Advanced book on Mathematics Olympiad

484 Real Analysis

Now we turn to the geometric mean. Applying the Cesàro–Stolz theorem to the sequences

un=ln

P( 1 )

1 m

+ln

P( 2 )

2 m

+···+ln

P (n)

nm

andvn=n,n≥1, we obtain

lim

n→∞

un

vn

=lim

n→∞

ln

Gn

(n!)m/n

=lim

n→∞

ln

P (n)

nm

=lnam.

We therefore have

nlim→∞

An

Gn

·

(√n

n!

n

)m

=

1

m+ 1

.

Now we can simply invoke Stirling’s formula

n!≈nne−n

√

2 πn,

or we can argue as follows. If we letun=nnn!, then the Cesàro–Stolz theorem applied to
lnunandvn=nshows that ifunu+n^1 converges, then so doesn

√

un, and to the same limit.
Because

nlim→∞

un+ 1

un

=nlim→∞

(

n

n+ 1

)n

=

1

e

,

we have

lim

n→∞

√nn!

n

=

1

e

.

Therefore,

nlim→∞

An

Gn

=

em

m+ 1

.

(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), 1937, proposed by T. Popo-
viciu)

347.Clearly,(an)n≥ 0 is an increasing sequence. Assume thatanis bounded. Then it must
have a limitL. Taking the limit of both sides of the equation, we have

lim

n→∞

an+ 1 = lim

n→∞

an+lim

n→∞

1

√kan,

orL=L+√k^1 L, contradiction. Thus limn→∞an=+∞and dividing the equation by

an, we get limn→∞ana+n^1 =1.