I thought of another possibilty, but i doubt its practicality
say we have function of O(n), say f(n)= O(n)
Now say we have another function g(n) = 1+2+3 +4+ 5+..upto n terms
so g(n) is O($n^2$).
so, f(n) =O(g(n)
But what about if n is infinity?
according to Ramanujan 1+2+3 +4+ 5+...... = -1/12
so g(n) = -1/12 , but f(n)= positive infinity
so this time , g(n) = O(f(n))
and it can be easily stated that they f(n) is not theta of g(n).
So this case also does satisfies-->
Doubt its practicality, but i think mathematicaly it is sound..