How to obtain the identity $\sum_k\frac{(m-k)!}{(n-k)!} = \frac{(m+1-k)!}{(m+1-n)(n-k)!}$

Question

Hereby $$\sum_k\frac{(m-k)!}{(n-k)!} = \frac{(m+1-k)!}{(m+1-n)(n-k)!} =: SOL(k)$$ is supposed to be similar to the way of writing antiderivatives. In complete form it should be like this:
$$\sum_{k=a}^b \frac{(m-k)!}{(n-k)!} = SOL(b)-SOL(a)$$

I'm not completely sure whether it actually holds for all $a,b$.

Okay, using Pascal's identity, I get: $$\sum_{k=a}^b \binom{m-k}{m-n}=\\ \binom{m-a}{m-n}+\sum_{k=a+1}^b \binom{m-k}{m-n}\\ = \binom{m-a-1}{m-n} +\binom{m-a-1}{m-n-1}+\sum_{k=a+1}^b \binom{m-k}{m-n} \\ = \binom{m-a-1}{m-n} +\binom{m-a-1}{m-n-1}+\binom{m-a-1}{m-n}+\sum_{k=a+2}^b \binom{m-k}{m-n}\\ = \binom{m-a-1}{m-n-1}+2\binom{m-a-1}{m-n}+ \sum_{k=a+2}^b \binom{m-k}{m-n}\\ = \binom{m-a-1}{m-n-1}+2\binom{m-a-2}{m-n}+2\binom{m-a-2}{m-n-1} \sum_{k=a+2}^b \binom{m-k}{m-n}\\ =(\sum_{i=a+1}^{b-1} i\binom{m-a-i}{m-n-1}) + b\binom{m-b}{m-n} $$

Answer 1

It looks that your question is relevant to the Indefinite Sum concept.

If we have that $$ \Delta _{\,x} \,F(x) = F(x + 1) - F(x) = f(x) $$ then we write $$ F(x) = \Delta _{\,x} ^{\, - \,1} \,f(x) = \sum\nolimits_{\;x\;} {f(x)} $$ in particular we have $$ F(b) - F(a) = \sum\nolimits_{\;x = \,a\,}^b {f(x)} = - \sum\nolimits_{\;x = \,b\,}^a {f(x)} $$

The relation with the standard definition of the sum of a function of discrete variable is $$ \sum\limits_{k = a}^{b - 1} {f(k)} = \sum\limits_{a\, \le \,k\, < \,b} {f(k)} = \sum\nolimits_{\;k = \,a\,}^b {f(k)} = F(b) - F(a)\quad \left| \matrix{ \;a,b \in \mathbb Z \hfill \cr \;a \le b \hfill \cr} \right. $$

The first two sum are "standard", the third is meant to indicate the Indef. one.
Note the $< $ in the second ! that is exactly what is needed to allow the chain
$\sum\limits_{a\, \le \,k\, < \,b}+\sum\limits_{b\, \le \,k\, < \,c}=\sum\limits_{a\, \le \,k\, < \,c}$ with $\sum\limits_{a\, \le \,k\, < \,a}=0$

corresponding to
$\sum\nolimits_{\;k = \,a\,}^{\;b} +\sum\nolimits_{\;k = \,b\,}^{\;c}=\sum\nolimits_{\;k = \,a\,}^{\;c} \quad;\quad \sum\nolimits_{\;k = \,a\,}^{\;b}+ \sum\nolimits_{\;k = \,b\,}^{\;a}=0\quad \to \quad \sum\nolimits_{\;k = \,a\,}^{\;b} = - \sum\nolimits_{\;k = \,b\,}^{\;a}$

That premised, for the binomials we have $$ \Delta _{\,x} \,\left( \matrix{ x \cr q \cr} \right) = \left( \matrix{ x + 1 \cr q \cr} \right) - \left( \matrix{ x \cr q \cr} \right) = \left( \matrix{ x \cr q - 1 \cr} \right)\quad \Rightarrow \quad \sum\nolimits_{\;x\;} {\left( \matrix{ x \cr q \cr} \right)} = \left( \matrix{ x \cr q + 1 \cr} \right) $$ and therefore $$ \eqalign{ & \Delta _{\,x} \,\left( \matrix{ r - x \cr q \cr} \right) = \left( \matrix{ r - x - 1 \cr q \cr} \right) - \left( \matrix{ r - x \cr q \cr} \right) = - \left( \matrix{ r - x - 1 \cr q - 1 \cr} \right) = \left( { - 1} \right)^q \left( \matrix{ q - r + x - 1 \cr q - 1 \cr} \right) \cr & \sum\nolimits_{\;x\;} {\left( \matrix{ r - x \cr q \cr} \right)} = - \left( \matrix{ r - x + 1 \cr q + 1 \cr} \right) \cr & \sum\limits_{k = 0}^b {\left( \matrix{ r - k \cr q \cr} \right)} = - \left( \matrix{ r - \left( {b - 1} \right) + 1 \cr q + 1 \cr} \right) + \left( \matrix{ r - 0 + 1 \cr q + 1 \cr} \right) \cr} $$ which applies also to complex values of $r,q,b$ if the binomial is defined through the Gamma function.

From here you should be able to conclude for your particular case.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{\infty}{\pars{m - k}! \over \pars{m - n}!}} = \pars{m - n}!\sum_{k = 0}^{\infty}{m - k \choose m - n} \\[5mm] = &\ \pars{m - n}!\sum_{k = 0}^{\infty}\bracks{z^{m - n}} \pars{1 + z}^{m - k} = \pars{m - n}!\bracks{z^{m - n}}\pars{1 + z}^{m}\sum_{k = 0}^{\infty} \pars{1 \over 1 + z}^{k} \\[5mm] = &\ \pars{m - n}!\bracks{z^{m - n}}\pars{1 + z}^{m}\, {1 \over 1 - 1/\pars{1 + z}} = \pars{m - n}!\bracks{z^{m - n + 1}}\pars{1 + z}^{m + 1} \\[5mm] = &\ \bbx{\pars{m - n}!{m + 1 \choose m - n + 1}} \end{align}