Advanced book on Mathematics Olympiad

Real Analysis 491

let

bnk+ 1 = ··· =bnk+ 2 rk=

1

2 rk

and

ank+ 1 = ··· =ank+ 2 rk=

1

2 k· 2 rk

.

Ifkis odd, we do precisely the same thing, with the roles of the sequences(an)nand
(bn)nexchanged. As such we have

∑

n

bn≥

∑

kodd

2 rk

1

2 rk

= 1 + 1 +···=∞,

∑

n

an≥

∑

keven

2 rk

1

2 rk

= 1 + 1 +···=∞,

which shows that both series diverge. On the other hand, if we letcn=min(an,bn), then

∑

n

cn=

∑

k

2 rk

1

2 k 2 rk

=

∑

k

1

2 k

,

which converges to 1. The example proves our claim.
(b) The answer to the second question is no, meaning that the situation changes if we
work with the harmonic series. Suppose there is a series

∑

nanwith the given property.
Ifcn=^1 nfor only finitely manyn’s, then for largen,an=cn, meaning that both series
diverge. Hencecn =^1 nfor infinitely manyn. Let(km)mbe a sequence of integers
satisfyingkm+ 1 ≥ 2 kmandckm=k^1 m. Then

k∑m+ 1

k=km+ 1

ck≥(km+ 1 −km)ckm+ 1 =(km+ 1 −km)

1

km+ 1

=

1

2

.

This shows that the series

∑

ncndiverges, a contradiction.

(short list of the 44th International Mathematical Olympiad, 2003)

359.Forn≥1, define the functionfn:( 0 , 1 )→R,fn(x)=x−nx^2. It is easy to see
that 0<fn(x)≤ 41 n, for allx∈( 0 , 1 ). Moreover, on( 0 , 21 n]the function is decreasing.
With this in mind, we prove by induction that

0 <xn<

2

n^2

,

forn≥2. We verify the first three cases: