Question 10.20: Prove that if f(n,N) increases with N, then the same is true...

Suppose that f(n,N) increases with N. We want to show that the same is
true for

y(n,N) =\frac{f(n, n) + f(n, n + 1) + · · · + f(n,N − 1)}{N − n}

This follows from the fact that if a sequence a_n increases, then so does the sequence of averages Sn=\frac{a_1+..+a_n}{n} To see this multiply the target inequality S_{n+1} > S_n by n(n+1) to get (after cancellations) na_{n+1} > a_1 +· · ·+a_n. The latter is true, since a_{n+1} > a_i for all i = 1, . . . , n.