IMO 1988 SL 24

Origin: SWE

Problem

Let {ak}\infty be a sequence of nonnegative real numbers such that ak −2ak+1 + ak+2 \geq0 and k j=1 aj \leq1 for all k = 1, 2, . . . . Prove that 0 \leq(ak −ak+1) < k2 for all k = 1, 2, . . . .

Solution

The first condition means in fact that ak−ak+1 is decreasing. In particular, if ak −ak+1 = −\delta < 0, then ak −ak+m = (ak −ak+1) + \cdot \cdot \cdot + (ak+m−1 − ak+m) < −m\delta, which implies that ak+m > ak + m\delta, and consequently ak+m > 1 for large enough m, a contradiction. Thus ak −ak+1 \geq0 for all k. Suppose that ak −ak+1 > 2/k2. Then for all i < k, ai −ai+1 > 2/k2, so that ai −ak+1 > 2(k + 1 −i)/k2, i.e., ai > 2(k + 1 −i)/k2, i = 1, 2, . . . , k. But this implies a1+a2+\cdot \cdot \cdot+ak > 2/k2+4/k2+\cdot \cdot \cdot+2k/k2 = k(k + 1)/k2, which is impossible. Therefore ak −ak+1 \leq2/k2 for all k.