Let $0=t_0<t_1<\cdots<t_n=T$ be a partition of the interval $[0,T]$. Denote $\Delta t_k\equiv t_{k+1}-t_k$ for every $0\leq k<n$. Assume $\lim_{n\to\infty}\Delta t_k=0$. Find the limit of $S_n$ as $n\to\infty$, where:
$$S_n\equiv\sum_{k=0}^{n-1}(\Delta t_k)^2$$
My intuition is that $\lim_{n\to\infty}S_n=0$. I'm not sure about this, but $S_n$ looks like the sum with which one defines the Stieltjes integral. Some sort of an integral, but with $dt^2$ instead of $dt$... And I was told once that $dt^2=0$ (I'm aware this is incredibly not rigorous, sorry about that). It is clear to me that the limit of $S_n$ can be written as an integral on $[0,T]$, but I'm not sure how. I might be totally wrong about this.
Thanks!
For arbitrary partitions Giulio has already answered this. If you take the $t_i$'s to be equidistant, you will find that
$$S_n = \sum_{k = 0}^{n - 1} \left(\frac{T}{n}\right)^2 = \left(\frac{T}{n}\right)^2 \sum_{k = 0}^{n - 1} 1 = \left(\frac{T}{n}\right)^2 n = \frac{T^2}{n},$$
which converges to $0$ as presumed.
$S_n$ however is not the Riemann-Stietjes sum with respect to $h:[0,T] \to \mathbb{R}$, $x \mapsto x^2$. Since this is defined to be
$$T_n := \sum_{k=0}^{n-1} 1 (h(t_{k+1}) - h(t_k)) = \sum_{k=0}^{n-1} 1 (t_{k+1}^2 - t_k^2).$$
The sum $T_n$ converges to $T^2$. Indeed, since $h$ is differentiable, the Riemann-Stieltjes integral can be written as
$$\int_0^T 1 \, \mu(dx) = \int_0^T h'(x) \, dx = \int_0^T 2x \, dx = T^2.$$
I am not so sure if $S_n$ is a Riemann-Stietjes integral w.r.t. some monotone function (or function of bounded variation) $h$. This is indeed an interesting question.