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Adv. Inequal. Appl. 2015, 2015:4 ISSN: 2050-7461

NOTE ON STIRLING’S FORMULA AND WALLIS’ INEQUALITY OF ORDER m

G. ZABANDAN

Department of Mathematics, Kharazmi University, Tehran, Iran.

Copyright c2015 G. Zabandan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we obtain Stirling’s formula and improvement of it. Also we get the Wallis’ inequality and Wallis’ formula of orderm.

Keywords:Stirling’s formula; Gamma function; Wallis’ inequality; Wallis’ formula.

2010 AMS Subject Classification:26D15, 26D10.

1.

Introduction

The Gamma functionΓ:(0,∞)−→Ris defined by the relation Γ(x) =

Z +∞

0

tx−1e−tdt (x>0).

The Gamma function has the following well know properties [1]:

Γ(1) =1.

(1.1)

Γ(x+1) =xΓ(x) (Γ(n+1) =n!, n∈N).

(1.2)

E-mail address: zabandan@khu.ac.ir Received November 24, 2014

(2)

Γis log-convex, that is lnΓ(x)is convex on(0,∞),

(1.3)

Γ(

x 2)Γ(

x+1 2 ) =

π

2x−1Γ(x)(Gauss-Legendre duplication formula),

(1.4)

m−1

k=0

Γ(x+

k

m) = (2π) m−1

2 m 1

2−mx,Γ(mx)(Gauss multiplication formula). (1.5)

Closely related to Gamma function is the Beta functionB, which is the real function of two variables defined by

B(x,y) = Z 1

0

tx−1(1−t)y−1dt (x,y>0).

The following connection between the Beta and Gamma function hold [1]

(1.6) B(x,y) = Γ(x)Γ(y)

Γ(x+y) (x,y>0).

Let f :[a,b]−→Rbe a convex function, then the inequality

(1.7) f(a+b

2 )≤ 1 b−a

Z b

a

f(x)dx≤ f(a) + f(b)

2

is known as the Hermite-Hadamard inequality. I previously obtained a new refinement of Hermite-Hadamard inequality [6]

f(a+b 2 )≤

1 n+1

n

k=0

f(a+2k+1

2n+2(b−a))≤ 1 b−a

Z b

a

f(x)dx (1.8)

≤ 1

n+1

n

k=0

f(a+k

n(b−a))≤

f(a) +f(b)

2 .

Wallis’ inequality is a well known and important inequality

(1.9) q 1

πn+π2

< 1.3.5. . . .(2n−1)

2.4.6. . . .(2n) < 1

πn

or

(1.10) q 1

πn+π

2

< (2n)!

(n!)222n <

1

πn

and Wallis’ formula is

(1.11)

r

π

2 =nlim→∞

(3)

In this paper by the log-convexity ofΓand the refinement of Hermite-Hadmad inequality, the

following inequalities will be obtained

aΓ(a+1

2)≤

2πa aae−a

≤√a m+1

s m

k=0

Γ(a+ k

m)

!

≤Γ(a+1)≤ m+1

s m

k=0

Γ(a+2k+1

m+1)

≤√2πa

a

2(a+1) a+1

2 e−a− 1 2

≤ r

a+1

a Γ(a+1)

and we will deduce the Stirling’s formula and improvement of it.

Moreover in this paper we will get the Wallis’ inequality and Wallis’ formula of orderm

mn

m−1

k=0

Γ(n+mk)

(n!)m−1p

(2π)m−1m

< (mn)!

(n!)mmmn <

m

p

(2πn)m−1

,

lim

n→∞

(n!)2mm2mn

((mn)!)2(mn+1)m−1 =

(2π)m−1

mm

and we show that the Wallis’ inequality and Waills’ formula of order 2 are classic Wallis’ inequality and formula.

Finally we confirm these results by the infinite product representation of the sine [1] sinx=x

n=1

1− x 2

π2n2

form=2,3,4.

2.

Main results

First we bring two easy and well known Lemmas.

(4)

(i)R1

0 lnΓ(x)dx=ln √

2π,

(ii)Ra+1

0 lnΓ(x)dx=−a+alna+ln √

2π.

