2016 is absolutely awesome 10

Let's start by forming a nested radical expession for $|x|$ : $|x| = \sqrt{x^2}\\ |x| = \sqrt{1+(x-1)\sqrt{1+x(x+2)}}\\ |x| = \sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}}\\ \vdots \\ |x| = \sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\cdots}}}}}$ Simply putting $x=2017$ gives us: $\boxed{2017=\sqrt{1+2016\sqrt{1+2017\sqrt{1+2018\sqrt{1+2019\sqrt{\cdots}}}}}}$

Ramanujan's nested radical formula (equation 21) is as follows:

$\small x+n+a = \sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt{...}}}}$

Substitute $x = 2016$ , $n=1$ and $a=0$ , we have:

$2016+1+0 = \sqrt{1+2016\sqrt{1+2017\sqrt{1+2018\sqrt{1+2019\sqrt{1+...}}}}}$

Therefore, $S = \boxed{2017}$

Ramanujan told us that: 3 = $\sqrt{1+2\sqrt{1+3\sqrt{...}}}$ . Through manipulation, it is easy to see that: $y = \sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}$ , where $y = x + 1$ . So then $x = \sqrt{1+2016\sqrt{1+2017\sqrt{...}}}$ $x = \boxed{2017}$

More of a hint than solution, refer Ramanujan's infinite nested radicals.