Here is the goal sum where the [Pochhammer Symbol](https://mathworld.wolfram.com/PochhammerSymbol.html) with the [Incomplete Beta function series](https://functions.wolfram.com/GammaBetaErf/Beta3/06/01/03/01/01/0003/)


$$\sum\_{m=0}^\infty \frac{\text B\_z(m+a,b-m)x^m}{m!}=\sum\_{m=0}^\infty z^{m+a}\sum\_{n=0}^\infty\frac{(1-(b-m))\_nz^n x^m}{(m+a+n)m!n!}=\frac{z^a}a\sum\_{m,n\ge0}\frac{(a)\_{m+n} (b-1)\_{m+n}(xz)^mz^n}{(a+1)\_{m+n}(b-1)\_m m! n!}$$


Now use the [Kampé de Fériet](https://www.researchgate.net/publication/341441364_Novel_reductions_of_Kampe_de_Feriet_function) function also [found on Wolfram Mathworld](https://mathworld.wolfram.com/KampedeFerietFunction.html):


$$\text F^{p,r,u}\_{q,s,v}\left(^{a\_1,…,a\_p;c\_1,…,c\_r;f\_1,…,f\_u}\_{b\_1,…,b\_q;d\_1,…,d\_s;g\_1,…,g\_v}\ x,y\right)\mathop=^\text{def}\sum\_{m=0}^\infty\sum\_{n=0}^\infty\frac{\prod\limits\_{j=1}^p(a\_j)\_{m+n} \prod\limits\_{j=1}^r(c\_j)\_m \prod\limits\_{j=1}^u (f\_j)\_n x^my^n}{\prod\limits\_{j=1}^q (b\_j)\_{m+n} \prod\limits\_{j=1}^s(d\_j)\_m \prod\limits\_{j=1}^v(g\_j)\_n m!n!}$$


Therefore:


$$\sum\_{m=0}^\infty \frac{\text B\_z(m+a,b-m)x^m}{m!}\mathop= ^{|z|,-a\not\in\Bbb N}\frac{z^a}a \text F^{2,0,0}\_{1,1,0}\left(^{a,b-1;;}\_{a+1;b-1;}\ xz,z\right)$$


The second sum of interest is:


$$\sum\_{m=0}^\infty \text B\_z(m+a,b-m)x^m\mathop= ^{|z|,-a\not\in\Bbb N}\frac{z^a}a \text F^{2,1,0}\_{1,1,0}\left(^{a,b-1;1;}\_{a+1;b-1;}\ xz,z\right)$$


Is there a simpler closed form or decomposition formula for this case of the Kampé de Fériet function? A solution verification is also needed, but the derivation should be correct. Please give feedback.



Simplification of $\sum_{m=0}^\infty \text B_z(m+a,b-m)x^m,\sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!}$ in terms of Kampé de Fériet function
I'm not sure if this is considered simpler, but still:
\begin{split}
\sum\_{m=0}^\infty \frac{\text B\_z(m+a,b-m)x^m}{m!} &= \sum\_{m=0}^\infty \frac{x^m}{m!}\int\_0^z y^{m+a-1}(1-y)^{b-m-1}\,{\rm d}y\\
&=\int\_0^z y^{a-1}(1-y)^{b-1} e^{\frac{xy}{1-y}}\,{\rm d}y.
\end{split}
Similarly,
$$\sum\_{m=0}^\infty \text B\_z(m+a,b-m)x^m =\int\_0^z \frac{y^{a-1}(1-y)^b}{1-y-xy}\,{\rm d}y.$$


