Hermite's identity for sum of floor function 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.
 A: 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.
