On a multisection of a series Let $x$ be a real number and $\lfloor{x}\rfloor$ denote the greatest integer less than or equal to $x$. Let $r$ be a positive integer. Let $a$ and $n$ be nonnegative integers such that $n-a+1\geq r$. Prove that
$$ \sum_{k=a}^{n} f(k) = \sum_{j=0}^{r-1} \sum_{k=\lfloor{\frac{a+r-1-j}{r}}\rfloor}^{\lfloor{\frac{n-j}{r}}\rfloor} f(rk+j).$$
I couldn't arrive at any proof.
Also it is noted that for a fixed $r$, this identity partitions the sum modulo remainder classes. The number of remainder classes is indexed by $j$. Could you explain this note?
I generally have trouble with series whose indices involves greatest integer function, are there any fast techniques to solve them?

This is an answer  based  on    series   manipulation.  In order to ease  things we make a two-step approach. At first we show the identity is valid in case $a=0$ and then the generalisation for integral $a\geq 0$  follows easily.

Step 1: $a=0$
The following is valid for integral $1\leq r\leq n$:
\begin{align*}
\sum_{k=0}^nf(k)=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n-j}{r}\right\rfloor}f(rk+j)\tag{1}
\end{align*}

Observe      the simple lower limit $k=0$ of   the inner sum. Setting  $a=0$ we  have
\begin{align*}
\left\lfloor\frac{a+r-1-j}{r}\right\rfloor=\left\lfloor\frac{r-1-j}{r}\right\rfloor=0
\end{align*}
since    $0\leq  \frac{r-1-j}{r}<1$ if $0\leq  j<r$.
At first we consider (1) when $n$ is an integer multiple of $r$. In this case we see
\begin{align*}
\sum_{k=0}^nf(k)&=\sum_{k=0}^{\frac{n}{r}-1}\sum_{j=0}^{r-1}f(rk+j)+f(n)\\
&=\sum_{j=0}^{r-1}\sum_{k=0}^{\frac{n}{r}-1}f(rk+j)+f(n)\tag{2}
\end{align*}
The multisection in (2) corresponds to a partitioning of $[0,n]\cap\mathbb{Z}$ into $r$ equally sized, right half-open intervals $$[rk,rk+r)\cap\mathbb{Z}, \qquad\qquad 0\leq k<\frac{n}{r}$$ of length $\frac{n}{r}$ together with the right-most point $\{n\}$.

Now we consider $0\leq r\leq n$. In this more general setting we have
  \begin{align*}
n=r\left\lfloor\frac{n}{r}\right\rfloor+l\qquad\qquad 0\leq l <\left\lfloor\frac{n}{r}\right\rfloor\tag{3}
\end{align*}
We  obtain from (2) and (3)
  \begin{align*}
\color{blue}{\sum_{k=0}^n f(k)}&=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor-1}f(rk+j)
+\sum_{j=0}^l f\left(r\left\lfloor\frac{n}{r}\right\rfloor+j\right)\tag{4}\\
&=\sum_{j=0}^{l}\sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor}f(rk+j)
+\sum_{j=l+1}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor-1}f(rk+j)\tag{5}\\
&=\sum_{j=0}^{r-1}\,\,\ \sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor+\left\lfloor\frac{l-j}{r}\right\rfloor}f(rk+j)\tag{6}\\
&=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{r\lfloor  n/r\rfloor +l-j}{r}\right\rfloor}  f(rk+j)\tag{7}\\
&\,\,\color{blue}{=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)}\tag{8}\\
\end{align*}
  and the claim (1) follows.

Comment:


*

*In (4) we use $\left\lfloor\frac{n}{r}\right\rfloor$ instead of $\frac{n}{r}$ and instead of the single value $f(n)$ we have now to consider a sum ranging over $0\leq j\leq l$ according to (3) resulting in the right-hand sum.

*In (5) we split the left-hand of (4) into two double sums with index region $0\leq j\leq l$ and $l<j\leq r-1$. This way we can merge the right-hand sum of (4) into the left-hand sum. Note the upper limit of the inner sum from the left-hand sum is $\left\lfloor\frac{n}{r}\right\rfloor$ (and no longer $\left\lfloor\frac{n}{r}\right\rfloor-1$).

*In (6) we collect both double sums of (5) into one double sum with a small trick. We add
\begin{align*}
\left\lfloor\frac{l-j}{r}\right\rfloor=
\begin{cases}
0&\quad 0\leq j\leq l\\
-1&\quad l< j\leq r-1
\end{cases}
\end{align*}

*In (7) we observe $m:=\left\lfloor\frac{n}{r}\right\rfloor$ is an integer and we use the identity
$$m+\left\lfloor\frac{l-j}{r}\right\rfloor=\left\lfloor\frac{rm+l-j}{r}\right\rfloor.$$

*In (8) we use $n=r\left\lfloor\frac{n}{r}\right\rfloor+l$ from (3).

Step 2: $a\geq 0$
We obtain from (8)
  \begin{align*}
\color{blue}{\sum_{k=a}^nf(k)}
&=\sum_{k=0}^n f(k)-\sum_{k=0}^{a-1} f(k)\\
&=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)
-\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{a-1-j}{r}\right\rfloor} f(rk+j)\\
&=\sum_{j=0}^{r-1}\ \sum_{k=\left\lfloor\frac{a-1-j}{r}\right\rfloor+1}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)\\
&\,\,\color{blue}{=\sum_{j=0}^{r-1}\ \sum_{k=\left\lfloor\frac{a+r-1-j}{r}\right\rfloor}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)}\\
\end{align*}
and OPs claim follows.

