# Easy proof for sum of squares $\approx n^3/3$

I'd like to prove to my (undergraduate, not math-major) students that $$\lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^n k^2 =\frac{1}{3},$$ to later show them that this can be interpreted as taking Riemann sums for the integral of $x^2$. Of course I could pull out of my hat the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, which makes it obvious, or start from the telescopic sum $$\sum_{k=1}^n ((k+1)^3-k^3)$$ and do some algebra. Neither of them sounds very convincing to me, since they will be new ideas for them, not immediate to grasp, and this should not be the central point of my lecture.

Is there a simpler way to work out this limit, without going through a proof for the value of the sum?

• A very rough heuristic would replace the $k^2$ by $\frac{n^2}{2}$ so that $$\sum_{k=1}^nk^2\approx\sum_{k=1}^n\frac{n^2}{2}= n\cdot\frac{n^2}{2}=\frac{n^3}{2}.$$ That would be the why the $n^3$ makes an appearance. Mar 19, 2014 at 13:26
• If you're not working on this being particularly rigorous, you could interpret the sum as a square pyramid, and then quote the volume for such a shape. Mar 19, 2014 at 13:27
• Also, you could just put numbers in - the limit converges pretty quickly. Mar 19, 2014 at 13:38
• @DanielLittlewood I like your suggestions. If you formulate them as an answer, you'll have my upvote. Mar 19, 2014 at 13:41
• @FedericoPoloni Done, glad to be of help. Mar 19, 2014 at 13:56

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• Nice. This also answers why formula of volumes of pyramids also include 1/3 in them instead of going through the integral. Easier to explain for secondary school! Mar 19, 2014 at 13:41

The limit is suitable for Stolz-Cesaro: $$\lim_{n\to\infty}{1^2+2^2+\cdots+n^2\over n^3}= \lim_{n\to\infty}{(n+1)^2\over (n+1)^3-n^3}= \lim_{n\to\infty}{n^2+2n+1\over 3n^2+3n+1}={1\over 3}.$$ In fact, repeating the trick with $\sum_{k=1}^n k^2 -\frac{n^3}{3}$, you can calculate the coefficient of $n^2$ in $\sum_{k=1}^n k^2$... until the full formula.

There are a couple of ways to make this limit seem intuitive. The sum of the squares can be interpreted as a square pyramid; the volume of such a pyramid is approximately $\frac{1}3 n^{3}$.

Another method could be to just calculate some of the partial sums. For example,
$$\frac{\sum_{k=1}^{10}k^{2}}{10^{3}}=0.385 \approx 1/3$$

• Other good answers in the thread, but this one seems the most practical approach to me, so I am accepting this one. I am all for creative proofs of that identity, but I'd like to have something that the bottom 50% of the class can understand. Mar 22, 2014 at 8:01

\begin{align*} 2\sum_{k=1}^n k^2 & = \sum_{k=1}^n k^2 + \sum_{k=1}^n(n-k+1)^2 \\ \\ & \approx \sum_{k=1}^n (n-k)^2 - (ik)^2 \\ & = \sum_{k=1}^n (n-(1+i)k)(n+(-i-1)k) \\ & \approx \sum_{k=1}^n(n-ik)^2 \end{align*}

Taking the real part

\begin{align*} 2\sum_{k=1}^n k^2 & \approx \sum_{k=1}^nn^2 - \sum_{k=1}^nk^2 \end{align*} and so \begin{align*} \sum_{k=1}^n k^2 & \approx \frac{n^3}{3} \end{align*}