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limn→∞12+22+32+⋯+n2n3= ?\large\displaystyle\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \cdots+ n^2}{n^3} = \, ? n→∞limn312+22+32+⋯+n2=?
Give your answer to 3 decimal places.
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