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abver · Nov 11, 2015, 12:51 PM

$If \delta=\frac 1m then<br /> \frac{n(n+\delta)(2n+\delta)}{6}=\frac1{m^3}\frac{ mn(mn+1)(2mn+1)}{6}=\frac1{m^3}\sum_{k=1}^{mn}k^2= \frac1{m}\sum_{k=1}^{mn}\left(\frac km\right)^2.$

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?

Curlyben · Nov 11, 2015, 12:55 PM

What do YOU think ?
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