<s>[INST] 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?
 [/INST] 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.

</s>