# The symbolic result does not give a proper answer when inputs are specified

Define F by

F[m_, k_, i_, j_] := (-1)^(m+k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k+1,j]


I am trying to find this sum:

Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}]


Mathematica gives a symbolic answer but I get an error message whenever I try to evaluate the result at any positive pair $$(i, j)$$. But it turns out that if I evaluate the sum with $$(i, j)$$ specified, say $$(2, 2)$$, then Mathematica gives a correct answer. What am I missing here? Thank you in advance for your help!

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


$$\frac{2^i (1-j)e^{-3}}{i! j! (-1)^{i+j}}$$

$Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global*"] F[m_, k_, i_, j_] := (-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j] sum[i_, j_] = Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}] // FullSimplify (* -2^i E^(-3 + I i π) (-1 + j) Binomial[0, i] Binomial[1, j] (Gamma[1 - i] + i Gamma[-i, -2]) (E + E^(I j π) (Gamma[2 - j] + E (-1 + j) Subfactorial[-j])) *) sum[2, 2]  Define the sum as a Limit sum2[i_, j_] := Limit[sum[m, n], {m, n} -> {i, j}]  Then sum2[2, 2] (* -(1/E^3) *)  Comparing with direct evaluation (this is quite slow) And @@ Flatten[ Table[Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}] == sum2[i, j], {i, 0, 5}, {j, 0, 5}]] (* True *)  I tried to Plot3D it. Plot3D[Evaluate@ Sum[(-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j], {m, 0, Infinity}, {k, 0, Infinity}], {i, -5, 5}, {j, -5, 5}, PlotPoints -> 20, PlotRange -> All]  Looks like a lot of discontinuity. But there are some points return values. /. {i -> 51/10, j -> 45/10}  -(1/(E^3))(-((544 (-2)^(1/10) E Binomial[0, 51/10] Gamma[-(51/10)])/( 15 π)) + ( 73984 I (-2)^(1/10) Binomial[0, 51/10] Gamma[-(51/10)])/( 2205 Sqrt[π]) - ( 17408 (-1)^(3/5) 2^(1/10) Binomial[0, 51/10] Gamma[-(51/10)])/( 1225 Sqrt[π]) + ( 544 (-2)^(1/10) E Binomial[0, 51/10] Gamma[-(51/10), -2])/( 15 π) - ( 73984 I (-2)^(1/10) Binomial[0, 51/10] Gamma[-(51/10), -2])/( 2205 Sqrt[π]) + ( 17408 (-1)^(3/5) 2^(1/10) Binomial[0, 51/10] Gamma[-(51/10), -2])/( 1225 Sqrt[π]) - ( 4624 I (-2)^(1/10) Binomial[0, 51/10] Gamma[-(51/10)] Gamma[-(7/2), -1])/( 21 π) + ( 3264 (-1)^(3/5) 2^(1/10) Binomial[0, 51/10] Gamma[-(51/10)] Gamma[-(7/2), -1])/( 35 π) + ( 4624 I (-2)^(1/10) Binomial[0, 51/10] Gamma[-(51/10), -2] Gamma[-(7/2), -1])/( 21 π) - ( 3264 (-1)^(3/5) 2^(1/10) Binomial[0, 51/10] Gamma[-(51/10), -2] Gamma[-(7/2), -1])/( 35 π)) ` • Okay but you may find that if you put any positive integer pair$(i, j)$into the summation part, then you will get a simple answer. For example,$(1, 1)$gives$0$and$(2, 2)$gives$-1/e^{3}\$. Commented Oct 14, 2020 at 5:11