Applied Mathematics E-Notes, 14(2014), 250-255 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
On The Arithmetic-Geometric Means Of Positive
Integers And The Number e
Mehdi Hassaniy
Received 31 July 2014
Abstract
Assume that An and Gn denote the arithmetic and geometric means of the
integers 1; 2; : : : ; n, respectively. It this paper, we obtain some sharp inequalities
and the asymptotic expansion of the ratio An =Gn .
1
Introduction
Assume that (an )n2N is a positive real sequence. Through the paper, we denote
the arithmetic and geometric means of the numbers a1 ; a2 ; : : : ; an , respectively, by
A(a1 ; : : : ; an ) and G(a1 ; : : : ; an ). A nice relation which connects the number e to the
mean values
An := An (1; 2; : : : ; n) and Gn := Gn (1; 2; : : : ; n)
asserts (see [2]) that
lim
n!1
An
e
= ;
Gn
2
which is a consequence of the Stirling’s approximation
n! =
n
e
n
p
2 n 1+O
1
n
:
(1)
Motivated by this fact, recently we obtained similar asymptotic result concerning the
sequence of prime numbers, by proving validity of
A(p1 ; : : : ; pn )
e
1
= +O
;
G(p1 ; : : : ; pn )
2
log n
where as usual pn denotes the nth prime number. More precisely, we computed the
value of constant of O-term for the case of prime numbers (see [1]).
In this paper, we obtain various properties of the ratio An =Gn , including sharp
and explicit lower and upper bounds, precise asymptotic expansion, and monotonicity.
More precisely, we show the following results.
Mathematics Subject Classi…cations: 26E60, 26D15, 05A10.
of Mathematics, University of Zanjan, University Blvd., 45371-38791, Zanjan, Iran.
The author was supported by a grant of the University of Zanjan, research project number 9247
y Department
250
M. Hassani
251
THEOREM 1. For any integers m > 1 and n > 1, let
J := Jm (n) =
m
X
r=1
B2r
(2r)(2r 1)n2r
1
and um (n) =
jB2m j
2m(2m 1)n2m
1
;
(2)
where Bn denote the Bernoulli numbers. Then, for any integers m > 1 and n > 1,
e
2
1+
1
n
e
1
n (log
p
2 n+J+um (n))
6
e
An
6
Gn
2
1+
1
n
e
1
n (log
p
2 n+J um (n))
COROLLARY 2. For any integer n > 1; we have
!
p
2 n
log2 n
An
e
1
+O
=
1
log
Gn
2
n
e
n2
and
An
Gn
n
=p
en+1
n
1
2n+ 2
1+O
1
n
:
:
(3)
(4)
(5)
COROLLARY 3. For any integer n > 1; we have
An
e
< :
Gn
2
(6)
The proof of the above results is hidden in heart of the following precise form of
Stirling’s approximation for n!.
LEMMA 4. For any integers m > 1 and n > 1; we have
n np
n np
2 n eJ um (n) 6 n! 6
2 n eJ+um (n) :
e
e
(7)
Our last result concerning the ratio An =Gn asserts that the sequence with general
term An =Gn is indeed strictly increasing.
THEOREM 5. For any integer n > 1; we have
An+1
An
>
:
Gn+1
Gn
(8)
Finally, we note that in our proofs we will use the notion of Bernoulli functions
Bn (fxg), where fxg denotes the fractional part of the real x. Among the proofs we
obtain an improper integral concerning the Bernoulli functions as follows.
COROLLARY 6. For any integer m > 1; we have
Z
m
X
1 1 B2m (fxg)
2
B2r
dx
=
log
+
:
2m
2
m 1
x
e
r(2r 1)
r=1
252
2
On the Arithmetic-Geometric Means of Some Numbers
Proofs
PROOF OF LEMMA 4. We apply Euler–Maclaurin summation formula (see [3]) by
letting g(k) = log k, from which we obtain
log n! = n log n
n+
m
X
1
log n + 1
2
B2r
(2r)(2r
r=1
where m > 1 is any …xed integer and
Z 1
B2m (fxg)
dx
Rm =
2mx2m
1
1)
Z
1
n
+
m
X
r=1
B2r
(2r)(2r 1)n2r
1
+ Rm ;
B2m (fxg)
dx:
2mx2m
Thus, we obtain
log n! = n log n
with
Cm = 1 +
Z
1
1
n+
1
log n + Cm + J
2
m
X
B2m (fxg)
dx
2mx2m
r=1
I
B2r
(2r)(2r
1)
(9)
;
(10)
a constant depending, at most, only on m. Also, the remainders J, de…ned as in (2),
and
Z 1
B2m (fxg)
I=
dx
(11)
2mx2m
n
satisfy J
1
n
and I
1
n
as n ! 1. So, if we let
n!
