Making Mathematica return the full simplified expression by default

Question

I have an expression defined as

Ν[Δ_, l_] = FunctionExpand[((2^l (Δ + l - 1) Gamma[Δ + l - 1]^2 Gamma[Δ])/
              (Gamma[Δ - 1] Gamma[(Δ + l)/2]^4 Gamma[1/8 - ((Δ - l)/2)]^2 *
               Gamma[1/8 - ((2 - Δ - l)/2)]^2)), 
             Assumptions -> {Δ ∈ Integers, l ∈ Integers, �� >= 0, l >= 0}];

The problem is that it diverges naively when I try to compute Ν[1,0] with the divergence coming from the fact that Mathematica evaluates Ν[k,0] to be

Ν[k, 0]
(* (2^(-4 + 2 k) (-1 + k)^2 Gamma[-(1/2) + k/2]^2)/(π Gamma[
    1/8 - k/2]^2 Gamma[-(7/8) + k/2]^2 Gamma[k/2]^2) *)

Which means that the numerator becomes a product 0 ComplexInfinity. But when I use FullSimplify on the above, I get

Ν[k, 0] // FullSimplify
(* (4^(-1 + k) Gamma[(1 + k)/2]^2)/(π Gamma[
      1/8 - k/2]^2 Gamma[-(7/8) + k/2]^2 Gamma[k/2]^2) *)

Is there a way to make Mathematica do this simplification by default? I am using this function in a sum, so I have to sum over $l = 0$ terms too, which is causing annoying errors.

Answer 1

How about taking the limit instead:

f[kin_, Lin_] := Module[{num, den, L, k, t},
  num = Limit[(2^(-4 + 3*L + 2*k)*(-1 + k)*(-1 + L + k)*
      Gamma[-1/2 + L/2 + k/2]^2), {L -> Lin, k -> kin}];
  den = Limit[(Pi*Gamma[1/8 + L/2 - k/2]^2*Gamma[-7/8 + L/2 + k/2]^2*
      Gamma[L/2 + k/2]^2), {L -> Lin, k -> kin}];
  t = num/den;
  FunctionExpand[t, 
   Assumptions -> {k ∈ Integers, L ∈ Integers, 
     k >= 0, L >= 0}
   ]
  ]

And now

f[1, 0]
(* 0 *)

And

 f[k, 0] // FullSimplify