IMO 2001 SL A2

Origin: POL | Category: Algebra

Problem

Let a0, a1, a2, . . . be an arbitrary infinite sequence of positive numbers. Show that the inequality 1 + an > an−1 n\sqrt 2 holds for infinitely many positive integers n.

Solution

It follows from Bernoulli’s inequality that for each n \inN,
1 + 1 n n \geq2, or n\sqrt 2 \leq1 + 1 n. Consequently, it will be enough to show that 1 + an >
1 + 1 n  an−1. Assume the opposite. Then there exists N such that for each n \geqN, 1 + an \leq  1 + 1 n  an−1, i.e., n + 1 + an n + 1 \leqan−1 n . Summing for n = N, . . . , m yields am m+1 \leqaN−1 N −  N+1 + \cdot \cdot \cdot + m+1

. However, it is well known that the sum N+1 + \cdot \cdot \cdot + m+1 can be arbi- trarily large for m large enough, so that am m+1 is eventually negative. This contradiction yields the result. Second solution. Suppose that 1 + an \leq n\sqrt 2an−1 for all n \geqN. Set bn = 2−(1+1/2+\cdot\cdot\cdot+1/n) and multiply both sides of the above inequality to obtain bn + bnan \leqbn−1an−1. Thus bNaN > bNaN −bnan \geqbN + bN+1 + \cdot \cdot \cdot + bn. However, it can be shown that  n>N bN diverges: in fact, since 1 + 1 2 + \cdot \cdot \cdot + 1 n < 1 + ln n, we have bn > 2−1−ln n = 1 2n−ln 2 > 2n, and we already know that  n>N 2n diverges. Remark. As can be seen from both solutions, the value 2 in the problem can be increased to e.