Problem A. 552. (January 2012)
A. 552. Prove that for an arbitrary sequence of nonnegative real numbers and >0 there exist infinitely many positive integers n for which .
Deadline expired on February 10, 2012.
Solution. We prove by contradiction. Suppose that there is a sequence and >0 for which
holds with finitely many exception. As a simple corollary of (1) we have
it follows that an>0 and an-1<1. Moreover, from the AM-GM inequality we get
so an>an-1; the sequence increases from the index n0.
By the monotonicity and boundedness the sequence converges. We show that its limit is . Let ; from (2) we see that
therefore . Hence, .
Define . This sequence decreases from the index n0, and it converges to 0. Transforming (1), for n>n0 we have
Summing up (3), for n>n0 we get a lower bound for xn:
In the rest of the solution we consider the properties of the sequence (n+1)xn. We distinguish two cases.
Case 1: the sequence (n+1)xn has a finite accumulation point. Since except for finitely many indices, all accumulation points lie in the interval . It is well-known that there is a minimal accumulation point; this point is called as the limes inferior of the sequence. Denote by c the smallest accumulation point. Then there is a sequence of indices such that .
Since there is no smallest accumulation point than c, for all >0, by the Bolzano-Weierstrass theorem, (n+1)xn>c- holds except for finitely many indices. Substuting this into (3) and summing up again, we obtain that for nk>n0 and c,
From the transition k we get
Then from +0, we obtain
Case 2: The sequence (n+1)xn has no accumulation point. By the Bolzano-Weierstrass theorem and the positivity this means that for all real K, (n+1)xn with finitely many exceptions, so (n+1)xn. Then there is an index n>n0 such that 1<nxn-1<(n+1)xn. Applyinig (3),
we get a contradiction, since both nxn-1-(n+1)xn and 1-2(n+1)xn are negative.
Remark. The term in the problem statement is important. Ommiting this quantity the statement would be false. For example, for the sequence ,
6 students sent a solution. 5 points: Ágoston Tamás, Mester Márton, Omer Cerrahoglu, Strenner Péter. 1 point: 2 students.