n - OEI

PROPOSED SOLUTION TO PROBLEMA 141
OVIDIU FURDUI
Sea
n
i
un numero combinatorio, n, i ∈ N n ≥ i. Demostrar entonces que
n n
n
···
1
2
n
1
2
y e 2 n son dos infinitos asintoticamente equivalentes, es decir,
1
lim
n→∞
2
e2n
n n
1 2 ···
n
n
= 1.
Solution. In what follows we are going to prove that
2
1
lim
n→∞
e2n
n n
1 2 ···
n
n
= ∞.
We need the following Lemma.
Lemma 0.1. Glaisher-Kinkelin constant. The following limit holds:
A = lim
n→∞
n
11 22 · · · nn
,
e−n2 /4
n2 /2+n/2+1/12
where A = 1.282712 is the Glaisher-Kinkelin constant.
1
Let xn =
2
e2n
.
)(n2 )···(nn)
n
1
(
ln xn =
We have
n
n
k=1
k=1
X
n!
n2
n2 X
−
ln
=
− (n + 1) ln n! + 2
ln k!.
2
k!(n − k)!
2
Date: May 7, 2007.
1
2
OVIDIU FURDUI
On the other hand, we note that
n
X
ln k! = n ln 1 + (n − 1) ln 2 + · · · + 2 ln(n − 1) + ln n =
k=1
n
X
(n + 1 − k) ln k
k=1
= (n + 1) ln n! −
n
X
k ln k.
k=1
Thus,
n
ln xn =
X
n2
+ (n + 1) ln n! − 2
k ln k
2
k=1
!
2
n
X
n
n
1
n2
n2
k ln k −
+ (n + 1) ln n! − 2
+ +
ln n +
=
2
2
2
12
4
k=1
1
n2
1
= n2 + (n + 1) ln n! − n2 + n +
ln n +
ln n − 2An ,
− n2 + n +
6
2
6
where
An =
n
X
k ln k −
k=1
Using Stirling’s formula n! ≈
√
2πn
n n
,
e
n2
n
1
+ +
2
2
12
ln n +
n2
.
4
we get that
1
n + (n + 1) ln n! − n + n +
6
2
2
√
ln n
ln n ≈ (n + 1) ln 2πn − n −
.
6
It follows, based on Lemma 0.1, that
√
ln n
ln xn ≈ (n + 1) ln 2πn − n −
− 2An → ∞ − 2 ln A = ∞,
6
and hence xn → ∞.
Western Michigan University, Kalamazoo, MI
E-mail address: [email protected], [email protected]
Revista Escolar de la Olimpíada Iberoamericana de
Matemática
http://www.campus-oei.org/oim/revistaoim/
Edita: