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?

Solution 1:

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, $n=k+1$, $$ \frac{\alpha\; {}_2F_2\big(2,1+z;2-m,2+z;\alpha (x-c) \big)}{(1-m)!(1+z)(x-c)^{m-1}} = \sum_{n=1}^\infty \frac{n}{(n-m)!(n+z)} \alpha^{n}(x-c)^{n-m} $$