Alternative method for evaluating double infinite sum I've come across the following double infinite sum through my current work:
$$\sum_{k=0}^\infty \sum_{m=0}^\infty \binom{m+k}{m} \frac{a^k b^m}{k! \; m!},$$
where a and b are real constants. I found a method to evaluate this sum using complex analysis, in particular Egorychev's method stating that
$$\binom{m+k}{m} = \frac{1}{2\pi j} \oint_C \frac{(1+z)^{m+k}}{z^{m+1}} \mbox{d}z,$$
where the countour is a unit circle centered around the origin
Substituting this and using the complex integral representation of the modified Bessel function gives the apparent result that
$$\sum_{k=0}^\infty \sum_{m=0}^\infty \binom{m+k}{m} \frac{a^k b^m}{k! \; m!} = \exp(a + b) I_0(2\sqrt{ab}).$$
Is there another way to evaluate this sum to a) check this result and b) to extend the result to higher dimensions e.g.
$$\sum_{k=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty  (k+m+n)!\frac{a^k b^m c^n}{k!^2 \; m!^2 \; n!^2}$$
for real a, b and c? Egorychev's method becomes significantly more complicated in 3 or more dimensions! If it's possible to reduce the dimensionality of the sum, that would still be useful.
As always, any help greatly appreciated!
 A: Here is an alternative method for the 2D case, which has an obvious extension to the 3D problem.  Use the Euler representation of the gamma function and the power series for the Bessel $I_0$ function
$$ n! = \int_0^\infty e^{-t} t^n dt \quad\text{ and }\quad I_0(2\sqrt{t})=\sum_{n=0}^\infty t^n/n!^2$$
Upon interchanging $\int$ and $\sum$ the 2D sum becomes
$$ \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(m+k)!}{k!\  m!} \frac{a^k \ b^m}{k!\  m!}= 
\int_0^\infty e^{-t} \sum_{k=0}^\infty \frac{(a\ t)^k}{k!^2}  \sum_{m=0}^\infty \frac{(b\ t)^m}{m!^2} dt$$
$$ = \int_0^\infty e^{-t} I_0(2\sqrt{a t})  I_0(2\sqrt{b t}) dt $$
The last integral can be solved by table lookup; e.g., with a simple modification of Gradshteyn and Ryzhik, eq. 6.633.4.
The 3D sum becomes
$$ \sum_{k=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(m+k+n)!}{k!\  m! \ n!} \frac{a^k \ b^m c^n}{k!\  m! \ n!}=$$
$$ = \int_0^\infty e^{-t} I_0(2\sqrt{a t})  I_0(2\sqrt{b t}) I_0(2\sqrt{c t}) dt $$
I am not aware of a simplification of this integral. G&R doesn't have such integrals.  If the function multiplying the triple product was a power (with certain restrictions) then in some cases there are closed-form solutions.  This integral has a Gaussian (upon $t \to t^2.$)  On the bright side, the integral is easily calculable numerically.