Proof.(i) By (1.4) we have

lnΓ(x) = (x−1)ln 2−ln

π+lnΓ(x

2) +lnΓ( x+1

2 ). By integrating, we obtain

Z 1

0

lnΓ(x)dx=ln 2

Z 1

0

(x−1)dx−ln√π+

Z 1

0

lnΓ(x

2)dx+ Z 1

0

lnΓ(x+1

2 )dx. Since

Z 1

0

lnΓ(x

2)dx=2 Z 12

0

lnΓ(t)dt

and

Z 1

0

lnΓ(x+1

2 )dx=2 Z 1

1 2

lnΓ(t)dt,

we deduce that

Z 1

0

lnΓ(x)dx=−1

2ln 2−ln

π+2

Z 1

0

lnΓ(x)dx.

Hence, we have

Z 1

0

lnΓ(x)dx= 1

2ln 2+ 1

2lnπ=ln

2π.

(ii) We have

Z a+1

a

lnΓ(x)dx

0

= (−a+alna)0=lna.

HenceRa+1

a lnΓ(x)dx+a−alnais constant. By part (i) we obtain

Z a+1

a

lnΓ(x)dx+a−alna=ln

2π.

Lemma 2.2.

lim

a→∞

aΓ(a+12)

(5)

Proof.By identity (1.6) we have

aΓ(a+12)

Γ(a+1) =

aB(a+12,12)

Γ(12)

= 1

Γ(12)

a Z 1

0

x12−1(1x)a+ 1 2−1dx

= 1

Γ(12)

a Z 1

0

x−12(1−x)a− 1 2dx.

Takex=at, sodt =adx. Since lim

a→∞(1−

t a)a−

1

2 =e−t, we get

lim

a→∞

aΓ(a+12)

Γ(a+1) =

1

Γ(12)

lim

a→∞

a Z a

0

t−12 a−12

(1−t

a)

a−1 21

adt = 1

Γ(12)

Z +∞

0

t−12e−tdt = 1

Γ(12) Γ(1

2) =1. In the following theorems, we obtain the Stirling’s formula and improvment of it.

Theorem 2.3. For all m∈Nand a>0the following inequalities hold:

(i) √aΓ(a+1

2)≤

a m+1

s m

k=0

Γ(a+ 2k+1

2m+2)

!

≤√2πa aae−a

≤√a m+1

s m

k=0

Γ(a+ k

m)

!

≤Γ(a+1)

and we have

(ii) lim

a→∞

a

m+1

r m

k=0

Γ(a+22mk++12)

Γ(a+1) =alim→∞

a

m+1

r m

k=0

Γ(a+mk)

Γ(a+1)

= lim

a→∞

2πa aae−a

Γ(a+1) =1.

Proof. Since f(x) =lnΓ(x)is convex on(0,∞), by the right side of inequalities (1.8), we have

Ra+1

a lnΓ(x)dx≤ m+11 m

k=0

lnΓ(a+mk)≤lnΓ(a)+2lnΓ(a+1) =lnpΓ(a)Γ(a+1).

By Lemma 2.1, we get

−a+alna+ln√2π≤ 1

m+1ln

m

k=0

Γ(a+ k

m)≤ln

q

(6)

It follows that

lnaa√2πe−a≤ln m+1

s m

k=0

Γ(a+ k

m)≤ln

aΓ(a).

Sinceex is increasing, we obtain

2πaae−a≤ m+1

s m

k=0

Γ(a+ k

m)≤

aΓ(a).

By multiplying both side of inequalities√a, we get

(∗) √2πa aae−a≤

a m+1

s m

k=0

Γ(a+ k

m)

!

≤aΓ(a) =Γ(a+1).

On the other hand, by the left side of inequalities (1.8) we have

ln(2a+1 2 )≤

1 m+1

m

k=0

lnΓ(a+ 2k+1

2m+2)≤ Z a+1

a

lnΓ(x)dx

so

lnΓ(a+1

2)≤ln m+1

s m

k=0

Γ(a+ 2k+1

2m+2)≤ Z a+1

a

lnΓ(x)dx=ln

2π aae−a.

Hence

Γ(a+1

2)≤ m+1

s m

k=0

Γ(a+ 2k+1

2m+2)≤

2πaae−a.

Multiplying both side of inequalities by√a, we get

(∗∗) √aΓ(a+1

2)≤

a m+1

s m

k=0

Γ(a+ 2k+1

2m+2)

!

≤√2πa aae−a

By (*) and (**) the proof of (i) is complete.

For the proof of (ii) divide all of inequalities (i) byΓ(a+1)

aΓ(a+12)

Γ(a+1) ≤

a

m+1

r m

k=0

Γ(n+22mk++12)

Γ(a+1) ≤

2πa aae−a

Γ(a+1)

≤ √

a

m+1

r m

k=0

Γ(a+mk)

Γ(a+1) ≤1.

