> The pattern in those cases is > p(f(n))/p(n) -> f'(n), > where f' is the derivative of f.
> I wonder if there's a nice way to generalize this, > e.g., to arbitrary polynomial f. (?)
Yes, if f is increasing on some (N, infinity). We then have f(n) ~ c n^d for some positive c and positive integer d as n -> infinity (~ meaning asymptotic to). Then p(f(n)) ~ f(n) log f(n), and log f(n) ~ d log n, so p(f(n)) ~ cd n^d log n p(f(n))/p(n) ~ c d n^d log n /(n log n) = c d n^(d-1) ~ f'(n).
Robert Israel firstname.lastname@example.org Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2