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

Question

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?

Answer 1

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,
$$\frac{\sum_{k=1}^{10}k^{2}}{10^{3}}=0.385 \approx 1/3$$

Answer 2

${ }$

Answer 3

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.

Answer 4

\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*}