fzero for sum of functions?

The actual equation is 0 = sum(ai/(ai-p)) where i is 1 to 4 and I have to look for p values.

This is what I put in the code,

 a=[1.2 3.5 1.9 1];

 func=@(p)sum((a/(a-p)))

 fsolve(func,0)

and the answer it is giving me is -154.25. I checked with my calculator and it does not make sense. Is there something wrong with this code possibly? I am little bit confused about putting function inside the sum().

Answer 1

[Walter Roberson]

func=@(p)sum((a./(a-p)))