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

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