Advanced book on Mathematics Olympiad

Real Analysis 501

This telescopes to

1

2 (n+ 1 )

[n!(n+ 1 )!+(− 1 )n( 2 n+ 1 )!].

(T. Andreescu, second solution by R. Stong)

375.The sequence is obviously strictly decreasing. Becauseak−ak+ 1 = 1 −ak^1 + 1 ,

we have

an=a 0 +(a 1 −a 0 )+···+(an−an− 1 )= 1994 −n+

1

a 0 + 1

+···+

1

an− 1 + 1

> 1994 −n.

Also, because the sequence is strictly decreasing, for 1≤n≤998,

1

a 0 + 1

+···+

1

an− 1 + 1

<

n

an− 1 + 1

<

998

a 997 + 1

< 1 ,

since we have seen above thata 997 > 1994 − 997 =997. Hencean= 1994 −n,as

desired.

(short list of the 35th International Mathematical Olympiad, 1994, proposed by

T. Andreescu)

376.Letx 1 =k+

√

k^2 +1 andx 2 =k−

√

k^2 +1. We have|x 2 |=x^11 < 21 k≤^12 ,so

−(^12 )^2 ≤xn 2 ≤(^12 )n. Hence

x 1 n+x 2 n− 1 <x 1 n+

(

1

2

)n

− 1 <an≤x 1 n−

(

1

2

)n

+ 1 <x 1 n+x 2 n+ 1 ,

for alln≥1. From

xn 1 +^1 +x 2 n+^1 =(x 1 +x 2 )(x 1 n+xn 2 )−x 1 x 2 (x 1 n−^1 +x 2 n−^1 )

= 2 k(x 1 n+xn 2 )+(xn 1 −^1 +xn 2 −^1 )

forn≥1, we deduce thatxn 1 +x 2 nis an integer for alln. We obtain the more explicit

formulaan =x 1 n+x 2 nforn≥ 0, and consequently the recurrence relationan+ 1 =

2 kan+an− 1 , for alln≥1. Then

1

an− 1 an+ 1

=

1

2 kan

·

2 kan

an− 1 an+ 1

=

1

2 k

·

an+ 1 −an− 1

an− 1 anan+ 1

=

1

2 k

(

1

an− 1 an

−

1

anan+ 1

)

.

It follows that

∑∞

n= 1

1

an− 1 an+ 1

=

1

2 k

(

1

a 0 a 1

− lim

N→∞

1

aNaN+ 1

)

=

1

2 ka 0 a 1

=

1

8 k^2

.