Advanced book on Mathematics Olympiad

Real Analysis 475

Passing to the limit in the recurrence relation, we obtainL=

√

L^2 +L−1, and therefore

L^2 =L^2 +L−1. But this equation has no solution in the interval[r, 1 ), a contradiction.

Hencea 1 cannot lie in the interval(− 2 , 1 ).

(Bulgarian Mathematical Olympiad, 2002)

326.This is the Bolzano–Weierstrass theorem. For the proof, let us call a term of the

sequence agiantif all terms following it are smaller. If the sequence has infinitely many

giants, they form a bounded decreasing subsequence, which is therefore convergent. If

the sequence has only finitely many giants, then after some rank each term is followed by

larger term. These terms give rise to a bounded increasing subsequence, which is again

convergent.

Remark.The idea can be refined to show that any sequence ofmn+1 real numbers has

either a decreasing subsequence withm+1 terms or an increasing subsequence with

n+1 terms.

327.Consider the truncations

sn=a 1 −a 2 +a 3 −···±an,n≥ 1.

We are to show that the sequence(sn)nis convergent. For this we verify that the sequence

(sn)nis Cauchy. Because(an)n≥ 1 is decreasing, for alln>m,

|sn−sm|=am−am+ 1 +am+ 2 −···±an

=am−(am+ 1 −am+ 2 )−(am+ 3 −am+ 4 )−···,

where the sum ends either inanor in−(an− 1 −an). All terms of this sum, except for the
first and maybe the last, are negative. Therefore,|sn−sm|≤am+an, for alln>m≥1.
Asan→0, this shows that the sequence(sn)nis Cauchy, and hence convergent.
(the Leibniz criterion)
328.For a triple of real numbers(x,y,z)define (x,y,z)=max(|x−y|,|x−z|,|y−z|).
Let (a 0 ,b 0 ,c 0 )=δ. From the recurrence relation we find that

    (an+ 1 ,bn+ 1 ,cn+ 1 )=

1

2

    (an,bn,cn), n≥ 0.

By induction    (an,bn,cn)= 21 nδ. Also, max(|an+ 1 −an|,|bn+ 1 −bn|,|cn+ 1 −cn|)=

1

2   (an,bn,cn). We therefore obtain that|an+^1 −an|,|bn+^1 −bn|,|cn+^1 −cn|are all less

than or equal to 21 nδ. So forn>m≥1, the absolute values|an−am|,|bn−bm|, and

|cn−cm|are less than

(

1

2 m

+

1

2 m+^1

+···+

1

2 n

)

δ<

δ

2 m

.