(7)

Theorem 2.4. For all m∈Nand a>0the following inequalities hold:

(i) Γ(a+1)≤ m+1

s m

k=0

Γ(a+2k+1

m+1)<

2πa

a

2(a+1) a+1

2 e−a− 1 2

≤ m+1

s m

k=0

Γ(a+2k

m)<

r

a+1

a Γ(a+1).

(ii) lim

a→∞

m+1

r m

k=0

Γ(a+2mk++11)

Γ(a+1) =alim→∞

m+1

r m

k=0

Γ(a+2mk)

Γ(a+1)

= lim

a→∞

2πaa2(a+1)a

+1 2 e−a−

1 2

Γ(a+1) =1.

Proof.(i) By Lemma 2.1 we have

1 2

Z a+2

a

lnΓ(x)dx= 1

2

Z a+1

a

lnΓ(x)dx+

Z a+2

a+1

lnΓ(x)dx

= 1 2

h

lnaa√2πe−a+ln(a+1)a+1

2πe−(a+1)

i

=ln

q

2πaa(a+1)a+1e−2a−1.

Now by the right side of inequalities (1.8), we get

1 2

Z a+2

a

lnΓ(x)dx=ln

q

2πaa(a+1)a+1e−2a−1≤ln m+1

s m

k=0

Γ(a+2k

m)

≤lnpΓ(a)Γ(a+2)

so

q

2πaa(a+1)a+1e−2a−1≤ m+1

s m

k=0

Γ(a+2k

m)

≤p(a+1)Γ(a)Γ(a+1) =

r

a+1

(8)

On the other hand, by the left side of inequalities (1.8), we have

lnΓ(2a+2

2 )≤ 1 m+1

m

k=0

lnΓ(a+2k+1

m+1)

≤ 1

2 Z n+2

n

lnΓ(x)dx

=ln

q

2πaa(a+1)a+1e−2a−1

so

(∗∗) Γ(a+1)≤ m+1

s m

k=0

Γ(a+2k+1

m+1)≤

q

2πaa(a+1)a+1e−2a−1.

By(∗)and(∗∗)the proof is complet. The proof of (ii) is clear by (i).

Corollary 2.5. For all m∈Nand a>0the following inequalities hold:

aΓ(a+1

2)≤

2πa aae−a≤

a m+1

s m

k=0

Γ(a+ k

m)

!

≤Γ(a+1)

≤ m+1

s m

k=0

Γ(a+2k+1

m+1)≤

2πa

a

2(a+1)a

+1 2 e−a−

1 2

≤ r

a+1

a Γ(a+1)

Remark 2.6. In the above corollary, letabe a natural numbern. Then we have

nΓ(n+1

2)≤

2πn nne−n≤√n m+1

s m

k=0

Γ(n+ k

m)

!

≤n!

≤ m+1

s m

k=0

Γ(n+2k+1

m+1)≤

2π n

n

2(n+1) n+1

2 e−n− 1 2

≤ r

(9)

We can write m s

m−1

k=0

Γ(n+2km+1)(m=1,2, . . .)instead of m+1

r m

k=0

Γ(n+2mk++11) (m=0,1,2, . . .).

So

nΓ(n+1

2)≤

2πn nne−n≤

n m+1

s m

k=0

Γ(n+

k m)

!

≤n!

≤ m v u u t

m−1

k=0

Γ(n+2k+1

m )≤

2πn

n

2(n+1) n+1

2 e−n− 1 2

≤ r

n+1 n n!.

Now letxm= √

n

m+1

r m

k=0

Γ(n+mk)

andym= m s

m−1

k=0

Γ(n+2km+1). Form=1,2,3 we have

x1=y1=n!

x2=√n3

r

Γ(n)Γ(n+1

2)Γ(n+1), y2=

r

n+1 2 Γ(n+

1 2) x3=√n4

r

Γ(n)Γ(n+1

3)Γ(n+ 2

3)Γ(n+1), y3= 3

r

n+2 3 3

r

Γ(n+1

3)Γ(n+ 2

3)Γ(n+1).

Since{xm}is decreasing and{ym}is increasing (see [5], [6]), we obtain the following inequal-ities

nΓ(n+1

2)<

2πn nne−n<· · ·<

n 4

r

Γ(n)Γ(n+1

3)Γ(n+ 2 3)

<√n 3

r

Γ(n)Γ(n+1

2)Γ(n+1)<n!

and

n!< r

n+1 2Γ(n+

1 2)<

3

r

n+2 3

3

r

Γ(n+1

3)Γ(n+ 2

3)Γ(n+1)

<· · ·<√2πn

n

2(n+1) n+1

2 e−n− 1 2 <

r

(10)

3.

