abver
Nov 11, 2015, 12:51 PM
If \delta=\frac 1m then
\frac{n(n+\delta)(2n+\delta)}{6}=\frac1{m^3}\frac{ mn(mn+1)(2mn+1)}{6}=\frac1{m^3}\sum_{k=1}^{mn}k^2= \frac1{m}\sum_{k=1}^{mn}\left(\frac km\right)^2.
We get *in the limit* as $m\to\infty$ is that
$$ \frac{n^3}{3}=\int_0^nx^2\,\mathrm dx.$$
so the question is this expression can be interpreted as a Riemann sum, and why?
\frac{n(n+\delta)(2n+\delta)}{6}=\frac1{m^3}\frac{ mn(mn+1)(2mn+1)}{6}=\frac1{m^3}\sum_{k=1}^{mn}k^2= \frac1{m}\sum_{k=1}^{mn}\left(\frac km\right)^2.
We get *in the limit* as $m\to\infty$ is that
$$ \frac{n^3}{3}=\int_0^nx^2\,\mathrm dx.$$
so the question is this expression can be interpreted as a Riemann sum, and why?