The polynomials p_{n}(x) are defined by p_{1}(x) = 1 + x, p_{2}(x) = 1 + 2x, p_{2n+1}(x) = p_{2n}(x) + (n + 1) x p_{2n-1}(x), p_{2n+2}(x) = p_{2n+1}(x) + (n + 1) x p_{2n}(x). Let a_{n} be the largest real root of p_{n}(x). Prove that a_{n} is monotonic increasing and tends to zero.

**Solution**

It is fairly obvious that a_{n} < 0 and that the sequence is strictly monotonic increasing. The problem is proving it tends to zero.

Trivial inductions show that p_{n}(0) = 1 and p_{n}(x) > 0 for x > 0, so a_{n} < 0. Also we note that a_{1} = -1, a_{2} = -1/2. If x > a_{n}, then p_{n}(x) > 0 (otherwise there would be a root greater than a_{n}).

p_{2n+1}(a_{2n}) = (n + 1) a_{2n} p_{2n-1}(a_{2n}). By induction p_{2n-1}(a_{2n}) > 0, so p_{2n+1}(a_{2n}) < 0 and hence a_{2n+1} > a_{2n}. Similarly a_{2n+2} > a_{2n+1}. So, as claimed, a_{n} is strictly monotonic increasing.

To show it tends to zero, it suffices to prove that: a_{2n-1} > -1/(n-1) and a_{2n} > -1/n. This is true for n = 1. Suppose it is true for n. We have already established monotonicity, so a_{2n+1} > -1/n. Also p_{2n-1}(-1/(n+1)) > 0. But p_{2n+2}(-1/(n+1)) = p_{2n+1}(-1/(n+1)) - p_{2n}(-1/(n+1)) = - p_{2n-1}(-1/(n+1)) < 0, so a_{2n+2} > -1/(n+1) and we have established the inductive hypothesis for n+1.

© John Scholes

jscholes@kalva.demon.co.uk

30 Nov 1999