Dn =
n n
e
1
n2
and D = lim Dn ,
n!1
then we have
Cm = lim
n!1
log n!
n log n
n+
1
log n
2
= lim log Dn = log D:
n!1
A simple computation shows that
2
(Dn ) =
n!2 e2n
n2n+1
and D2n =
Hence, we obtain
2
(Dn )
n!2 22n
=
D2n
(2n)!
r
(2n)!e2n
1
(2n)2n+ 2
:
2
:
n
We recall the Wallis product formula for (see [5] for an elementary proof), which
asserts that
n
Y
2k
2k
lim
= :
n!1
2k 1 2k + 1
2
k=1
M. Hassani
253
We note that
n
Y
k=1
2k
2k 1
2k
2k + 1
2
n!2 22n
(2n)!
1
2n + 1
!2
2r
n
(Dn )
1
=
D2n
2
2n + 1
=
=
Hence, we get
2
D2
= lim
n!1
4
(Dn )
D2n
!2
(Dn )
D2n
2
!2
n
:
2(2n + 1)
n
= :
2(2n + 1)
2
p
2 , and consequently
p
Cm = log D = log 2 for any integer m > 1:
Thus, we obtain D =
(12)
Therefore, by using (9), we imply that
n
e
n! =
n
p
n eCm eJ
I
n
e
=
n
p
2 n eJ
I
:
(13)
In particular, we obtain Stirling’s approximation for n! as in (1). More precisely, we
have
Z
Z 1
jB2m j 1 dx
jB2m (fxg)j
dx 6
= um (n):
jIj 6
2mx2m
2m n x2m
n
This completes the proof of Lemma 4.
We apply the relations (10) and (12) to obtain Corollary 6.
PROOF OF COROLLARY 2. By using (13), we obtain
An
e
=
Gn
2
1+
1
n
Thus, we have
An
Gn
(2 n)
n
=
1
2n
e
2
J
e
I
n
n
1+
=
1
n
e
2
1+
n
e
(log
1
n
p
e
1
n (log
2 n+J I)
p
2 n+J I)
:
(14)
:
n
We use the expansion 1 + n1 = e 1 + O( n1 ) to conclude the proof of (5). To prove
(4), we use (14) with the approximation
e
1
n (log
p
2 n+J I)
=1
p
1
log2 n
log 2 n + O
:
n
n2
This completes the proof of Corollary 2.
PROOF OF THEOREM 1. We start from the fact that
n+1
An
=
1 ;
Gn
2 n! n
254
On the Arithmetic-Geometric Means of Some Numbers
and then, we use the sharp inequalities in (7) to complete the proof.
PROOF OF COROLLARY 3. The assertion is valid for n = 1. We consider the
right hand side of the inequalities in (3) with m = 5. In order to prove (6), we require
to have
p
1
1
e n (log 2 n+J5 (n) u5 (n)) < 1:
(15)
1+
n
n
Considering the inequality 1 + n1
< e, which is valid for any integer n > 1, we
observe that the inequality (15) holds true, provided
f (n) := J5 (n)
u5 (n)
1 + log
p
2 n > 0:
The function f (x); de…ned over x 2 [1; 1), is strictly increasing and f (1)f (2) < 0.
Thus, f (n) > 0 for n > 2, from which we imply validity of (6) for n > 2. This
completes the proof of Corollary 3.
PROOF OF THEOREM 5. The inequality (8) is equivalent to
n+1
n+2
n! > (n + 1)n
n(n+1)
:
We prove the last inequality by induction on n. Clearly, it is ture for n = 1. To deduce
the (n + 1)th step from the nth step, we require to have
(n + 1)n+1
n+1
n+2
n(n+1)
> (n + 2)n+1
n+2
n+3
(n+1)(n+2)
;
or equivalently, we should have
(n + 1)n+1 (n + 3)n+2 > (n + 2)2n+3 ;
(16)
for any integer n > 1. Now, we note that (16) is equivalent by the assertion en+1 < en+2
for any integer n > 1, where
n
1
:
(17)
en = 1 +
n
The sequence with general term en is strictly increasing, because if we apply the
Arithmetic–Geometric mean inequality (see [4] for a very fast and elementary proof)
on the numbers
n tim es
z
}|
{
1
1
1;
;:::;
;
n+1
n+1
we imply that
1+n 1+
n+1
1
n
>
s
n+1
1+
1
n
n
;
M. Hassani
255
or equivalently
1
1+
>
n+1
1
1+
n
n
n+1
;
and the later inequality is en+1 > en . The proof is complete.
Acknowledgment. The author wishes to express his thanks to the referees for
studying the paper carefully and giving very valuable comments, more precisely on the
proofs of Lemma 4 and Thoreom 5.
References
[1] M. Hassani, On the ratio of the arithmetic and geometric means of the prime
numbers and the number e, Int. J. Number Theory, 9(2013), 1593–1603.
[2] B. J. McCartin, e: The master of all, Math. Intelligencer, 28(2006), 10–21.
[3] A. M. Odlyzko, Asymptotic Enumeration Methods, Handbook of combinatorics,
Vol. 1, 2, 1063–1229, Elsevier, Amsterdam, 1995.
[4] O. A. S. Karamzadeh, One-line proof of the AM-GM inequality, Math. Intelligencer,
33(2011), Page 3.
[5] J. Wästlund, An elementary proof of the Wallis product formula for pi, Amer.
Math. Monthly, 114(2007), 914–917.