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⌊limn→∞n−3/2∑j=16nj⌋= ?\large \displaystyle \left \lfloor \lim_{n \to \infty} { n^{{-3}/{2}} \sum_{j=1}^{6n} {\sqrt{j}}} \right \rfloor = \, ?⎣⎢⎢⎢n→∞limn−3/2j=1∑6nj⎦⎥⎥⎥=?
Notation: ⌊⋅⌋ \lfloor \cdot \rfloor ⌊⋅⌋ denotes the floor function.
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