Formula for the sum of the squares of numbers We have the well-known 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, we obtain, for example 
$$\frac{n \times (n + 0.1)  (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$ 
It is easy to check. Now if the difference between the closest numbers becomes smallest possible, we will obtain 
$$ \frac{n \cdot (n + 0.0..1) \cdot (2 n  + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$ 
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \frac{0.0..1 ^ 2 + 0.0..2 ^ 2 + \cdots + n ^ 2}{0.0..1}.$$ 
Is this conclusion correct?
 A: Suppose we want to calculate the sum of squares with successive differences $\epsilon$ from $0$ to some fixed $n$ (we require $\frac{n}{\epsilon}\in\mathbb{N}$ for this particular calculation, however for the general formulation of integrals and Riemann sums, this is not required), that is
$$S_\epsilon = \sum_{i=0}^{\frac{n}{\epsilon}}(i\epsilon)^2$$
letting $m = \frac{n}{\epsilon}$ this sum is equivalent to
$$\sum_{i=0}^{m}i^{2}\left(\frac{n}{m}\right)^2$$
which we can write as
$$=\left(\frac{n^2}{m^2}\right)\sum_{i=0}^{m}i^2=\left(\frac{n^2}{m^2}\right)\frac{m(m+1)(2m+1)}{6} $$
$$= \frac{\left(\frac{n}{m}\right)m\left(\frac{n}{m}\right)(m+1)\left(\frac{n}{m}\right)(2m+1)}{6\left(\frac{n}{m}\right)}=\frac{n(n+\epsilon)(2n+\epsilon)}{6\epsilon}$$
taking the limit $\epsilon\rightarrow 0$ this is equivalent to $m\rightarrow\infty$ and $S_{\epsilon\rightarrow 0}$ is easily seem to be divergent to $+\infty$. However, $S_{\epsilon\rightarrow 0}\cdot \epsilon$ is convergent (which easily evaluated by simply substituting $\epsilon = 0$, which we can do by the continuity of the expression) and is of certain interest. In particular, we can write
$$S = S_{\epsilon\rightarrow 0}\cdot\epsilon=\lim_{m\rightarrow\infty}\ \sum_{i=0}^{m}\left(i\frac{n}{m}\right)^2\left(\frac{n}{m}\right)$$
we recognize this as the Riemann Sum which defines the integral
$$\lim_{m\rightarrow\infty}\ \sum_{i=0}^{m}\left[f\left(x_0 + i\frac{n}{m}\right)\frac{n}{m}\right]=\int_{x_0}^{x_0 + n}f(x)\ dx$$
for $f(x) = x^2$ and $x_0 = 0$. (In particular this is the left Riemann sum). By the Fundamental Theorem of Calculus, 
$$\int_{0}^{n}x^2\ dx = \frac{x^3}{3}\bigg|_{0}^{n} = \frac{n^3}{3}$$
which is exactly the quantity you cite.
