Let $p,q,\in \Bbb N$ such that $q<p$ then set $$\sum_{n=1}^\infty\frac{1}{{(n+q)(n+q+1)…(n+p)}}$$ I want the explicit formula for $S_{q,p}?$
I know that by telescoping sum we have, $$S_{0,p} = \sum_{n=1}^\infty\frac{1}{{n(n+1)(n+2)…(n+p)}}=\frac{1}{p!p}$$
See here, Calculate the infinite sum $\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)} $
What could be the suitable formula for $S_{q,p}?$
Another approach is to note that the summand can be expressed as $$ {1 \over {\left( {n + q} \right)\left( {n + q + 1} \right) \cdots \left( {n + p} \right)}} = {1 \over {\left( {n + q} \right)^{\;\overline {\,p - q + 1} } }} = \left( {n + q - 1} \right)^{\,\underline {\, - \,\left( {p - q + 1} \right)} } $$ where $x^{\;\overline {\,m\,} } $ is the Rising Factorial and $x^{\,\underline {\,m\,} } $ is the Falling Factorial.
Then the Indefinite Sum (anti-difference) of the Falling factorial is $$ \sum\nolimits_{\;x} {x^{\,\underline {\,m\,} } } = \left\{ {\matrix{ {{1 \over {m + 1}}\;x^{\,\underline {\,m + 1\,} } + c} & { - 1 \ne m} \cr {\psi (x + 1) + c} & { - 1 = m} \cr } } \right. $$ which is not difficult to demonstrate.
Therefore $$ \eqalign{ & \sum\limits_{n = 1}^\infty {{1 \over {\left( {n + q} \right)\left( {n + q + 1} \right) \cdots \left( {n + p} \right)}}} = \cr & = \sum\limits_{n = 1}^\infty {\left( {n + q - 1} \right)^{\,\underline {\, - \,\left( {p - q + 1} \right)\,} } } = \sum\limits_{k = q}^\infty {k^{\,\underline {\, - \,\left( {p - q + 1} \right)\,} } } = \cr & = {1 \over {p - q}}\;q^{\,\underline {\, - \,\left( {p - q} \right)\,} } = {1 \over {\left( {p - q} \right)\left( {q + 1} \right)^{\overline {\,\left( {p - q} \right)\,} } }} = {1 \over {\left( {p - q} \right)\left( {q + 1} \right)^{\overline {\, - \,q\,} } 1^{\overline {\,p\,} } }} = \cr & = {{q!} \over {\left( {p - q} \right)p!}}\quad \left| {\;p \le q} \right. \cr} $$
The same telescoping technique works for $S_{q,p}$ as well: $$ \frac{1}{{(n+q)(n+q+1)\cdots(n+p)}} = \frac{1}{p-q}\cdot\frac{(n+p)-(n+q)}{{(n+q)(n+q+1)\cdots(n+p)}} \\ = \frac{1}{p-q} \left( \frac{1}{{(n+q)\cdots(n+p-1)}} - \frac{1}{{(n+q+1)\cdots(n+p)}}\right) $$ which leads to $$ S_{q,p} = \frac{1}{p-q} \cdot \frac{1}{(q+1)\cdots p} = \frac{q!}{(p-q) p!} $$
One can take a hypergeometric view with the following calculation.
\begin{align} \sum_{n=1}^\infty\frac{1}{{(n+q)(n+q+1)…(n+p)}} &= \sum_{n=1}^{\infty} \frac{\Gamma(n+q)}{\Gamma(n+p+1)} = \sum_{n=0}^{\infty} \frac{\Gamma(n+q+1)}{\Gamma(n+p+2)} \\ &= \sum_{n=1}^{\infty} \frac{\Gamma(q+1)}{\Gamma(p+2)} \, \frac{(1)_{n} \, (q+1)_{n}}{n! \, (p+2)_{n}} \\ &= \frac{\Gamma(q+1)}{\Gamma(p+2)} \, {}_{2}F_{1}(q+1, 1; p+2; 1) \\ &= \frac{\Gamma(q+1)}{\Gamma(p+2)} \, \frac{\Gamma(p+2) \, \Gamma(p-q)}{\Gamma(p+1) \, \Gamma(p-q+1)} \\ &= \frac{\Gamma(q+1)}{(p-q) \, \Gamma(p+1)}, \end{align} where $p \neq q$and $(a)_{n}$ is the Pochhammer symbol.
This sum also telescopes: $$\frac1{n(n+1)\cdots(n+p-1)}-\frac1{(n+1)(n+2)\cdots(n+p)} =\frac p{n(n+1)\cdots(n+p)}$$ etc.
