A series question whose indices involves greatest integer function

Question

Let $x$ be a real number and $\lfloor{x}\rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $n\geq a+1$. Prove that

$$ \sum_{k=a}^{n} f(k) - \sum_{k=a}^{n} (-1)^{k} f(k) = 2 \sum_{k=\lfloor{\frac{a+2}{2}}\rfloor}^{\lfloor{\frac{n+1}{2}}\rfloor} f(2k-1).$$

I proved this by considering the situations where $a$ and $n$ takes even or odd values respectively. Are there any other way to show the equation is true? I generally have trouble with series whose indices involves greatest integer function, are there fast techniques to solve them?

Answer 1

The index of summation is ranging as $$ a \le k \le n $$

Replace it with its even and odd components $$ k = 2j - i\quad \left| {\,i = 0,1} \right. $$

Then for the even component it shall be $$ \eqalign{ & i = 0\quad \Rightarrow \quad a \le 2j \le n\quad \Rightarrow \quad \left\lceil {{a \over 2}} \right\rceil \le j \le \left\lfloor {{n \over 2}} \right\rfloor \quad \Rightarrow \cr & \Rightarrow \quad \left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor \cr} $$ and for the odd $$ \eqalign{ & i = 1\quad \Rightarrow \quad a \le 2j - 1 \le n\quad \Rightarrow \quad \left\lceil {{{a + 1} \over 2}} \right\rceil \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor \quad \Rightarrow \cr & \Rightarrow \quad \left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor \cr} $$ Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.

The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component $$ \eqalign{ & \sum\limits_{a\, \le \,k\, \le \;n} {f(k)} = \sum\limits_{a\, \le \,k\, \le \;n} {\left. {f(k)\,} \right|_{\,k = 2j} + \left. {f(k)\,} \right|_{\,k = 2j - 1} } = \cr & = \sum\limits_{a\, \le \,2j\, \le \;n} {f(2j)} + \sum\limits_{a\, \le \,2j - 1\, \le \;n} {f(2j - 1)} = \cr & = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {f(2j)} + \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {f(2j - 1)} \cr} $$ and same for that of $(-1)^k f(k)$ $$ \eqalign{ & \sum\limits_{a\, \le \,k\, \le \;n} {\left( { - 1} \right)^{\,k} f(k)} = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {\left( { - 1} \right)^{\,2j} f(2j)} + \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {\left( { - 1} \right)^{\,2j - 1} f(2j - 1)} = \cr & = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {f(2j)} - \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {f(2j - 1)} \cr} $$

Now, just subtract the two above to confirm the thesis.

--- addendum ----

Concerning your request on how to transform ceiling <-> floor, consider that $$ \eqalign{ & n = 2\left\lceil {{n \over 2}} \right\rceil + 2\left\{ {{n \over 2}} \right\} = 2q + i\quad \left| \matrix{ \;n,q \in Z \hfill \cr \,i = 0,1 \hfill \cr} \right. \cr & \left\lceil {{n \over 2}} \right\rceil = \left\lceil {{{2q + i} \over 2}} \right\rceil = \left\lceil {q + {i \over 2}} \right\rceil = q + \left\lceil {{i \over 2}} \right\rceil = q + i \cr & \left\lfloor {{{n + 1} \over 2}} \right\rfloor = \left\lfloor {{{2q + 1 + i} \over 2}} \right\rfloor = \left\lfloor {q + {{1 + i} \over 2}} \right\rfloor = q + \left\lfloor {{{1 + i} \over 2}} \right\rfloor = q + i \cr} $$ and, in general $$ \left\lceil {{n \over m}} \right\rceil = \left\lfloor {{{n + m - 1} \over m}} \right\rfloor \quad \left| \matrix{ \;n,m \in Z \hfill \cr \;1 \le m \hfill \cr} \right. $$ re. for details to this article.