Inverse Trigonometric Series Product!

First note that $27k^3 + 54k^2 + 36k + 8 = (3k+2)^3$ .

For a fixed $k$ , let $\theta = \tan^{-1} \left( \frac{1}{\sqrt{3k+1}} \right)$

So, by simple trigonometry, we get $\theta = \sin^{-1} \left( \frac{1}{\sqrt{3k+2}} \right)$ .

Now $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) = \dfrac{9k+2}{\sqrt{27k^3 + 54k^2 + 36k + 8}}$ .

Therefore, we simply obtain:

$S_n = \prod_{k=1}^n \ \dfrac{\arcsin \left( \frac{9k+2}{\sqrt{27k^3 + 54k^2 + 36k + 8}} \right) }{\arctan \left( \frac{1}{\sqrt{3k+1}} \right)} = \prod_{k=1}^n \dfrac{3\theta}{\theta} = \prod_{k=1}^n 3 = 3^n$

Thus $\ln(S_{2015}) = \ln(3^{2015}) = 2015\ln(3) \approx \boxed{2213.703}$

$Log({ S }_{ n })=nLog(3)$

That complicated trigonometric expression on the right is just $3$ . Looks like a difficult problem until it's solved.