Telescopic Products' Triads.

$\begin{aligned} P_1 & = \prod_{n=3}^\infty \left(1-\frac 4{n^2}\right) = \prod_{n=3}^\infty \left(\frac {n^2-4}{n^2}\right) = \prod_{n=3}^\infty \left(\frac {(n-2)(n+2)}{n^2}\right) = \frac {\color{#3D99F6}\prod_{n=1}^\infty n\color{#D61F06}\prod_{n=5}^\infty n }{\color{#3D99F6}\prod_{n=3}^\infty n \color{#D61F06}\prod_{n=3}^\infty n } = \frac {\color{#3D99F6}1\cdot 2}{\color{#D61F06}3\cdot 4} = \frac 16 \end{aligned}$

$\begin{aligned} P_2 & = \prod_{n=2}^\infty \frac {n^3-1}{n^3+1} = \prod_{n=2}^\infty \frac {(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} = \frac {\prod_{n=2}^\infty (n-1) \prod_{n=2}^\infty\color{#3D99F6}(n^2+n+1)}{\prod_{n=2}^\infty (n+1) \prod_{n=2}^\infty (n^2-n+1)} & \small \color{#3D99F6} \text{Note that }n^2+n+1 = (n+1)^2 - (n+1) + 1 \\ & = \frac {\prod_{n=1}^\infty n \prod_{\color{#3D99F6}n=3}^\infty\color{#3D99F6}(n^2-n+1)}{\prod_{n=3}^\infty n \prod_{n=2}^\infty (n^2-n+1)} = \frac {1\cdot 2}{2^2-2+1} = \frac 23 \end{aligned}$

$\begin{aligned} P_3 & = \prod_{n=3}^\infty \frac {\left(1+\frac 1n\right)^2}{1+\frac 2n} = \prod_{n=3}^\infty \frac {(n+1)^2}{n(n+2)} = \frac {\prod_{n=3}^\infty (n+1) \prod_{n=3}^\infty (n+1)}{\prod_{n=3}^\infty n \prod_{n=3}^\infty (n+2)} = \frac {\prod_{n=4}^\infty n \prod_{n=4}^\infty n}{\prod_{n=3}^\infty n \prod_{n=5}^\infty n} = \frac 43 \end{aligned}$

Therefore, $P_1+P_2+P_3 = \dfrac 16+\dfrac 23 + \dfrac 43 = \dfrac {13}6 \approx \boxed{2.167}$ .

The first partial product is $\begin{aligned} \prod_{n=3}^N \frac{(n-2)(n+2)}{n^2} & = \; \frac{(N-2)! \times \frac{1}{24}(N+2)!}{\frac14 (N!)^2} \\ & = \; \frac{(N+2)(N+1)}{6N(N-1)} \end{aligned}$ and hence the first infinite product is $\tfrac16$ .

The second partial product is $\begin{aligned} \prod_{n=2}^N \frac{n^3-1}{n^3+1} & = \; \prod_{n=2}^N \frac{(n-1)(n^2 + n + 1)}{(n+1)(n^2-n+1)} \; = \; \prod_{n=2}^N \frac{(n-1)\big[n(n+1)+1\big]}{(n+1)\big[(n-1)n + 1\big]} \\ & = \; \frac{(N-1)!}{\frac12(N+1)!} \times \frac{\prod_{n=2}^N [n(n+1)+1]}{\prod_{n=1}^{N-1}[n(n+1)+1]} \\ & = \; \frac{2}{N(N+1)} \times \frac{N(N+1)+1}{3} \end{aligned}$ and hence the second infinite product is $\tfrac23$ .

The third partial product is $\begin{aligned} \prod_{n=3}^N \frac{(1 + \frac{1}{n})^2}{1 + \frac{2}{n}} & = \; \prod_{n=3}^N \frac{(n+1)^2}{n(n+2)} \\ & = \; \frac{\frac{1}{36}((N+1)!)^2}{\frac12N! \times \tfrac{1}{24}(N+2)!} \; = \; \frac{4(N+1)}{3(N+2)} \end{aligned}$ making the third infinite product $\tfrac43$ . The sum of the three products is $\tfrac16 + \tfrac23 + \tfrac43 = \boxed{2\tfrac16}$