Advanced book on Mathematics Olympiad

Real Analysis 485

Let us write

lim

n→∞

ank+^1

nk

=

⎛

⎝lim

n→∞

a

k+k 1

n

n

⎞

⎠

k

.

Using the Cesàro–Stolz theorem, we have

lim

n→∞

a

k+k 1

n

n

=lim

n→∞

a

k+k 1

n+ 1 −a

k+k 1

n

n+ 1 −n

=lim

n→∞

k

√

akn++^11 −k

√

akn+^1

=nlim→∞

akn++^11 −akn+^1

(

k

√

akn++^11

)k− 1

+

(

k

√

akn++^11

)k− 2

√kak+ 1

n +···+

(

√kak+ 1

n

)k− 1

=lim

n→∞

(an+ 1 −an)(ank+ 1 +akn+−^11 an+···+ank)

(

k

√

akn++^11

)k− 1

+

(

k

√

akn++^11

)k− 2

√kak+ 1

n +···+

(

√kak+ 1

n

)k− 1

=nlim→∞

akn+ 1 +akn−+^11 an+···+ank

√kan

((

k

√

ank++ 11

)k− 1

+

(

k

√

ank++ 11

)k− 2

√kak+ 1

n +···+

(

√kak+ 1

n

)k− 1

).

Dividing both sides byank, we obtain

lim

n→∞

a

k+k 1

n

n

= lim

n→∞

(

an+ 1

an

)k

+

(

an+ 1

an

)k− 1

+···+ 1

(

an+ 1

an

)(k+^1 )(kk−^1 )

+

(

an+ 1

an

)(k+^1 )(kk−^2 )

+···+ 1

.

Since limn→∞ana+n^1 = 1 ,we obtain

nlim→∞

a

k+k 1

n

n

=

k+ 1

k

.

Hence

lim

n→∞

ank+^1

nk

=

(

1 +

1

k

)k

.

(67th W.L. Putnam Mathematical Competition, proposed by T. Andreescu; the special
casek=2 was the object of the second part of a problem given at the regional round of
the Romanian Mathematical Olympiad in 2004)

348.Assume no suchξexists. Thenf(a)>aandf(b)<b. Construct recursively the
sequences(an)n≥ 1 and(bn)n≥ 1 witha 1 =a,b 1 =b, and