2016 is absolutely awesome 20

Calculus Level 5

\[ \large\displaystyle \mathop{\lim}_{n\to\infty} \dfrac{1}{n^2}\sum_{k=1}^{n-1} k\left\lfloor\sqrt{2016}+\dfrac{n-k-1}{n}\right\rfloor =\, ? \]


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