\[\large{ \lim_{n \to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n-1} \ k \cdot \left \lfloor x + \dfrac{n-k-1}{n} \right \rfloor = \ ? }\]

Find the value of the above limit upto 3 decimal places when \(x = \sqrt{2015}\).

**Bonus** - Generalize the above limit in terms of \(x\).

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