# 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

4a_{n}(1-a_{n-1})>1; | (2) |

it follows that *a*_{n}>0 and *a*_{n-1}<1. Moreover, from the AM-GM inequality we get

so *a*_{n}>*a*_{n-1}; the sequence increases from the index *n*_{0}.

By the monotonicity and boundedness the sequence converges. We show that its limit is . Let ; from (2) we see that

(2*A*-1)^{2}0,

therefore . Hence, .

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

(3) |

Summing up (3), for *n*>*n*_{0} we get a lower bound for *x*_{n}:

(4) |

In the rest of the solution we consider the properties of the sequence (*n*+1)*x*_{n}. We distinguish two cases.

*Case 1:* the sequence (*n*+1)*x*_{n} 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)*x*_{n}>*c*- holds except for finitely many indices. Substuting this into (3) and summing up again, we obtain that for *n*_{k}>*n*_{0} and *c*,

From the transition *k* we get

Then from +0, we obtain

contradiction.

*Case 2:* The sequence (*n*+1)*x*_{n} has no accumulation point. By the Bolzano-Weierstrass theorem and the positivity this means that for all real *K*, (*n*+1)*x*_{n} with finitely many exceptions, so (*n*+1)*x*_{n}. Then there is an index *n*>*n*_{0} such that 1<*nx*_{n-1}<(*n*+1)*x*_{n}. Applyinig (3),

we get a contradiction, since both *nx*_{n-1}-(*n*+1)*x*_{n} and 1-2(*n*+1)*x*_{n} are negative.

**Remark.** The term in the problem statement is important. Ommiting this quantity the statement would be false. For example, for the sequence ,

### Statistics:

6 students sent a solution. 5 points: Ágoston Tamás, Mester Márton, Omer Cerrahoglu, Strenner Péter. 1 point: 2 students.

Problems in Mathematics of KöMaL, January 2012