What type of Hypergeometric series is this? 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?
 A: [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.
A: $\sum\limits_{n=0}^\infty\dfrac{1}{n!}\dfrac{1}{n+1}(-z)^n{}_2F_2\left(m,n+1;\dfrac{1}{2},n+2;bz\right)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(n+1)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_k(n+2)_k(n+1)!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_kn!k!(n+k+1)}$
$=\dfrac{1}{z}\int_0^z\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_kn!k!}~dz$
$=\dfrac{1}{z}\int_0^z\sum\limits_{k=0}^\infty\dfrac{(m)_kb^kz^ke^{-z}}{\left(\dfrac{1}{2}\right)_kk!}~dz$
$=-\dfrac{1}{z}\left[\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(m)_kb^kz^ne^{-z}}{\left(\dfrac{1}{2}\right)_kn!}\right]_0^z$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\dfrac{1}{z}\sum\limits_{k=0}^\infty\dfrac{(m)_kb^k}{\left(\dfrac{1}{2}\right)_k}-\dfrac{1}{z}\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(m)_kb^kz^ne^{-z}}{\left(\dfrac{1}{2}\right)_kn!}$
$=\dfrac{1}{z}{}_2 F_1\left(1,m;\dfrac{1}{2};b\right)-\dfrac{1}{z}\sum\limits_{n=0}^\infty\sum\limits_{k=n}^\infty\dfrac{(m)_kb^kz^ne^{-z}}{\left(\dfrac{1}{2}\right)_kn!}$
$=\dfrac{1}{z}{}_2 F_1\left(1,m;\dfrac{1}{2};b\right)-\dfrac{1}{z}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_{n+k}b^{n+k}z^ne^{-z}}{\left(\dfrac{1}{2}\right)_{n+k}n!}$
$=\dfrac{1}{z}{}_2 F_1\left(1,m;\dfrac{1}{2};b\right)-\dfrac{e^{-z}}{z}\Phi_1\left(m,1,\dfrac{1}{2};b,bz\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)
A: 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.
