Evaluate the following expression: $$\frac{2016}{\pi}[\sum_{n=1}^{\infty}\cos^{-1}\frac{n^2+n+3}{\sqrt{n^2+3}\sqrt{n^2+2n+4}}]$$
note- I can only tell that the presence of cosine inverse function can imply that we can use any geometric ideas and the dot product to arrive at the solution but certainly no geometric figure comes to my mind to solve this problem. So please can anyone suggest any ideas? Thank you!
Hint:
First, we notice that $$\begin{align} \frac{n^2+n+3}{\sqrt{n^2+3}\sqrt{n^2+2n+4}} &= \frac{n(n+1)+3}{\sqrt{n^2+3}\sqrt{(n+1)^2+3}}\\ &=\frac{n}{\sqrt{n^2+3}}\cdot \frac{n+1}{\sqrt{(n+1)^2+3}}+\frac{\sqrt{3}}{\sqrt{n^2+3}}\cdot \frac{\sqrt{3}}{\sqrt{(n+1)^2+3}} \end{align}$$ Second, try to apply the formula $$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$ with $$\cos(x) = \frac{n}{\sqrt{n^2+3}}$$ $$\cos(y) = \frac{n+1}{\sqrt{(n+1)^2+3}}$$
and you deduce easily that $$\begin{align} S &= \frac{2016}{\pi}\left(\arccos\left(\frac{1}{\sqrt{1^2+3}} \right) -\arccos\left(\frac{2}{\sqrt{2^2+3}}\right)+\arccos\left(\frac{2}{\sqrt{2^2+3}}\right) -... \right)\\ &= \frac{2016}{\pi} \cdot \arccos\left(\frac{1}{\sqrt{1^2+3}} \right)=\frac{2016}{\pi} \cdot \frac{\pi}{3} = 672 \end{align}$$