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Evaluate the limit $$\lim_{n\to \infty} \sum_{j=1}^n \frac{4j^{2}}{n^3}.$$

I was under the assumption that this would just tend to $0$ after expanding everything because the denominator will grow quicker than the numerator, but apparently this is not the case.

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Note that $$\sum_{j=1}^n j^2=\frac{n(n+1)(2n+1)}{6}$$ we have \begin{align} \lim_{n\to\infty}\sum_{j=1}^n\frac{4j^2}{n^3}&=\lim_{n\to\infty}\frac{2(n+1)(2n+1)}{3n^2}\\ &=\frac{2}{3}\lim_{n\to\infty}(1+\frac{1}{n})(2+\frac 1n)\\ &=\frac{4}{3}. \end{align}

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Recognize the limit as the integral $$\int _0^1 4x^2 dx $$

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