Proving Hermite's Identity in a different approach To prove 
$$S=\left [x \right]+\left [x+\frac{1}{n} \right]+\left [x +\frac{2}{n}\right]+\cdots+\left [x +\frac{n-1}{n}\right]=\left [nx \right]$$
using and starting with $$x-1 \lt \left [x \right]\le x \tag{1}$$ we have
$$x+\frac{1}{n}-1 \lt \left [x +\frac{1}{n}\right] \le x+\frac{1}{n} \tag{2}$$
$$x+\frac{2}{n}-1 \lt \left [x +\frac{2}{n}\right] \le x+\frac{2}{n} \tag{3}$$
$$x+\frac{3}{n}-1 \lt \left [x +\frac{3}{n}\right] \le x+\frac{3}{n} \tag{4}$$
and so on till
$$x+\frac{n-1}{n}-1 \lt \left [x +\frac{n-1}{n}\right] \le x+\frac{n-1}{n} \tag{n}$$
Adding all we get
$$nx+\frac{n(n-3)}{2} \lt S \le nx+\frac{n(n-1)}{2}$$
Now how can we prove that between $nx+\frac{n(n-3)}{2}$ and $nx+\frac{n(n-1)}{2}$ there is only on integer which is $\left [nx \right]$?

This method doesn't work.
Adding all of them we get $$nx+\frac{\frac{n(n-1)}2}{n}-n<S\le nx+\frac{\frac{n(n-1)}2}{n}$$ or $$nx-\frac{n+1}2<S\le nx+\frac{n-1}2\implies -\frac n2<S-nx+\frac12\le\frac n2$$ and this interval gets wider and wider as $n$ gets larger.
