Advanced book on Mathematics Olympiad

Real Analysis 473

The last two relations imply

nlim→∞(byn−yn+^1 )=nlim→∞

bn+^2

byn+yn+ 1

=

b

√

b− 1

2

.

Here we used the fact that

lim

n→∞

bn+^2

byn

= lim

n→∞

bn+^2

yn+ 1

=b

√

b− 1.

Sincebyn−yn+ 1 is an integer, if it converges then it eventually becomes constant. Hence

there existsN>Msuch thatbyn−yn+ 1 =b

√b− 1

2 forn>N. This means thatb−1is

a perfect square. Ifbis odd, then

√b− 1

2 is an integer, and sobdivides

b√b− 1

2. Since the

latter is equal tobyn−yn+ 1 forn>N, and this dividesbn+^2 + 3 b^2 − 2 b−5, it follows

thatbdivides 5. This is impossible.

Ifbis even, then by the same argumentb 2 divides 5. Henceb=10. In this case we

have indeed thatxn=(^10

n+ 5

3 )

(^2) , and the problem is solved.
(short list of the 44th International Mathematical Olympiad, 2003)
323.Recall the double inequality
(
1 +

1

n

)n

<e<

(

1 +

1

n

)n+ 1

,n≥ 1.

Taking the natural logarithm, we obtain

nln

(

1 +

1

n

)

< 1 <(n+ 1 )ln

(

1 +

1

n

)

,

which yields the double inequality

1

n+ 1

<ln(n+ 1 )−lnn<

1

n

.

Applying the one on the right, we find that

an−an− 1 =

1

n

−ln(n+ 1 )+lnn> 0 , forn≥ 2 ,

so the sequence is increasing. Adding the inequalities

1 ≤ 1 ,

1

2

<ln 2−ln 1,

1

3

<ln 3−ln 2,