Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
 Already signed up? New to KöMaL?

# 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 .

(Schweitzer-competition, 2011)

(5 pont)

Deadline expired on February 10, 2012.

Solution. We prove by contradiction. Suppose that there is a sequence and >0 for which

 (1)

holds with finitely many exception. As a simple corollary of (1) we have

 4an(1-an-1)>1; (2)

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

(2A-1)20,

therefore . Hence, .

Define . This sequence decreases from the index n0, and it converges to 0. Transforming (1), for n>n0 we have

 (3)

Summing up (3), for n>n0 we get a lower bound for xn:

 (4)

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