HMMT 二月 2011 · TEAM1 赛 · 第 9 题

英文解析

9. [ 15 ] We would now like to examine the behavior of p ( k ) as m becomes arbitrarily large; specifimcally, we would like to discern whether lim p (0) exists and, if it does, to determine its value. Letmm →∞
   lim p ( k ) = A .
   m km →∞
   (a) [ 5 ] Prove that p ( k ) ≥ p ( k + 2) for all m and k .2
   m m
   (b) [ 10 ] Prove that A exists and that A > 0. Feel free to assume the result of analysis that a 3
   0 0
   non-increasing sequence of real numbers that is bounded below by a constant c converges to alimit that is greater than or equal to c .
   Solution:
   (a) We proceed by induction.
   2 2 3 1 1
   Base Case: When n = 1, we get p (0) = · = > = p (2). And since p (2 k ) = 0 for k ≥ 2,
   1 1 1
   3 3 4 2 4
   we get p ( k ) ≥ p ( k + 2) for all k .2
   1 1
   2 23
   Induction Step: Assume that p ( k ) ≥ p ( k + 2) for all i ≤ n . This clearly implies p ( k ) ≥
   i i n +1
   3 3
   p ( k + 2) for all k ≥ 4 by the formula (c) from the previous problem.
   n +1
   Now, for k = 2,
   ( )
   2 2 1 3 3 1
   p (2) = p (0) + p (2) + p (4) + p (6)
   n +1 n n n n
   3 3 4 8 8 8
   ≥ p (0) + p (4)1
   n n +1
   ≥ p (4)12
   n +1
   by the formula (b) from the previous problem and our induction hypothesis.
   For k = 0, we write out formula (a) from the previous problem.
   ( )
   2 2 3 1 1
   p (0) = p (0) + p (2) + p (4)
   n +1 n n n
   3 3 4 2 8
   1 3 1 1
   ≥ p (0) + p (2) + p (4) + p (6)
   n n n n
   4 8 2 8
   Team Round A
   1 3 3 1
   by our induction hypothesis. But p (2) = p (0) + p (2) + p (4) + p (6), which is clearlyn +1 n n n n
   4 8 8 8
   less than the last line above. Thus p (0) ≥ p (2). This completes the induction step.2
   n +1 n +1
   Remark: This is nowhere near the strongest bound. We intentionally asked for a bound strong 3
   enough to be helpful in part b, but not strong enough to help with other problems. The strongest
   √
   √
   2+ 5
   bound is p (0) ≥ p (2) and p (2 k ) ≥ (2 + 5) p (2 k + 2) for k ≥ 1. This is intuitive becausen n n n
   √2
   √
   2+ 5
   A = A and for k ≥ 1, A = (2 + 5) A .
   0 2 2 k 2 k +2
   2 k 2
   (b) For any n and k , p (2 k ) ≤ ( ) p (0). Nown n
   ( )3
   ∞ ∞
   ∑ ∑k
   1 = p (2 k ) ≤ p (0) = 3 p (0) .2
   n n nk =0 k =03
   So p (0) ≥ for all n . This means p (0) is a non-increasing sequence that is bounded below by 1
   n n
   1 13
   . So A exists and A ≥ > 0.
   0 0
   3 3