Advanced book on Mathematics Olympiad

Real Analysis 471

>(n+ 1 )xn+ 1 + 2

∑n

k= 1

kxk−

∑n

k= 1

kxk=

n∑+ 1

k= 1

kxk,

as desired. Furthermore,x 1 >0 by definition andx 2 ,x 3 ,...,xnare also positive by the

induction hypothesis. Therefore,xn+ 2 >(n+ 1 )xn+ 1 >(n+ 1 )(a·n!)=a·(n+ 1 )!.

This completes the induction, proving the claim.

(Romanian Team Selection Test for the International Mathematical Olympiad, 1999)

320.Denoteλ=infn≥ 1 xnnand for simplicity assume thatλ>−∞. Fix>0. Then

there existsn 0 such thatxnn 00 ≤λ+. LetM=max 1 ≤i≤n 0 xi.

An integermcan be written asn 0 q+n 1 , with 0≤n 1 <qandq=nm 0 . From the

hypothesis it follows thatxm≤qxn 0 +xn 1 ; hence

λ≤

xm

m

≤

qxn 0

m

+

xn 1

m

≤

qn 0

m

(λ+)+

M

m

.

Therefore,

λ≤

xm

m

≤

⌊

m

n 0

⌋

m

n 0

(λ+)+

M

m

.

Since

mlim→∞

nm 0 

m

n 0

=1 and mlim→∞

M

m

= 0 ,

it follows that for largem,

λ≤

xm

m

≤λ+ 2 .

Sincewas arbitrary, this implies

lim

n→∞

xn

n

=λ=inf

n≥ 1

xn

n

,

as desired.

321.We use the fact that

lim

x→ 0 +

xx= 1.

As a consequence, we have

lim

x→ 0 +

xx+^1

x

= 1.