Difference between Sum and NSum?

Questions
,
1

I have to compute this,

WhatsAppImage2023-03-22at16.24.07.jpg
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I have written the Mathematica code, keeping Ta, and Tb as variables. Later we will compute prob[1,0,1,0,Ta,Tb]

Open in Wolfram Cloud ↗ 
In[75]:= prob[i_,j_,k_,l_,Ta_,Tb_]:= Sum[Binomial[ip,i]*Ta^i*(1-Ta)^(ip-i)*Binomial[jp,j]*Ta^j*(1-Ta)^(jp-j)*Binomial[kp,k]*Tb^k*(1-Tb)^(kp-k)*Binomial[lp,l]*Tb^l*(1-Tb)^(lp-l),{ip,1,Infinity},{jp,1,Infinity},{kp,1,Infinity},{lp,1,Infinity}];
In[80]:= prob[1,0,1,0,Ta,Tb]
Out[80]= ((-1+Ta) (-1+Tb))/(Ta^2 Tb^2)

My question is,

  1. I have seen that if I use NSum in place of sum it takes too much to compute prob[1,0,1,0,Ta,Tb]
    what is the difference between sum and NSum here?
  2. what will be a good idea to use here sum or NSum as I think Sum is not giving correct answer.(or maybe I am wrong that it is giving the correct answer)
2

Why do you think using Sum gives the wrong answer? There’s no need to use NSum.

The way you’ve constructed the sum can be simplified and you have the product of 4 separate terms:

p[i_, t_] := Sum[Binomial[n, i]*t^i*(1 - t)^(n - i), {n, 1, \[Infinity]}]
prob[i_, j_, k_, l_, Ta_, Tb_] := p[i, Ta] p[j, Ta] p[k, Tb] p[l, Tb]
prob[1, 0, 1, 0, Ta, Tb]
(* ((1 - Ta) (1 - Tb))/(Ta^2 Tb^2) *)

Addition:

It appears that p[i, t] equals -1+1/t for i=0 and 1/t for i>0 (with i being a non-negative integer). That simplifies things considerably.

3

Thanks for your response