Page 41, Exercise 3, Prove Theorem 1.3
From Theorem 1.2, we know that there exisit positive c and n0 such that sum (0 <= i < m) |ai|ni is less than or equal to cnm-1 for ever n greater than or equal to n0.
So, f(n) is greater than or equal to amnm - cnm-1 for every n greater than or equal to n0. Hence, f(n) is greater than or equal to (am-c/n)nm for every n greater than or equal to n0.
Let n1 be max{n_0, ceiling(c/n0)}+1 and let c1 be am-c/n1. It is easy to see that c1 is positive and that
f(n) is greater than or equal to c1nm for every n greater than or equal to n1.