Use Z-transform on the sequence $x(n) = \frac 1{(n+1)(n+2)...(n+p)}$

Question

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.

Answer 1

Let us define

$$x_p(n)=\prod_{k=1}^p\frac1{n+k}$$

See then that we have

$$x_p(n)-x_p(n+1)=px_{p+1}(n)$$

and thus,

\begin{align}pZ_{p+1}(z)&=\sum_{n=0}^\infty px_{p+1}(n)z^{-n}\\&=\sum_{n=0}^\infty[x_p(n)-x_p(n+1)]x^{-n}\\&=Z_p(z)-zZ_p(z)+x_p(0)z\\&=(1-z)Z_p(z)+\frac z{p!}\end{align}

Now define

$$pZ_{p+1}(z)=\frac{(1-z)^{p+1}}{(p-1)!}f_{p+1}(z)=\frac{(1-z)^{p+1}}{(p-1)!}f_p(z)+\frac z{p!}$$

Which can then be solved:

$$f_{p+1}(z)=f_p(z)+\frac z{(1-z)^{p+1}p}$$

$$f_p(z)=f_1(z)+\sum_{n=1}^{p-1}\frac z{(1-z)^{n+1}n}$$

$$Z_p(z)=\frac{(1-z)^p}{(p-1)!}\left[f_1(z)+\sum_{n=1}^{p-1}\frac zn(1-z)^{p-n-1}\right]$$

Answer 2

Hint: First solve the problem of finding the $\mathcal{Z}$-transform for ${x}_k(n)=\dfrac{\alpha_k}{n+k}$.

Then determine the partial fraction decomposition of $x(n)$ as

$$x(n)=\prod_{k=1}^p\dfrac{1}{n+k}=\sum_{k=1}^p\dfrac{\alpha_k}{n+k}.$$

The coefficients of the partial fractions decomposition can be obtained by multiplying by $n+k_0$ and calculating the limit for $n\to -k_0$. This will result in

$$\alpha_{k_0} = \prod_{k=1, k\neq k_0}^{p}\dfrac{1}{k-k_0}.$$

From this point, it should be clear how to determine the $\mathcal{Z}$-transform by simply determining the transform of each term in the partial fractions decomposition that we determined.

Answer 3

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 ?