Infant Monster

First, let's simplify that product:

$\prod_{x=1}^{n} \left| \sqrt{2 \int x \ dx} \right| = \prod_{x=1}^{n} \left| \sqrt{2 \times \frac{1}{2} x^{2}} \right| = \prod_{x=1}^{n} x = n!$

Hence, we have our sum equal to

$\sum_{n=2}^{\infty} \frac{1}{n!} = -\frac{1}{0!}-\frac{1}{1!}+\sum_{n=0}^{\infty} \frac{1}{n!} = -2+e$

which can be derived from the Maclaurin series for $e^{x}$ . Now, putting this altogether we have

$-2+e+2 = \boxed{e \approx 2.718}$

$\large{{ \left( \frac { 1 }{ \left( \prod _{ x=1 }^{ n }{ \left| \sqrt { 2\int { xdx } } \right| } \right) } \right) }}$

$\large{{ \left( \frac { 1 }{ \left( \prod _{ x=1 }^{ n }{ \left| \sqrt { 2\times \frac { { x }^{ 2 } }{ 2 } } \right| } \right) } \right) }\\ ={ \left( \frac { 1 }{ \left( \prod _{ x=1 }^{ n }{ \left| \sqrt { { x }^{ 2 } } \right| } \right) } \right) }\\ =\frac { 1 }{ \left( \prod _{ x=1 }^{ n }{ x } \right) } \\ =\frac { 1 }{ x! }}$

We know that

$\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } =e$

$\Rightarrow \sum _{ x=2 }^{ \infty }{ \frac { 1 }{ x! } } =\quad e-(\frac { 1 }{ 0! } +\frac { 1 }{ 1! } )\quad =\quad e-2$

Hence, our answer is $e-2+2 \approx 2.718$