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I'm preparing for an exam, and one of the review problems is to sort functions by order of growth, and this was the only summation in it. I know that

$\sum \limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6},$

But what if I did not know the closed form? How, then, would I prove

$\sum \limits_{i=1}^n i^2 \in \Theta (n^3).$

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    Hint: compare with the integral of x^22011-12-11

1 Answers 1

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There is a trick to compute easily the growth of such a sum at first order. Indeed, we have :

$\frac{1}{n^3} \sum_{i=1}^n i^2 = \frac{1}{n} \sum_{i=1}^n \left(\frac{i}{n}\right)^2,$

which is a Riemann sum for the function $x \to x^2$ on $[0,1]$. Hence :

$\lim_{n \to + \infty} \frac{1}{n^3} \sum_{i=1}^n i^2 = \int_0^1 x^2 dx = \frac{1}{3}.$

Not only does this answer your problem, but it also gives you an asymptotic equivalent of the series.