I am trying to find a closed form for the series
$$ \sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right)$$
$m$ is a nonzero positive integer, and $b$, $z$ are positive real numbers. I to rewrite the sum as
$$ \sum^\infty_{n=0} \sum^\infty_{q=0} \frac{1}{n!} \frac{1}{q!} \frac{(m)_q}{(\frac{1}{2})_q} \frac{(1)_{q+n}}{(2)_{q+n}} (-z)^n (b z)^q$$
any idea what type of multi-variable hypergeometric function is the last equation?
On
Here is a suggestion, not a complete answer.
Since $n+q$ seems prevalent, let $k = n+q$.
$\begin{align} \sum^\infty_{n=0} \sum^\infty_{q=0} \frac{1}{n!} \frac{1}{q!} \frac{(m)_q}{(\frac{1}{2})_q} \frac{(1)_{q+n}}{(2)_{q+n}} (-z)^n (b z)^q &=\sum^\infty_{n=0} \sum^\infty_{q=0} \frac{1}{n!} \frac{1}{q!} \frac{(m)_q}{(\frac{1}{2})_q} \frac{(1)_{q+n}}{(2)_{q+n}} z^{n+q} (-1)^n b^q\\ &=\sum^\infty_{k=0} \sum^k_{q=0} \frac{1}{(k-q)!} \frac{1}{q!} \frac{(m)_q}{(\frac{1}{2})_q} \frac{(1)_{k}}{(2)_{k}} z^{k} (-1)^{k-q} b^q\\ &=\sum^\infty_{k=0} \frac{(-1)^k(1)_{k}}{(2)_{k}} z^{k} \sum^k_{q=0} \frac{1}{(k-q)!} \frac{1}{q!} \frac{(m)_q}{(\frac{1}{2})_q} (-1)^{q} b^q\\ \end{align} $
The next step would be to do something with the inner sum, but I'll stop here, since I'm not much of a hypergeometric function expert.
[Too long for a comment]. If you denote $$ G(z)=\sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right),$$ then certain combination of $G(z)$ and $G'(z)$ is a simpler single function: $$\frac{d}{dz}\Bigl[zG(z)\Bigr]=e^{-z}{}_1F_1\left(m,\frac12;bz\right).$$ Therefore, $$G(z)=\frac{1}{z}\int_0^ze^{-t}{}_1F_1\left(m,\frac12;bt\right)dt.$$ Also, for integer $m$ the $_1F_1$ function can be written in terms of error function.