Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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). – JavaMan Dec 11 '11 at 21:46
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. – cardinal Dec 11 '11 at 21:48
Hint: compare with the integral of x^2 – Bruno Joyal Dec 11 '11 at 21:49
up vote 8 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.