KöMaL - Középiskolai Matematikai és Fizikai Lapok
 English
Információ
A lap
Pontverseny
Cikkek
Hírek
Fórum

Rendelje meg a KöMaL-t!

ELTE

VersenyVizsga portál

Kísérletek.hu

Matematika oktatási portál

A. 552. Prove that for an arbitrary sequence a_1,a_2,\ldots of nonnegative real numbers and \varepsilon>0 there exist infinitely many positive integers n for which n^2
\big(4a_n(1-a_{n-1})-1 \big) \le \frac14+\varepsilon.

(Schweitzer-competition, 2011)

(5 points)

Deadline expired on 10 February 2012.


Solution. We prove by contradiction. Suppose that there is a sequence a_1,a_2,\ldots and \varepsilon>0 for which

 n^2 \big(4a_n(1-a_{n-1})-1 \big) > \frac14+\varepsilon. (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


1 < 2\sqrt{a_n(1-a_{n-1})} \le a_n + (1-a_{n-1}),

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 \frac12. Let A=\lim a_n; from (2) we see that


4A(1-A) = \lim\big(4a_n(1-a_{n-1})\big) \ge 1

(2A-1)2\le0,

therefore A=\frac12. Hence, a_n\to\frac12.

Define x_n=\frac12-a_n. This sequence decreases from the index n0, and it converges to 0. Transforming (1), for n>n0 we have


\frac{\frac14+\varepsilon}{n^2} < 4a_n(1-a_{n-1})-1
= 4\left(\frac12-x_n\right)\left(\frac12+x_{n-1}\right) - 1
= 2(x_{n-1}-x_n)-4x_{n-1}x_n


x_{n-1}-x_n > \frac{\frac14+\varepsilon}{2n^2} + 2x_{n-1}x_n
> \frac1{8n^2}. (3)

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


x_n 
= \sum_{i=n+1}^\infty (x_{i-1}-x_i)
> \sum_{i=n+1}^\infty \frac1{8i^2}
> \sum_{i=n+1}^\infty
\frac18 \left(\frac1i-\frac1{i+1}\right)
= \frac1{8(n+1)}. (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 (n+1)x_n>\frac18 except for finitely many indices, all accumulation points lie in the interval [\frac18,\infty). 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 n_1<n_2<\ldots of indices such that (n_k+1) x_{n_k}\to c \ge\frac18.

Since there is no smallest accumulation point than c, for all \delta>0, by the Bolzano-Weierstrass theorem, (n+1)xn>c-\delta holds except for finitely many indices. Substuting this into (3) and summing up again, we obtain that for nk>n0 and \delta\lec,


(n_k+1) x_{n_k} 
= (n_k+1) \sum_{i=n_k+1}^\infty (x_{i-1}-x_i)
> (n_k+1) \sum_{i=n_k+1}^\infty
\left(\frac{\frac14+\varepsilon}{2i^2} + 2x_{i-1}x_i \right) >


> (n_k+1) \sum_{i=n_k+1}^\infty \left(\frac{\frac14+\varepsilon}{2i(i+1)}
  + 2\frac{(c-\delta)^2}{i(i+1)}\right) =


= \left(\frac{\frac14+\varepsilon}2+2(c-\delta)^2\right) (n_k+1)
\sum_{i=n_k+1}^\infty \left(\frac1i-\frac1{i+1}\right)
= \frac{\frac14+\varepsilon}2+2(c-\delta)^2.

From the transition k\to\infty we get


c = \lim (n_k+1) x_{n_k}
\ge \frac{\frac14+\varepsilon}2+2(c-\delta)^2.

Then from \delta\to+0, we obtain


c \ge \frac{\frac14+\varepsilon}2+2c^2


\left(2c-\frac12\right)^2 + \varepsilon \le 0,

contradiction.

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\to\infty with finitely many exceptions, so (n+1)xn\to\infty. Then there is an index n>n0 such that 1<nxn-1<(n+1)xn. Applyinig (3),


0< x_{n-1}-x_n - 2x_{n-1}x_n
= \frac1{n+1}\Big(nx_{n-1}-(n+1)x_n\Big)
+ \frac{x_{n-1}}{n+1}\Big(1-2(n+1)x_n\Big),

we get a contradiction, since both nxn-1-(n+1)xn and 1-2(n+1)xn are negative.

Remark. The term \varepsilon in the problem statement is important. Ommiting this quantity the statement would be false. For example, for the sequence a_n=\frac{n}{2n+1},


n^2 \big(4a_n(1-a_{n-1})-1 \big)
= n^2 \left(\frac{4n}{2n+1}\bigg(1-\frac{n-1}{2n-1}\bigg)-1\right)
= \frac{n^2}{4n^2-1} > \frac14.


Statistics on problem A. 552.
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

  • Támogatóink:   Ericsson   Cognex   Emberi Erőforrás Támogatáskezelő   Emberi Erőforrások Minisztériuma   Nemzeti Tehetség Program    
    MTA Energiatudományi Kutatóközpont   MTA Wigner Fizikai Kutatóközpont     Nemzeti
Kulturális Alap   ELTE   Morgan Stanley