Use Z-transform on the sequence $x(n) = \frac 1{(n+1)(n+2)...(n+p)}$ Use Z-transform on the sequence $x(n) = \frac 1{(n+1)(n+2)...(n+p)}, $ $n\geq 0$. I know that the Z-transform on a sequence is:
$$Z\{x(n)\}=\sum_{n=0}^{\infty}x(n)z^{-n}$$
But with this exercise I should arrive at something like:
$$Z\{x(n)\}=\frac {(1-z)^p}{(p-1)!}(\frac 1{1-z}z\ln |\frac {z}{z-1}|+z\sum_{k=2}^{p}\frac 1{(k - 1)(1-z)^k})$$
I should do some algebraic manipulation but I have no idea where to start.

Note that $x(n)$ can be expressed in terms of the Falling and Rising Factorial
as
$$
x(n) = {1 \over {\left( {n + 1} \right)^{\overline {\,p\,} } }} = n^{\,\underline {\, - \,p\,} } 
$$
The Falling and Rising Factorials have a simple expression for the finite difference and summation
$$
\left\{ \matrix{
  \Delta _{\,n} \;n^{\,\underline {\,q\,} }  = \left( {n + 1} \right)^{\,\underline {\,q\,} }  - n^{\,\underline {\,q\,} }  = qn^{\,\underline {\,q - 1\,} }  \hfill \cr 
  \sum\limits_{a\, \le \,k\, \le \,b} {k^{\,\underline {\,q\,} } }  = {{\left( {b + 1} \right)^{\,\underline {\,q + 1\,} } } \over {q + 1}} - {{a^{\,\underline {\,q + 1\,} } } \over {q + 1}}
\quad \left| {\;a,b \in \mathbb Z} \right. \hfill \cr}  \right.
$$
To get the Z transform it is better to consider the partial fraction decomposition of the Falling factorial with negative integral exponent, which is
easy to demonstrate to be
$$
{1 \over {x^{\,\overline {\,q + 1\,} } }}\quad \quad \left| {\;0 \le {\rm integer }q} \right.\quad  = {1 \over {q!}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  q \cr 
  k \cr}  \right){1 \over {x + k}}} 
$$
Therefore
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {{{z^{\, - \,n} } \over {\left( {n + 1} \right)^{\,\overline {\,p\,} } }}} \quad \quad \left| {\;1 \le {\rm integer }p} \right.\quad  =   \cr 
  &  = {1 \over {\left( {p - 1} \right)!}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,p - 1} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  p - 1 \cr 
  k \cr}  \right)\sum\limits_{0\, \le \,n} {{{z^{\, - \,n} } \over {n + 1 + k}}} }  =   \cr 
  &  = {1 \over {\left( {p - 1} \right)!}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,p - 1} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  p - 1 \cr 
  k \cr}  \right)z^{\,k} \sum\limits_{k\, \le \,m} {{{z^{\, - \,m} } \over {m + 1}}} }  =  \cr} 
$$
Can you continue from here ?
