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

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

  • 3
    $\begingroup$ Note that the hard part is showing $\sum_{i=1}^n i^2 \geq c n^3$, since trivially: $\sum_{i=1}^n i^2 \leq n \cdot n^2 = n^3$ (bounding the sum by its largest term times the number of terms). $\endgroup$ – JavaMan Dec 11 '11 at 21:46
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    $\begingroup$ Well, $\sum_{i=1}^n i^2 \leq n \cdot n^2$ (trivially) and $\sum_{i=1}^n i^2 \geq \frac{n}{2} \cdot (n/2)^2$ by taking the last half of the series. $\endgroup$ – cardinal Dec 11 '11 at 21:48
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    $\begingroup$ Hint: compare with the integral of x^2 $\endgroup$ – Bruno Joyal Dec 11 '11 at 21:49

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.


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