<s>[INST] The value of $\prod_2^\infty \left(1-2/(n(n+1))\right)$ I'm trying to find the value of $\prod_2^\infty \left(1-2/(n(n+1))\right)$. So far I have the following.
\begin{align*}
\prod_2^\infty \left(1-2/(n(n+1))\right) &= (1-2/6)*(1-2/12)*(1-2/20)*(1-2/30)... \\&= (1-1/3)*(1-1/6)*(1-1/10)*(1-1/15)... \\&= (2/3)*(5/6)*(9/10)*(14/15)... \\&= (2/3) * ((1*5)/(2*3)) * ((3*3)/(5*2)) * ((2*7)/(3*5))...
\end{align*}
The partial product with just the first four terms equals $(1/3)*(7/5)$ because almost everything cancels out. I claim that the term which $(1/3)$ gets multiplied by goes to zero since it goes from $5/3$ to $3/2$ to $7/5$... (making the value $1/3$), but I don't know the explicit formula for each factor of each term so I can't prove that. Does anyone know the explicit formula for the factors?
 [/INST] $$\prod _{ 2 }^{ \infty  } \left( 1-\frac { 2 }{ n\left( n+1 \right)  }  \right) =\prod _{ 2 }^{ \infty  } \left( \frac { { n }^{ 2 }+n-2 }{ n\left( n+1 \right)  }  \right) =\prod _{ 2 }^{ \infty  } \left( \frac { \left( n+2 \right) \left( n-1 \right)  }{ n\left( n+1 \right)  }  \right) =\\ =\prod _{ 2 }^{ \infty  } \frac { n+2 }{ n+1 } \cdot \prod _{ 2 }^{ \infty  } \frac { n-1 }{ n } =\\ =\left( \frac { 4 }{ 3 } \cdot \frac { 5 }{ 4 } \cdot \frac { 6 }{ 5 } .... \right) \left( \frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } \cdot \frac { 3 }{ 4 } ... \right) =\color{blue}{\frac { 1 }{ 3 }} $$
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