Cfrgtkky

In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 to b$. Using the fact that,

begin{align}
&1^2+2^2+...+(n-1)^2 < frac{n^3}{3} < 1^2+2^2+...+n^2label{1} \
&Rightarrow frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < frac{b^3}{3} < frac{b^3}{n^3}(1^2+2^2+...+n^2) nonumber \
&Rightarrow s_{n} < frac{b^3}{3} < S_{n} nonumber
end{align}

where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=frac{b^3}{3}$ is the only number that satisfies

begin{equation}
s_{n}<A<S_{n} label{2}
end{equation}

for every $ngeq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $frac{b^3}{n^3}$,

begin{align*}
&frac{b^3}{n^3}(1^2+2^2+...+n^2)<frac{b^3}{3}+frac{b^3}{n^2} \
&Rightarrow S_{n}<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}

for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $frac{b^3}{n^3}$,

begin{align*}
&frac{b^3}{3}-frac{b^3}{n^2}<frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\
&Rightarrow frac{b^3}{3}-frac{b^3}{n^3}<s_{n}
end{align*}

this implies,

begin{align*}
frac{b^3}{3}-frac{b^3}{n^2}<A<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}

This is where it gets confusing. He says the only possibilities are:

$$begin{array}{ccc}
A>frac{b^3}{3},& A<frac{b^3}{3},& A=frac{b^3}{3}
end{array}$$

and proceeds to show that $A=frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $frac{b^3}{n^2}$?