# Deducing a chained inequality from two equations

From the two equations below:

$1^{2}+2^{2}+...+n^{2} = {n^3 \over 3}+{n^2 \over 2}+{\frac n6}$

$1^{2}+2^{2}+...+(n-1)^{2} = {n^3 \over 3}-{n^2 \over 2}+{\frac n6}$

How can the following inequalities be deduced:

$1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3} < 1^{2}+2^{2}+...+n^{2}$ for $n\ge 1$

I understand that it can be proved via induction, but how can we deduce the answer before we test that it is correct via induction.

This is part of a proof from the introduction of Tom M Apostol's book Calculus 1.

${n^2 \over 2}+{\frac n6} > 0$
$-{n^2 \over 2}+{\frac n6} < 0$
to get to the inequality you want. These inequalities are both true for $n\ge 1$
We don't even need the explicit formula. If we set $a_n=\sum_{k=1}^{n}k^2$, we may prove $$a_{n-1} < \frac{n^3}{3} < a_n$$ by induction. That is obviously true for $n=1$, and $$\frac{(n+1)^3}{3}-\frac{n^3}{3} = n^2+n+\frac{1}{3}$$ is clearly between $n^2$ and $(n+1)^2$.
• @JamieFearon: the correct answer to which question? $\sum_{k=1}^{n}k^2$ has to behave like $\frac{n^3}{3}$ since $\int_{0}^{n}x^2\,dx = \frac{n^3}{3}.$ – Jack D'Aurizio Nov 4 '15 at 15:21