Simplifying $\sum_{n=1}^{\infty}\frac{n\alpha^n}{(n-m)!(n+z)}(x-c)^{n-m}$ We would like to simplify the following summation
$$\sum_{n=1}^{\infty}\frac{n\alpha^n}{(n-m)!(n+z)}(x-c)^{n-m}.$$
What we know is that $x, z, \alpha, c > 0$ and $z$ is integer.
Here's what I did so far
\begin{align*}
\sum_{n=1}^{\infty}\frac{n\alpha^n}{(n-m)!(n+z)}(x-c)^{n-m}&=\sum_{n=1}^{\infty}\frac{n\alpha^n}{(n-m)!}(x-c)^{n-m}\int_{0}^{1}t^{n+z-1}dt \\
&=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{n\alpha^n}{(n-m)!}(x-c)^{n-m}t^{n}t^{z-1}dt \\
&=\int_{0}^{1}t^{z-1}\sum_{n=1}^{\infty}\frac{n\alpha^n}{(n-m)!}(x-c)^{n-m}t^{n}dt \\
&= \int_{0}^{1}t^{z-1}\sum_{n=1}^{\infty}\frac{n(\alpha t)^n}{(n-m)!}(x-c)^{n-m}dt \\
\end{align*}
But I don't know where to go from there. Some parts of the sum look familiar but I'm not sure what to do. Anyone could give a hint please?
 A: It may be expressed as a general hypergeometric:
$$
\sum _{n=1}^{\infty }{\frac {n{\alpha}^{n} \left( x-c \right) ^{n-m}}{
 \left( n-m \right) !\, \left( n+z \right) }}
={\frac {\alpha\;
{\mbox{$_2$F$_2$}(2,1+z;\,2-m,2+z;\,\alpha\, \left( x-c \right) )}}{
 \left( 1-m \right) !\, \left( 1+z \right)  \left( x-c \right) ^{m-1}
}}
$$

It is simply the definition of the hypergeometric series.
Use notation $x^{(n)}$ for the "rising factorial"
$x^{(n)} = \prod_{j=0}^{n-1}(x+j)$.
By definition,
$$
{}_2F_2\big(2,1+z;2-m,2+z;\alpha (x-c) \big)
= \sum_{k=0}^\infty\frac{2^{(k)} (1+z)^{(k)}}{(2-m)^{(k)}(2+z)^{(k)}k!}\big(\alpha(x-c)\big)^k
$$
But note $2^{(k)}/k! = (k+1)$ and
$(1+z)^{(k)}/(2+z)^{(k)} = (1+z)/(1+k+z)$, so
$$
{}_2F_2\big(2,1+z;2-m,2+z;\alpha (x-c) \big) =
\sum_{k=0}^\infty \frac{(k+1)(1+z)}{(2-m)^{(k)}(1+k+z)}\big(\alpha(x-c)\big)^k
\\
\frac{\alpha\; {}_2F_2\big(2,1+z;2-m,2+z;\alpha (x-c) \big)}{(1-m)!(1+z)(x-c)^{m-1}}
=
\sum_{k=0}^\infty \frac{(k+1)}{(1+k-m)!(1+k+z)}
\alpha^{k+1}(x-c)^{k-m+1}
$$
Change the index, $n=k+1$,
$$
\frac{\alpha\; {}_2F_2\big(2,1+z;2-m,2+z;\alpha (x-c) \big)}{(1-m)!(1+z)(x-c)^{m-1}}
=
\sum_{n=1}^\infty \frac{n}{(n-m)!(n+z)}
\alpha^{n}(x-c)^{n-m}
$$
