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Prove that according to http://www.iosrjournals.org/iosr-jm/papers/Vol8-issue6/L0866770.pdf?id=7287, this special case of Kampé de Fériet function can reduce to Generalized hypergeometric function:

$\mathrm{F}^{p:0;...;0}_{q:0;...;0}\Bigg(\begin{matrix}a_1,...,a_p&:&-&;&...&;&-&\\b_1,...,b_q&:&-&;&...&;&-&\end{matrix}\Bigg|x_1,...,x_n\Bigg)={_pF_q}(a_1,...,a_p;b_1,...,b_q;x_1+...+x_n)$

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By the definition from your paper $$\begin{align*}&\mathrm{F}^{p:0;...;0}_{q:0;...;0}\Bigg(\begin{matrix}a_1,...,a_p&:&-&;&...&;&-&\\b_1,...,b_q&:&-&;&...&;&-&\end{matrix}\Bigg|x_1,...,x_n\Bigg)=\sum_{s_1,\ldots,s_n=0}^\infty \Lambda(s_1,\ldots s_n)\frac{x_1^{s_1}}{s_1!}\ldots \frac{x_n^{s_n}}{s_n!}\\&=\sum_{s_1,\ldots,s_n=0}\frac{\prod_{j=1}^p(a_j)_{s_1+\ldots s_n}}{\prod_{l=1}^q(b_l)_{s_1+\ldots s_n}}\frac{x_1^{s_1}}{s_1!}\ldots \frac{x_n^{s_n}}{s_n!} \end{align*}.$$ In the $1$-variable case, we see from your link to wikipedia that $${_pF_q}\Bigg(\begin{matrix}a_1,...,a_p\\b_1,...,b_q\end{matrix}\Bigg|x\Bigg)=\sum_{k=0}^\infty\frac{\prod_{j=1}^p(a_j)_{k}}{\prod_{l=1}^q(b_l)_{k}} \frac{x^k}{k!}.$$ Thus it's natural that the definition of the right hand side is $${_pF_q}(a_1,...,a_p;b_1,...,b_q;x_1+...+x_n)=\sum_{s_1,\ldots,s_n=0}\frac{\prod_{j=1}^p(a_j)_{s_1+\ldots s_n}}{\prod_{l=1}^q(b_l)_{s_1+\ldots s_n}}\frac{x_1^{s_1}}{s_1!}\ldots \frac{x_n^{s_n}}{s_n!}$$

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  • $\begingroup$ But how to prove $\sum\limits_{s_1,\ldots,s_n=0}\dfrac{\prod\limits_{j=1}^p(a_j)_{s_1+\ldots s_n}}{\prod\limits_{l=1}^q(b_l)_{s_1+\ldots s_n}}\dfrac{x_1^{s_1}}{s_1!}\ldots\dfrac{x_n^{s_n}}{s_n!}=\sum\limits_{k=0}^\infty\dfrac{\prod\limits_{j=1}^p(a_j)_{k}}{\prod\limits_{l=1}^q(b_l)_{k}} \dfrac{(x_1+...+x_n)^k}{k!}$ ? Natural is not the reasonable reason in maths! $\endgroup$ – Harry Peter Apr 25 '15 at 11:56
  • $\begingroup$ @HarryPeter Just consider those with same $s_1+\ldots+ s_n$. And notice the multinomial theorem! See here $\endgroup$ – Golbez Apr 25 '15 at 12:12

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