How to compute: $S_{q,p} =\sum\limits_{n=1}^\infty\frac{1}{{(n+q)(n+q+1)…(n+p)}}$ 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 $S_{q,p}?$

I have completely change the problem this is more general.
 A: 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
$$
\eqalign{
  & \sum\limits_{n = 1}^\infty  {{1 \over {\left( {n + q} \right)\left( {n + q + 1} \right) \cdots \left( {n + p} \right)}}}  =   \cr 
  &  = \sum\limits_{n = 1}^\infty  {\left( {n + q - 1} \right)^{\,\underline {\, - \,\left( {p - q + 1} \right)\,} } }  = \sum\limits_{k = q}^\infty  {k^{\,\underline {\, - \,\left( {p - q + 1} \right)\,} } }  =   \cr 
  &  = {1 \over {p - q}}\;q^{\,\underline {\, - \,\left( {p - q} \right)\,} }  = {1 \over {\left( {p - q} \right)\left( {q + 1} \right)^{\overline {\,\left( {p - q} \right)\,} } }} = {1 \over {\left( {p - q} \right)\left( {q + 1} \right)^{\overline {\, - \,q\,} } 1^{\overline {\,p\,} } }} =   \cr 
  &  = {{q!} \over {\left( {p - q} \right)p!}}\quad \left| {\;p \le q} \right. \cr} 
$$
A: 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
$$
S_{q,p} = \frac{1}{p-q} \cdot \frac{1}{(q+1)\cdots p} = \frac{q!}{(p-q) p!}
$$
A: 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. 
A: 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.
