series $\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$ Can anyone please help me with the computation of following series:
$$\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$
My thoughts: Since $$\displaystyle \frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = \frac{\binom{m+n-1}{m}\binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may  be the  probability of a hypergeometric distribution.   
Any help would be appreciated, thanks!
 A: I am going to outline a nasty trick which allows to prove
$$ \sum_{m,n\geq 0}\frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2\sqrt{ab}) \tag{1}$$
and similarly
$$ \sum_{m,n\geq 1}\frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = \sqrt{ab}\, e^{a+b} I_1(2\sqrt{ab})\tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ \frac{1}{(s+t-st)^2}=\frac{1}{\left[1-(1-s)(1-t)\right]^2}=\sum_{h\geq 0}\frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}\tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}\sum_{h\geq 0}\frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{\Gamma(h+2)^2} = \sqrt{ab}\, e^{a+b} I_1(2\sqrt{ab}).\tag{4}$$
If both $a$ and $b$ are $\gg 1$ this is approximately
$$ \frac{1}{\sqrt{4\pi}} e^{(\sqrt{a}+\sqrt{b})^2}\left[(ab)^{1/4}-\tfrac{3}{16}(ab)^{-1/4}\right]. \tag{5} $$
A: One feasible way to simplify it:
since $$\sum_{m=1}^\infty \frac{(m+n-1)!}{m!(m-1)!} a^m = \sum_{m=1}^\infty\frac{m(m+1)...(m+n-1)}{m!}a^m,$$
when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n. 
