<s>[INST] the sum of the squares I think it is interesting, if we have the  formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$ 
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n  (n + 0.1)  (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$ 
or, as another example if  a = 0.01 .$$\frac{n  (n + 0.01)  (2 n + 0.01) }{6 \cdot 0.01} = 0.01^2 + 0.02^2 + \cdots + n^2 .$$  
Now if we follow the same logic, I suppose if the difference between the closest numbers becomes smallest possible (a = 0.0...1), we will obtain 
$$ \frac{n (n + 0.0..1) (2 n  + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$ 
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \ (0.0..1 ^ 2 + 0.0..2 ^ 2 .. + n ^ 2) * {0.0..1}$$ If we take that,  a = 0.0...1 and,  b = $\frac {n}{0.0...1}$ and we rearrange the formula we will get this:

$$\frac{n ^ 3}{3} = 1^2a^3 + 2^2a^3 + \dots+b^2 a^3$$

following the same logic we prove that the formula holds for each degree (it is easy to check), so that we could generalize, If we take that degree call m, we will get

$$\frac{n ^ m}{m} = 1^{m-1}a^m + 2^{m-1}a^m + \dots+b^{m-1} a^m$$

so the question is this expression can be interpreted as a Riemann sum, and why?
 [/INST] The quantity on the right of your last equation can be expressed as 
$$ R(n, \epsilon) = 
\sum_{k=1}^{n/\epsilon} k^2 \epsilon^3 $$ where $\epsilon = 0.00\ldots 01$ approaches zero.
Yore conclusion is correct in the following sense:
$$\lim_{\epsilon \to 0^+} \frac{R(n, \epsilon)}{n^3/3} = 1 $$
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