The symbolic result does not give a proper answer when inputs are specified
$Version
(* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *)
Clear["Global`*"]
F[m_, k_, i_, j_] := (-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j]
sum[i_, j_] =
Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}] // FullSimplify
(* -2^i E^(-3 + I i π) (-1 + j) Binomial[0, i] Binomial[1,
j] (Gamma[1 - i] + i Gamma[-i, -2]) (E +
E^(I j π) (Gamma[2 - j] + E (-1 + j) Subfactorial[-j])) *)
sum[2, 2]
Define the sum as a Limit
sum2[i_, j_] := Limit[sum[m, n], {m, n} -> {i, j}]
Then
sum2[2, 2]
(* -(1/E^3) *)
Comparing with direct evaluation (this is quite slow)
And @@ Flatten[
Table[Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}] ==
sum2[i, j], {i, 0, 5}, {j, 0, 5}]]
(* True *)
You can get the sum for 169 values of $(i,j)$ by taking the limit of the what you get from Sum
:
expr = FullSimplify[Sum[F[m, k, i, j], {m, 0, ∞}, {k, 0, ∞}]];
ss2[iv_, jv_] := ss2[iv, jv] = Limit[expr, {i, j} -> {iv, jv}]
mat = Table[ss2[iv, jv], {iv, 0, 12}, {jv, 0, 12}];
From these values FindSequenceFunction
can suggest a closed form that works on integers:
FindSequenceFunction[#, j + 1] & /@ (mat E^3)
{((-1)^(1 + j) (-1 + j))/(2 Pochhammer[3, -2 + j]), ((-1)^(2 + j) (-1 + j))/Pochhammer[3, -2 + j], ((-1)^(1 + j) (-1 + j))/Pochhammer[3, -2 + j], -((2 (-1)^(1 + j) (-1 + j))/(3 Pochhammer[3, -2 + j])), ((-1)^(1 + j) (-1 + j))/(3 Pochhammer[3, -2 + j]), -((2 (-1)^(1 + j) (-1 + j))/(15 Pochhammer[3, -2 + j])), ..........}
which suggest a closed form except for a factor depending on i
:
MapThread[FindSequenceFunction[#2, j + 1]/(((-1)^(# + j) (1 - j))/Pochhammer[3, j - 2]) &, {Range[0, 12], (mat E^3)}] // Simplify
(* {1/2, 1, 1, 2/3, 1/3, 2/15, 2/45, 4/315, 1/315, 2/2835, 2/14175, 4/155925, 2/467775} *)
FindSequenceFunction
recognize the factor:
FindSequenceFunction[{1/2, 1, 1, 2/3, 1/3, 2/15, 2/45, 4/315, 1/315, 2/2835, 2/14175, 4/155925, 2/467775}, # + 1]
(* 2^(-1 + #1)/Pochhammer[1, #1] *)
Combining these we get:
2^(-1 + #1)/Pochhammer[1, #1] ((-1)^(# + j) (1 - j))/Pochhammer[3, -2 + j] &[i] // FullSimplify
$$\frac{2^i (1-j)e^{-3}}{i! j! (-1)^{i+j}}$$
I tried to Plot3D
it.
Plot3D[Evaluate@ Sum[(-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j], {m, 0, Infinity}, {k, 0, Infinity}], {i, -5, 5}, {j, -5, 5}, PlotPoints -> 20, PlotRange -> All]
Looks like a lot of discontinuity.
But there are some points return values.
/. {i -> 51/10, j -> 45/10}
-(1/(E^3))(-((544 (-2)^(1/10) E Binomial[0, 51/10] Gamma[-(51/10)])/(
15 π)) + (
73984 I (-2)^(1/10) Binomial[0, 51/10] Gamma[-(51/10)])/(
2205 Sqrt[π]) - (
17408 (-1)^(3/5) 2^(1/10) Binomial[0, 51/10] Gamma[-(51/10)])/(
1225 Sqrt[π]) + (
544 (-2)^(1/10) E Binomial[0, 51/10] Gamma[-(51/10), -2])/(
15 π) - (
73984 I (-2)^(1/10) Binomial[0, 51/10] Gamma[-(51/10), -2])/(
2205 Sqrt[π]) + (
17408 (-1)^(3/5) 2^(1/10) Binomial[0, 51/10] Gamma[-(51/10), -2])/(
1225 Sqrt[π]) - (
4624 I (-2)^(1/10)
Binomial[0, 51/10] Gamma[-(51/10)] Gamma[-(7/2), -1])/(
21 π) + (
3264 (-1)^(3/5) 2^(1/10)
Binomial[0, 51/10] Gamma[-(51/10)] Gamma[-(7/2), -1])/(
35 π) + (
4624 I (-2)^(1/10)
Binomial[0, 51/10] Gamma[-(51/10), -2] Gamma[-(7/2), -1])/(
21 π) - (
3264 (-1)^(3/5) 2^(1/10)
Binomial[0, 51/10] Gamma[-(51/10), -2] Gamma[-(7/2), -1])/(
35 π))