Wallis’ inequality and Wallis’ formula

Theorem 3.1. For all m∈Nand n∈Nthe following inequalities hold: mnm−1

k=0

Γ(n+mk)

(n!)m−1p(2π)m−1m m s

m−1

k=0

Γ(n+2km+1)

!<

(mn)! (n!)mmmn <

m

p

(2πn)m−1

.

ProofBy formula (1.5) and remark 2.6, we have

n m+1

q

(2π)

m−1 2 m

1

2−mnΓ(mn)Γ(n+1) =

n m+1

s m

k=0

Γ(n+ k

m)

!

<n!.

So

nm+21(2π) m−1

2 m 1

2−mnΓ(mn)n!<(n!)m+1=

nm+21(2π) m−1

2 m 1

2−mnΓ(mn)<(n!)m=⇒

Γ(mn)

(n!)mmmn <

1 nm+21(2π)

m−1 2 m

1 2

=⇒

mnΓ(mn)

(n!)mmmn <

mn nm+21(2π)

m−1 2 m 1 2 = √ m nm−21(2π)

m−1 2

=

m

p

(2πn)m−1

.

Hence

(mn)! (n!)mmmn <

m

p

(2πn)m−1

.

For the proof of left side again by formula (1.5) we have

m−1

k=0

Γ(n+ k

m) = (2π) m−1

2

mΓ(mn)

mmn = (2π) m−1

2

m(mn)! mmn(mn).

So

(mn)! mmn =

mn (2π)

m−1 2 m

1 2

m−1

k=0

Γ(n+ k

m) =

mn

p

(2π)m−1m

m−1

k=0

Γ(n+ k

m). Hence by remark 2.6 and above identity we obtain

mn

m−1

k=0

Γ(n+mk)

(n!)m−1p

(2π)m−1m m

s m−1

k=0

Γ(n+2km+1)

!=

(mn)!

(n!)m−1mmn m s

m−1

k=0

Γ(n+2km+1)

!

< (mn)!

(n!)m−1mmnn! =

(11)

The proof is complete.

Definition 3.2. We call the inequality

(3·1)

mn

m−1

k=0

Γ(n+mk)

(n!)m−1p

(2π)m−1m m

s m−1

k=0

Γ(n+2km+1)

! <

(mn)! (n!)mmmn <

m

p

(2πn)m−1

Wallis’ inequality of orderm. Form=2 we have

2n

1

k=0

Γ(n+k2)

(n!)√2π·2

s 1

k=0

Γ(n+2k2+1)

! <

(2n)! (n!)222n <

2

2πn

.

Thus

1

q

πn+π2

< (2n)!

(n!)222n <

1

πn,

which is the Wallis’ inequality. In other words, the Wallis’ inequality of order 2 is the classic Wallis inequality.

In the following theorem we extend Wallis’ formula for orderm.

Theorem 3.3. For all m∈Nwe have lim

n→∞

(mn)!(mn+1)m−21 (n!)mmmn =

mm2 (2π)

m−1 2

.

Proof.multiply both sides of Wallis’ inequality of ordermby(mn−1)m−21, mn(mn+1)m−21

m−1

k=0

Γ(n+mk)

(n!)m−1p

(2π)m−1m m

s m−1

k=0

Γ(n+2km+1)

! <

(mn)!(mn+1)m−21 (n!)mmmn

< √

m(mn+1)m−21

p

(2πn)m−1

.

It is obvious that

(∗) lim

n→∞

m(mn+1)m−21

p

(2πn)m−1

=

mmm−21 (2π)

m−1 2

= m m 2

(2π)

m−1 2

(12)

On the other hand, we have

lim

n→∞

mn(mn+1)m−21

m−1

k=0

Γ(n+mk)

(n!)m−1p

(2π)m−1.m m

s m−1

k=0

Γ(n+2km+1)

! =nlim

nm+21

m

k=0

Γ(n+mk)

(n!)m+1

lim

n→∞

n! m

s m−1

k=0

Γ(n+2km+1)

lim

n→∞

mn(mn+1)m−21

p

(2π)m−1m.n

m+1 2

.

