Welcome 2016! Part 17

$\begin{aligned} S & = \sum_{n=2}^{2016} \frac {n^2+1}{n^2 - 1}\\ & = \sum_{n=2}^{2016} \frac {n^2 - 1+2}{n^2 - 1}\\&= \sum_{n=2}^{2016}\left(1+\frac {2}{n^2 - 1}\right) \\ & = \sum_{n=2}^{2016}\left(1+\frac{1}{n-1}-\frac {1}{n+1}\right) \\ & = \sum_{n=2}^{2016} 1+ \sum_{n=2}^{2016} \frac{1}{n-1} - \sum_{n=2}^{2016} \frac {1}{n+1} \\ & = \sum_{n=1}^{2015} 1+ \sum_{n=1}^{2015} \frac{1}{n} - \sum_{n=3}^{2017} \frac {1}{n} \\ &= 2015+\frac{1}{1} +\frac{1}{2} - \frac {1}{2016}-\frac {1}{2017} \approx 2016.5 \\ \Rightarrow \lfloor S \rfloor +2016&=2016+2016= \boxed {4032} \end{aligned}$

$\begin{aligned} S & = \frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+ \ldots + \frac{2016^2+1}{2016^2-1} \\ & = \displaystyle \sum^{2015}_{n=1}\frac{(n+1)^2+1}{(n+1)^2-1} \\ & = \displaystyle \sum^{2015}_{n=1}\frac{(n+1)^2+1}{n(n+2)} \\ & = \displaystyle \sum^{2015}_{n=1}\left(\frac{1}{n}-\frac{1}{n+2}+1\right) \\ & = \displaystyle \sum^{2015}_{n=1}\frac{1}{n}- \displaystyle \sum^{2017}_{n=3}\frac{1}{n}+ \displaystyle \sum^{2015}_{n=1}1 \\ & = 1+\frac{1}{2}+\displaystyle \sum^{2015}_{n=3}\frac{1}{n}- \displaystyle \sum^{2017}_{n=3}\frac{1}{n}+ \displaystyle \sum^{2015}_{n=1}1 \\ & = 1+\frac{1}{2}-\frac{1}{2016}-\frac{1}{2017}+2015 \\ & = 1.499+2015=2016.499 \\ & \Rightarrow \lfloor S \rfloor + 2016 = 2016 +2016=\boxed{4032}\end{aligned}$