Simplify summation of factorials

Question

Hello I guess this equality is true but I don't know how to solve it.

$$\sum_{x=0}^{m(1-\text{sel})} (m-1-x)! (m \cdot \text{sel}) \frac{(m(1-\text{sel}))!}{(m(1-\text{sel})-x)!}(x+1) = \frac{(m+1)!}{1+m \cdot \text{sel}}$$

Answer 1

I am assuming that $sel=\frac{n}{m}$, then $$ \begin{align} &\sum_{k=0}^{m-n}(m-1-k)!n\frac{(m-n)!}{(m-n-k)!}(k+1)\\ &=\sum_{k=0}^{m-n}\binom{m-1-k}{n-1}(m-n)!n!(k+1)\\ &=\sum_{k=0}^{m-n}\binom{m-1-k}{n-1}(m-n)!n!((m+1)-(m-k))\\ &=\sum_{k=0}^{m-n}\binom{m-1-k}{n-1}(m+1)(m-n)!n!\\ &-\sum_{k=0}^{m-n}\binom{m-k}{n}n(m-n)!n!\\ &=\binom{m}{n}(m+1)(m-n)!n!\\ &-\binom{m+1}{n+1}n(m-n)!n!\\ &=\binom{m}{n}(m+1)(m-n)!n!\\ &-\binom{m}{n}(m+1)(m-n)!n!\frac{n}{n+1}\\ &=\binom{m}{n}(m+1)(m-n)!n!\frac1{n+1}\\ &=\frac{(m+1)!}{n+1} \end{align} $$

Answer 2

If $N$ is a non-negative integer, then the $N$th rising power of $x$ is defined as the following product of $N$ factors: $x^\overline{N}=x(x+1)\cdots(x+N-1)$. For a positive integer $K$, the calculus of rising powers gives $$\sum_{k=1}^K k^\overline{N}={K^\overline{N+1}\over N+1}.$$

Therefore, if $M\leq N$ are non-negative integers $$\begin{eqnarray*} \sum_{x=0}^M {(N-x)!\over (M-x)!}(x+1) &=& \sum_{j=0}^M \sum_{x=j}^M {(N-x)!\over (M-x)!}\\[5pt] &=& \sum_{j=0}^M \sum_{k=1}^{M+1-j} k^\overline{N-M} \\[5pt] &=& {\sum_{j=0}^M (M+1-j)^\overline{N-M+1}\over (N-M+1)} \\[5pt] &=& {\sum_{k=1}^{M+1} k^\overline{N-M+1}\over (N-M+1)} \\[5pt] &=& {(M+1)^\overline{N-M+2}\over (N-M+1)(N-M+2)}.\tag1 \end{eqnarray*}$$

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Assume that $j$ and $m$ are positive integers and that "sel" is $j/m$. Writing $s$ for "sel", your expression is $$\sum_{x=0}^{m(1-s)} {(m-1-x)!\over (m(1-s)-x)!}(x+1)\cdot ms\,(m(1-s))! \tag2$$ Setting $N=m-1$ and $M=m(1-s)=m-j$ in (1), we see that (2) reduces to $$ {(m-j+1)^\overline{j+1}\over (ms)(ms+1)}\cdot ms\,(m-j)!={(m+1)!\over ms+1},$$ as you expected.