IMO 2003 SL A3

Origin: GEO | Category: Algebra

Problem

Consider pairs of sequences of positive real numbers a1 \geq a2 \geqa3 \geq\cdot \cdot \cdot , b1 \geqb2 \geqb3 \geq\cdot \cdot \cdot and the sums An = a1 + \cdot \cdot \cdot + an, Bn = b1 + \cdot \cdot \cdot + bn, n = 1, 2, . . . . For any pair define ci = min{ai, bi} and Cn = c1 + \cdot \cdot \cdot + cn, n = 1, 2, . . .. (a) Does there exist a pair (ai)i\geq1, (bi)i\geq1 such that the sequences (An)n\geq1 and (Bn)n\geq1 are unbounded while the sequence (Cn)n\geq1 is bounded? (b) Does the answer to question (1) change by assuming additionally that bi = 1/i, i = 1, 2, . . .? Justify your answer.

Solution

(a) Given any sequence cn (in particular, such that Cn converges), we shall construct an and bn such that An and Bn diverge. First, choose n1 such that n1c1 > 1 and set a1 = a2 = \cdot \cdot \cdot = an1 = c1: this uniquely determines b2 = c2, . . . , bn1 = cn1. Next, choose n2 such that (n2 −n1)cn1+1 > 1 and set bn1+1 = \cdot \cdot \cdot = bn2 = cn1+1; again an1+1, . . . , an2 is hereby determined. Then choose n3 with (n3 − n2)cn2+1 > 1 and set an2+1 = \cdot \cdot \cdot = an3 = cn2+1, and so on. It is plain that in this way we construct decreasing sequences an, bn such that  an and  bn diverge, since they contain an infinity of subsums that exceed 1; on the other hand, cn = min(an, bn) and Cn is convergent. (b) The answer changes in this situation. Suppose to the contrary that there is such a pair of sequences (an) and (bn). There are infinitely many indices i such that ci = bi (otherwise all but finitely many terms of the sequence (cn) would be equal to the terms of the sequence (an), which has an unbounded sum). Thus for any n0 \inN there is j \geq2n0 such that cj = bj. Then we have j  k=n0 ck \geq j  k=n0 cj = (j −n0)1 j \geq1 2. Hence the sequence (Cn) is unbounded, a contradiction.