On The Arithmetic-Geometric Means Of Positive Integers And The Number e

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On The Arithmetic-Geometric Means Of Positive Integers And The Number e

Mehdi Hassani

^y

Received 31 July 2014

Abstract

Assume that An and Gn denote the arithmetic and geometric means of the integers1;2; : : : ; n, respectively. It this paper, we obtain some sharp inequalities and the asymptotic expansion of the ratioAn=Gn.

1 Introduction

Assume that (an)_n_2N 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) andG(a1; : : : ; an). A nice relation which connects the number e to the mean values

An:=An(1;2; : : : ; n)andGn :=Gn(1;2; : : : ; n) asserts (see [2]) that

nlim!1

An

G_n = e 2; which is a consequence of the Stirling’s approximation

n! = n e

np

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) G(p1; : : : ; pn) =e

2 +O 1 logn ;

where as usual pn denotes the nth prime number. More precisely, we computed the value of constant ofO-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 Sub ject Classi…cations: 26E60, 26D15, 05A10.

yDepartment 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

250

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THEOREM 1. For any integersm>1 andn>1, let J :=J_m(n) =

Xm

r=1

B2r

(2r)(2r 1)n^2r ¹ and u_m(n) = jB2mj

2m(2m 1)n^2m ¹; (2) where Bn denote the Bernoulli numbers. Then, for any integersm>1 andn>1,

e

2 1 + 1

n e ¹ⁿ^(log^p² ^n+J+u^m⁽ⁿ⁾⁾6 An

G_n 6 e

2 1 + 1

n e ¹ⁿ^(log^p² ^n+J ^u^m⁽ⁿ⁾⁾: (3) COROLLARY 2. For any integern>1;we have

An

G_n = e

2 1 1

nlog

p2 n

e +O log²n n²

!

(4)

and A_n

Gn n

= eⁿ⁺¹

p n2ⁿ⁺¹² 1 +O 1

n : (5)

COROLLARY 3. For any integern>1;we have An

G_n < e

2: (6)

The proof of the above results is hidden in heart of the following precise form of Stirling’s approximation forn!.

LEMMA 4. For any integersm>1andn>1; we have n

e

np

2 n e^J ^u^m⁽ⁿ⁾6n!6 n e

np

2 n e^J+u^m⁽ⁿ⁾: (7) Our last result concerning the ratioA_n=G_n asserts that the sequence with general termA_n=G_n is indeed strictly increasing.

THEOREM 5. For any integern>1; we have A_n+1

Gn+1

> A_n 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 integerm>1; we have 1

m Z ₁

1

B2m(fxg)

x^2m dx= log 2 e² +

Xm

r=1

B2r

r(2r 1):

(3)

2 Proofs

PROOF OF LEMMA 4. We apply Euler–Maclaurin summation formula (see [3]) by letting g(k) = logk, from which we obtain

logn! =nlogn n+1

2logn+ 1 Xm

r=1

B2r

(2r)(2r 1) + Xm

r=1

B2r

(2r)(2r 1)n^2r ¹ +Rm;

where m>1is any …xed integer and R_m=

Z ₁

1

B_2m(fxg) 2mx^2m dx

Z ₁

n

B_2m(fxg) 2mx^2m dx:

Thus, we obtain

logn! =nlogn n+1

2logn+C_m+J I (9)

with

C_m= 1 + Z ₁

1

B_2m(fxg) 2mx^2m dx

Xm

r=1

B_2r

(2r)(2r 1); (10)

a constant depending, at most, only on m. Also, the remaindersJ, de…ned as in (2), and

I= Z ₁

n

B_2m(fxg)

2mx^2m dx (11)

satisfyJ _n¹ andI _n¹ asn! 1. So, if we let Dn= n!

n e

nn¹² and D= lim

n!1Dn, then we have

C_m= lim

n!1 logn! nlogn n+1

2logn = lim

n!1logD_n= logD:

A simple computation shows that (D_n)²=n!²e²ⁿ

n²ⁿ⁺¹ and D_2n= (2n)!e²ⁿ (2n)²ⁿ⁺¹²: Hence, we obtain

(D_n)² D2n

= n!²2²ⁿ (2n)!

r2

n:

We recall the Wallis product formula for (see [5] for an elementary proof), which asserts that

nlim!1

Yn

k=1

2k 2k 1

2k 2k+ 1 =

2:

(4)

We note that Yn

k=1

2k 2k 1

2k

2k+ 1 = n!²2²ⁿ (2n)!

2 1

2n+ 1

= (Dn)² D_2n

rn 2

!2

1

2n+ 1 = (Dn)² D_2n

!2

n 2(2n+ 1): Hence, we get

D² 4 = lim

n!1

(Dn)² D_2n

!2

n 2(2n+ 1) =

2: Thus, we obtain D=p

2 , and consequently C_m= logD= logp

2 for any integerm>1: (12) Therefore, by using (9), we imply that

n! = n e

np

n e^C^me^J ^I = n e

np

2 n e^J ^I: (13)

In particular, we obtain Stirling’s approximation for n! as in (1). More precisely, we have

jIj6Z ₁

n

jB2m(fxg)j

2mx^2m dx6jB2mj 2m

Z ₁

n

dx

x^2m =u_m(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

G_n = e 2 1 + 1

n (2 n) ²ⁿ¹ e ^Jⁿ^I = e

2 1 + 1

n e ⁿ¹^(log^p² ^n+J ^I⁾: (14) Thus, we have

A_n Gn

n

= e 2

n

1 + 1 n

n

e ^(log^p² ^n+J ^I):

We use the expansion 1 + _n¹ ⁿ=e 1 +O(_n¹) to conclude the proof of (5). To prove (4), we use (14) with the approximation

e ⁿ¹^(log^p² ^n+J ^I)= 1 1 nlogp

2 n+O log²n n² : This completes the proof of Corollary 2.

PROOF OF THEOREM 1. We start from the fact that An

G_n = n+ 1 2n!ⁿ¹;

(5)

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

1 + 1

n e ⁿ¹^(log^p² ^n+J⁵⁽ⁿ⁾ ^u⁵⁽ⁿ⁾⁾<1: (15) Considering the inequality 1 + ¹_n ⁿ < 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 + logp

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!>(n+ 1)ⁿ n+ 1

n+ 2

n(n+1)

:

We prove the last inequality by induction onn. Clearly, it is ture forn= 1. To deduce the(n+ 1)^th step from then^th 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 integern>1. Now, we note that (16) is equivalent by the assertionen+1< en+2

for any integern>1, where

e_n= 1 + 1 n

n

: (17)

The sequence with general term e_n is strictly increasing, because if we apply the Arithmetic–Geometric mean inequality (see [4] for a very fast and elementary proof) on the numbers

1;

ntim es

z }| { 1

n+ 1; : : : ; 1 n+ 1; we imply that

1 +n 1 + ¹_n

n+ 1 > ⁿ⁺¹ s

1 + 1 n

n

;

(6)

or equivalently

1 + 1

n+ 1 > 1 + 1 n

n n+1

; and the later inequality isen+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 numbere, 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.