More fun in 2016

$\displaystyle\lim_{n\to\infty}\int_0^1x^{2016}(\big\{nx\big\})^{2016}dx=\lim_{n\to\infty}\frac1{n^{2017}}\int_0^\infty x^{2016}\big\{x\big\}^{2016}\text{ Now rewrite the fractional part:}$

$\displaystyle\frac1{n^{2017}}\sum_{k=0}^\infty(t+k)^{2016}t^{2016}dt=\frac1{n}\sum_{k=0}^\infty(\frac{t}{n}+\frac{k}{n})^{2016}t^{2016}dt\quad\text{(as }n\to\infty, \frac{t}{n}\to0\text{ )}$

$\displaystyle \lim_{n\to\infty}\frac1{n}\sum_{k=0}^\infty(\frac{k}{n})^{2016}\int_0^1t^{2016}dt=\int_0^1x^{2016}dx\int_0^1t^{2016}dt=\Huge\boxed{\frac1{2017^2}}$