# Method of Exhaustion: intuition behind inequalities

In Apostol's One-Variable Calculus, with an Introduction to Linear Algebra, when discussing the method of exhaustion for solving for the area under a curve (specifically $$x^2$$), Apostol sets up the following inequality:

$$1^2 + 2^2 + ... + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + ... + n^2$$

He goes on to state that it is valid for every integer $$n\geq1$$ and that they can be deduced easily as consequences of the following formulas:

1.3) $$1^2 + 2^2 + ... + n^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$

1.4) $$1^2 + 2^2 + ... + (n-1)^2 = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$$

I'm a bit confused at how he is able to deduce this from just 1.3 and 1.4 alone. Is it because since $$n\geq1$$, $$\frac{n^2}{2}$$ is certainly greater than $$\frac{n}{6}$$ and so we know 1.3 is greater than $$\frac{n^3}{3}$$ and vice versa for 1.4? Is there a more concrete explanation for how 1.3 and 1.4 setup the inequality? An induction proof is shared later on but I'm specifically interested in the intuition of the above!

• 1.3) $\dfrac {n^2}{2} + \dfrac n 6 < \dfrac {n^2}{2}+\dfrac {n^2}{2}= n^2$ Commented Mar 25, 2022 at 13:35

RHS of 1.3) is greater than $$n^3/3$$, this is really obvious because the RHS is just $$n^3/3$$ plus something that is positive. That something is $$(n^2/2 + n/6)$$.
RHS of 1.4) is smaller than $$n^3/3$$. OK, this is not so obvious but still quite obvious. That's true because $$(-n^2/2 + n/6) \lt 0$$ for every natural number $$n$$. You can prove this easily yourself e.g. by induction. Or, if you don't want to use induction (to prove this inequality), you can just notice that it is equivalent to $$n(3n-1)\gt0$$ which of course is true for every natural number $$n$$.
These two observations prove that the sum in 1.3) is greater than $$n^3/3$$ while the sum in 1.4) is smaller than $$n^3/3$$, and that is exactly what you want to prove here.