[✓] Calculate a double sum?

Hello, everyone!
I’m currently working on a university homework and I stuck with this problem. I’m trying to evaluate this sum, but Mathematica just rewrites it, after a little computation on it.

Sum[(n/m)*(l^n/(m! (n - m)!))*Exp[-l]*x^m*(1 - x)^(n - m), {n, 0, Infinity}, {m, 1, Infinity}]

Is there anyone that can help me, please?

Thank you in advance!!

Do you have any reason to think there is a closed form?

Most Sum don’t have one. Maybe the best you can do is numerical methods.

Maybe You put the question here more likely you will get someone to help you.

  Sum[(E^-l l^n n (1 - x)^(-m + n) x^m)/(
   m m! (-m + n)!), {m, 1, Infinity}, 
   Assumptions -> {0 < x < 1, 0 < l < 1, n >= 0}]
  (* -((E^-l l^n n (1 - x)^
    n x HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, 
     x/(-1 + x)])/((-1 + x) (-1 + n)!))*)

 Sum[-((E^-l l^n n (1 - x)^
    n x HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, 
     x/(-1 + x)])/((-1 + x) (-1 + n)!)), {n, 0, Infinity}, 
  Assumptions -> {0 < x < 1, 0 < l < 1}] (*Can't Find !*)

Regards M.I.

With Maple help I solve the problem:

HoldForm[Sum[(E^-l l^n n (1 - x)^(-m + n) x^m)/(
    m m! (-m + n)!), {n, 0, Infinity}, {m, 1, 
     Infinity}] == -Exp[-l*x]*(-1 + l*(-1 + x))*l*x*
    HypergeometricPFQ[{1, 1, -l*x + l + 2}, {2, 2, -l*x + l + 1}, 
     l*x]] // TeXForm
\sum _{n=0}^{\infty } \sum _{m=1}^{\infty } \frac{e^{-l} l^n n (1-x)^{-m+n} x^m}{m m! (-m+n)!}=-\exp (-l x) (-1+l (-1+x)) l x \, _3F_3(1,1,-l x+l+2;2,2,-l x+l+1;l x)

For: \{-1<x<1,l\in \mathbb{R}\}

ff[x_, l_, M_] := 
 Sum[(E^-l l^n n (1 - x)^(-m + n) x^m)/(
   m m! (-m + n)!), {n, 0, M}, {m, 1, M}] // N

f[x_, l_, M_] := Sum[-((E^-l l^n n (1 - x)^n x HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, 
                        x/(-1 + x)])/((-1 + x) (-1 + n)!)), {n, 0, M}] // N;

g[x_, l_] := -Exp[-l*x]*(-1 + l*(-1 + x))*l*x*HypergeometricPFQ[{1, 1, -l*x + l + 2}, {2, 2, -l*x + l + 1}, l*x]

ff[1/2, -1, 20]
(*-0.282835*)
f[1/2, -1, 20]
(*-0.282835*)
g[1/2, -1] // N
(*-0.282835*)

Thank you very much!!