Advanced book on Mathematics Olympiad

472 Real Analysis

For our problem, let>0 be a fixed small positive number. There existsn()such
that for any integern≥n(),

1 −<

(k

n^2

)k

n^2 +^1

k

n^2

< 1 +, k= 1 , 2 ,...,n.

From this, using properties of ratios, we obtain

1 −<

∑n

k= 1

(k

n^2

)k

n^2 +^1

∑n

k= 1

k

n^2

< 1 +, forn≥n().

Knowing that

∑n

k= 1 k=

n(n+ 1 )

2 , this implies

( 1 −)

n+ 1

2 n

<

∑n

k= 1

(

k

n^2

)nk 2 + 1

<( 1 +)

n+ 1

2 n

, forn≥n().

It follows that

nlim→∞

∑n

k− 1

(

k

n^2

)nk 2 + 1

=

1

2

.

(D. Andrica)

322.Assume thatxnis a square for alln>M. Consider the integersyn=

√

xn, for
n≥M. Because in baseb,

b^2 n

b− 1

= (^11) ︸ ︷︷... (^1) ︸
2 n

. 111 ...,

it follows that

nlim→∞

b^2 n

b− 1

xn

= 1.

Therefore,

nlim→∞

bn

yn

=

√

b− 1.

On the other hand,

(byn+yn+ 1 )(byn−yn+ 1 )=b^2 xn−xn+ 1 =bn+^2 + 3 b^2 − 2 b− 5.