In Hermite's 1884 paper "Sur quelques conséquences arithmétiques des formules de la théorie des fonctions elliptiques", volume 5 of Acta Mathematica, pages 310-315, he proves what is often called "Hermite's identity" differently than the usual proof you'll find by Googling. Hermite's identity is, for real x and positive integer n,
$\sum_{k=0}^{n-1} E(x+k/n) = E(nx)$,
where $E(x)$ is the greatest integer $\leq x$. Hermite first establishes that for nonnegative integers $a$ and $b$,
$\frac{z^b}{(1-z)(1-z^a)} = \sum_{n=0}^\infty \left( 1 + \left[\frac{n}{a} \right]\right) z^{n+b}$.
Then using $\frac{z^a}{(1-z)(1-z^a)} = \frac{z^a(1+z^a+z^{2a}+\cdots+z^{(n-1)a})}{(1-z)(1-z^{na})}$ he says that the above identity follows. I'm not seeing how to get this and I'd be glad to hear if anyone sees how the deduction works.
Assume that a function $f$ is given as a power series $$ f(x) = A_0 + A_1x + A_2x^2 + \ldots; $$ then $$ \frac{f(x)}{1-x} = A_0 + (A_0+A_1)x + (A_0 + A_1 + A_2)x^2 + \ldots $$ as can be verified easily by multiplying through by $1-x$.
Next $$ f(x^a) = A_0 + A_1x^a + A_2x^{2a} + \ldots, $$ hence $$ \frac{f(x^a)}{1-x} = \sum_{n \ge 0} (A_0 + A_1 + \ldots + A_\nu) x^n, $$ where $\nu = \lfloor \frac na \rfloor$. For $f(x) = \frac1{1-x}$ this implies \begin{equation}\label{Her1} \frac1{(1-x)(1-x^a)} = \sum_{n \ge 0} \Big[ 1 + \Big\lfloor \frac na \Big\rfloor \Big] x^n . \end{equation} This implies $$ \frac1{(1-x)(1-x^{ma})} = \sum_{n \ge 0} \Big\lfloor \frac {n+ma}{ma} \Big\rfloor x^n $$ since $1 + \lfloor \frac n{ma} \rfloor = \lfloor \frac {n+ma}{ma} \rfloor$.
Multiplying this last equation through by $x^{ka}$ we obtain $$ \frac{x^{ka}}{(1-x)(1-x^{ma})} = \sum_{n \ge 0} \Big\lfloor \frac {n+ma}{ma} \Big\rfloor x^{n + ka} = \sum_{n \ge ka} \Big\lfloor \frac {n+(m-k)a}{ma} \Big\rfloor x^n. $$ Now the identity $$ \frac{1 - x^{ma}}{1-x^a} = 1 + x^a + x^{2a} + \ldots + x^{(m-1)a} $$ implies \begin{align*} \frac{x^a}{(1-x)(1-x^a)} & = \frac{x^a(1 + x^a + a^{2a} + \ldots + x^{m-1})}{(1-x)(1-x^{ma})} \\ & = \sum_{k=1}^m \frac{x^{ka}}{(1-x)(1-x^{ma})} \\ & = \sum_{k=1}^m \sum_{n \ge ka} \Big\lfloor \frac {n+(m-k)a}{ma} \Big\rfloor x^n. \end{align*} Replacing the summation index $k$ by $m-k$ we find \begin{align*} \frac{x^a}{(1-x)(1-x^a)} & = \sum_{k=0}^{m-1}\sum_{n \ge ka} \Big\lfloor \frac {n+ka}{ma} \Big\rfloor x^n. \end{align*} On the other hand we know \begin{align*} \frac{x^a}{(1-x)(1-x^a)} & = \sum_{n \ge 0} \Big\lfloor \frac {n+a}a \Big\rfloor x^{n+a} = \sum_{n \ge a} \Big\lfloor \frac {n}a \Big\rfloor x^n. \end{align*} Comparing the coefficient of $x^n$ and setting $z = \frac n{ma}$ we find $$ \lfloor mz \rfloor = \Big \lfloor \frac na \Big\rfloor = \sum_{k=0}^{m-1} \Big\lfloor \frac {n+ka}{ma} \Big\rfloor = \sum_{k=0}^{m-1} \Big\lfloor z + \frac {k}{m} \Big\rfloor. $$ This proves Hermite's identity for rational values of $z$. By piecewise continuity it holds for all real numbers.