n \to\infty}\left(1-\frac{1}{1 \cdot 2}\right)\left(1-\frac{1}{2 \cdot 3}\right) \cdots\left(1-\frac{1}{n(n+1)}\right)$$
My attempt:
We have 
$$\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^{\infty}\left(\frac{n^{2}+n-1}{n^{2}+n}\right)=\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n \left(n+1\right)},$$
where $a_{1}=\dfrac{-1+\sqrt{5}}{2}$, $a_{2}=\dfrac{-1-\sqrt{5}}{2}$.
So I would just like a hint as to how to proceed. Any help would be appreciated.
 A: I continue from where you left. The infinite product representation of the sine function can be used to finish the calculation:
\begin{align*}
& \mathop {\lim }\limits_{N \to  + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{ - 1 - a_2 }}{{n + 1}}} \right)}  = \mathop {\lim }\limits_{N \to  + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{a_1 }}{{n + 1}}} \right)} 
\\ & = \mathop {\lim }\limits_{N \to  + \infty } \frac{{1 + \frac{{a_1 }}{{N + 1}}}}{{1 + a_1 }}\prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{a_1 }}{n}} \right)}  = \frac{1}{{1 + a_1 }}\prod\limits_{n = 1}^\infty  {\bigg(  1 - \frac{{a_1^2 }}{{n^2 }}  \bigg)} \\ & = \frac{1}{{1 + a_1 }}\frac{{\sin (\pi a_1 )}}{{\pi a_1 }} = \frac{{\sin (\pi a_1 )}}{{\pi }}.
\end{align*}
A: By means of  Weierstrass product of the gamma function，observe that
$$\frac{1}{\Gamma(z)}=e^{\gamma z} z \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}$$
where $\gamma $ is the Euler–Mascheroni constant.
Then
$$\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n(n+1)}=\frac{\Gamma\left(1\right)\Gamma\left(2\right)}{\Gamma\left(1-a_{1}\right)\Gamma\left(1-a_{2}\right)}=\frac{\Gamma(1) \Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}.$$
Since $\Gamma(x+1)=x \Gamma(x)$ and  $\Gamma(n)=(n-1) !$ for any positive integer $n$, 
$$\frac{\Gamma(1) \Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{  2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}.$$
Using the relation 
$\Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin \pi x}$, thus we obtain 
$$\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{(\frac{1-\sqrt{5}}{2})(\frac{1+\sqrt{5}}{2})}\frac{1}{\Gamma\left(\frac{1-\sqrt{5}}{2}\right) \Gamma\left(\frac{1+\sqrt{5}}{2}\right)}=-\frac{\sin \left(\frac{(1+\sqrt{5})\pi}{2}\right)}{\pi}$$
