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Any ideas how to find a closed form for the sum given by: $$ \sum^\infty_{n=0} \frac{1}{n!} \frac{a^n b^{n+m}}{(m+n)^2 \Gamma(m+n)} {}_2F_2 \left(m+n,m+n;m+n+1,m+n+1;-b\right) $$

Given that both $a$ and $b$ are positive real numbers, and $m$ is a nonzero positive integer.

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$\sum\limits_{n=0}^\infty\dfrac{1}{n!}\dfrac{a^nb^{n+m}}{(m+n)^2\Gamma(m+n)}{_2F_2}\left(m+n,m+n;m+n+1,m+n+1;-b\right)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^nb^{n+m}\Gamma(m+n)(m+n)_k(m+n)_k(-b)^k}{\Gamma(m+n+1)\Gamma(m+n+1)n!(m+n+1)_k(m+n+1)_kk!}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(n+k+m)\Gamma(n+k+m)(-1)^ka^nb^{n+k+m}}{\Gamma(n+m)\Gamma(n+k+m+1)\Gamma(n+k+m+1)n!k!}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{m\Gamma(m+1)(m)_{n+k}(m)_{n+k}(-1)^ka^nb^{n+k+m}}{(m)_n(m+1)_{n+k}(m+1)_{n+k}n!k!}$

$=b^mm!m\mathrm{F}^{2:0;0}_{2:1;0}\Bigg(\begin{matrix}m,m&:&-&;&-&\\m+1,m+1&:&m&;&-&\end{matrix}\Bigg|ab,-b\Bigg)$

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  • $\begingroup$ Thanks Harry ... I see you have used Kampé de Fériet. $\endgroup$
    – Sophie
    Jun 16, 2015 at 15:38

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