Why does a linear change of variable in a double sum enable Mathematica to obtain a symbolic answer?

Question

I perform two equivalent sums. Mathematica returns a symbolic expression in one case but not the other. Shouldn't Mathematica be able to do either sum?

First sum

summand1 = Exp[a1 x1 + a2 x2]/((a1 - a2)! a2! (eT - (a1 - a2))! (sT - a1)!)
Z1=Sum[summand1, {a1, 0, sT}, {a2, 0, Infinity}]

for which Mathematica returns a DifferenceRoot.

With the following change of variable

a1 = a2 + es;
summand2 = Exp[a1 x1 + a2 x2]/((a1 - a2)! a2! (eT - (a1 - a2))! (sT - a1)!)
Z2=Sum[summand2, {es, 0, eT}, {a2, 0, Infinity}]

In the latter case Mathematica is able to find a symbolic answer. Shouldn't it return the same answer either way?

I've edited this top demonstrate that the two sums are in fact equal for specific values of eT and sT. Set eT and sT to something small, say sT=3;eT=4; Then evaluate Z1 and Z2. Then take the difference.

EDIT TO SHOW THAT THE ORDER OF THE SUMMATION WITH at and es DOESN"T ALTER THE OUTCOME when integer values of eT and sT are used.

 Z3 = Sum[summand2, {a2, 0, Infinity}, {es, 0, eT}];
 Print["Z3=", Z3]

 eT = 8; sT = 11;

 Print["Z1-Z2=", FullSimplify[Z1 - Z2]]

 Print["Z2-Z3=", FullSimplify[Z2 - Z3]]