TITLE: Infinite series of Hypergeometric function
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: $\sum\limits_{n=0}^\infty\dfrac{1}{n!}\dfrac{a^nb^{n+m}}{(m+n)^2\Gamma(m+n)}{_2F_2}\left(m+n,m+n;m+n+1,m+n+1;-b\right)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^nb^{n+m}\Gamma(m+n)(m+n)_k(m+n)_k(-b)^k}{\Gamma(m+n+1)\Gamma(m+n+1)n!(m+n+1)_k(m+n+1)_kk!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(n+k+m)\Gamma(n+k+m)(-1)^ka^nb^{n+k+m}}{\Gamma(n+m)\Gamma(n+k+m+1)\Gamma(n+k+m+1)n!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{m\Gamma(m+1)(m)_{n+k}(m)_{n+k}(-1)^ka^nb^{n+k+m}}{(m)_n(m+1)_{n+k}(m+1)_{n+k}n!k!}$
$=b^mm!m\mathrm{F}^{2:0;0}_{2:1;0}\Bigg(\begin{matrix}m,m&:&-&;&-&\\m+1,m+1&:&m&;&-&\end{matrix}\Bigg|ab,-b\Bigg)$