Find the value of 2

$\begin{aligned} &\left( 1 - \dfrac{1}{2^2} \right) \left( 1 - \dfrac{1}{3^2} \right) \left( 1 - \dfrac{1}{4^2} \right) ... \left( 1 - \dfrac{1}{2018^2} \right) \\ \\ = \ &\left( 1 - \dfrac{1}{2} \right) \left( 1 + \dfrac{1}{2} \right) \left( 1 - \dfrac{1}{3} \right) \left( 1 + \dfrac{1}{3} \right) \left( 1 - \dfrac{1}{4} \right) \left( 1 + \dfrac{1}{4} \right) ... \left( 1 - \dfrac{1}{2018} \right) \left( 1 + \dfrac{1}{2018} \right) \\ \\ = \ &\left( 1 - \dfrac{1}{2} \right) \left( 1 - \dfrac{1}{3} \right) \left( 1 - \dfrac{1}{4} \right) .... \left( 1 - \dfrac{1}{2018} \right) \ \times \ \left( 1 + \dfrac{1}{2} \right) \left( 1 + \dfrac{1}{3} \right) \left( 1 + \dfrac{1}{4} \right) ... \left( 1 + \dfrac{1}{2018} \right) \\ \\ = \ &\left( \dfrac{1}{2} \right) \left( \dfrac{2}{3} \right) \left( \dfrac{3}{4} \right) ... \left( \dfrac{2017}{2018} \right) \ \times \ \left( \dfrac{3}{2} \right) \left( \dfrac{4}{3} \right) \left( \dfrac{5}{4} \right) ... \left( \dfrac{2019}{2018} \right) \\ \\ = \ &\dfrac{1}{2018} \times \dfrac{2019}{2} \\ \\ = \ &\dfrac{2019}{4036} \approx \boxed{0.5002477701} \end{aligned}$

$\begin{aligned} P & = \left(1-\frac 1{2^2}\right)\left(1-\frac 1{3^2}\right)\left(1-\frac 1{4^2}\right)\cdots \left(1-\frac 1{2018^2}\right) \\ & = \prod_{k=2}^{2018} \left(1 - \frac 1{k^2}\right) \\ & = \prod_{k=2}^{2018} \left(\frac {k^2-1}{k^2}\right) \\ & = \prod_{k=2}^{2018} \left(\frac {(k-1)\color{#3D99F6}(k+1)}{k^2}\right) \\ & = \left(\frac {2017! \times \color{#3D99F6}2019!}{{\color{#3D99F6}2}\times (2018!)^2}\right) \\ & = \frac {2019}{4036} \\ & \approx \boxed{0.5002} \end{aligned}$