What would o(nlogn) + O(n) be ?

My guess is O(nlogn) ?

f1(n) = o(nlogn) means for each constant c, there is an n0 such that 0 <= f1(n) < cnlogn . 

And f2(n) = O(n) means there exists some constant c1, such that for n > n1, 0 <= f2(n) <= c1n .

so , all I can get from this is that, there is some constant c1, such that for n > max(n0,n1), 0 <= f1(n) + f2(n) <= c1(nlogn) .