The first and second limits are equal to 1 by theorems 2.3 and 2.4. For the third limit we have

lim

n→∞

mn(mn+1)m−21 (2π)m−

1 2 .m

1 2.nm

+1 2

= lim

n→∞

m12(mn+1) m−1

2

(2π)m−

1 2 .nm−

1 2

= m 1 2.m

m−1 2

(2π)

m−1 2

= m m 2

(2π)

m−1 2

.

So

(∗∗) lim

n→∞

mn(mn+1)m−21

m−1

k=0

Γ(n+mk)

(n!)m−1p

(2π)m−1m m

s m−1

k=0

Γ(n+2km+1)

!=

mm2 (2π)

m−1 2

.

Now the assertion is obvious by (*) and (**).

Definition 3.4. We call the following formulas

(3·2) lim

n→∞

(n!)mmmn (mn)!(mn+1)m−21

=(2π) m−1

2

mm2 or

(3·3) lim

n→∞

(n!)2mm2mn

((mn)!)2(mn+1)m−1 =

(2π)m−1

mm

Wallis’ formula of orderm.

Remark 3.5. Takingm=2 in (3.3) we obtain

lim

n→∞

(n!)424n ((2n)!)2(2n+1) =

22 =

π

(13)

It is the classic Wallis’ formula. On the other hand, we know the Wallis’ formula follows from the infinite product representation of the sine [1].

(3·4) sinx=x

n=1

(1− x 2

(πn)2)

Takingx= π

2 in (3.4) gives

1=sinπ 2 =

π

2

n=

(1−

π2

4

π2n2) = π

2

n=1

4n2−1 4n2

so

2

π =

n=1

4n2−1 4n2 or

π

2 =

n=1

(2n)2 (2n−1)(2n+1). Hence

π

2 =nlim→∞

(2·4·6·. . .2n)2

(1·3)(3·5). . .(2n−1)(2n+1) =nlim→∞

24n(n!)2 ((2n)!)2(2n+1)

so (3.3) and (3.4) have the same results.

Now we verify (3.2) and (3.4) have the same results form=3 andx= π

3, respectivly. Taking

m=3 in (3.2) andx= π

3 in (3.4) we get

lim

n→∞

(n!)333n (3n)!(3n+1)=

3√3

and

3 2 =sin

π

3 =

π

3

n=0

(1−

π2

9

π2n2) = π

3

n=1

9n2−1 9n2 .

So

3√3 2π =

n=1

9n2−1 9n2 or

3√3=

n=1

(3n)2 (3n−1)(3n+1). Hence

3√3=nlim→∞

(3·6·9. . .3n)2

(2·4)(5·7)(8·10). . .(3n−1)(3n+1) = lim

n→∞

32n(n!)2(3·6·9. . .3n)

(3n)!(3n+1) =nlim→∞

33n(n!)3 (3n)!(3n+1). Finally we show that (3.3) and (3.4) have the same results form=4 andx= π

4, respectively.

Takingm=4 in (3.3) andx=π

4 in (3.4) we obtain

lim

n→∞

(n!)848n

((4n)!)2(4n+1)3 =

(2π)3

44 =

π3

(14)

and 2 2 =sin

π

4 =

π

4

n=1

16n2−1 16n2 . So

2√2

π =

n=1

16n2−1 16n2 or

π

2√2 =

n=1

(4n)2 (4n−1)(4n+1). Hence

π

2√2 =nlim→∞

(4.8.12. . .4n)2

(3.5)(7.9)(11.13). . .(4n−1)(4n+1) = lim

n→∞

(4n.n!)2(2.4.6. . .4n)

(4n)!(4n+1) =nlim→∞

42n(n!)222n(2n)! (4n)!(4n+1) . So

π2

8 =nlim→∞

44n(n!)424n((2n)!)2 ((4n)!)2(4n+1)2

= lim

n→∞

(n!)444n (4n)!(4n+1)32

lim

n→∞

24n((2n)!)2 (4n)!√4n+1 = lim

n→∞

(n!)444n (4n)!(4n+1)32

lim

p→∞

22p(p!)2 (2p)!√2p+1 =

r

π

2nlim→∞

(n!)444n (4n)!(4n+1)32

.

Since the second limit is equal toqπ

2 by Wallis’ formula of order 2. Therefore, we have

lim

n→∞

(n!)444n (4n)!(4n+1)32

=

π2

8

π

2

= π 3 2

4√2. It follows that limn→∞

(n!)848n

((4n)!)2(4n+1)3 = π 3

32.

Conflict of Interests

The author declares that there is no conflict of interests.

[1] Emil Artin, The Gamma function, Holt, Rinehart and Winston, Inc. 1964.

[2] S.S. Dragomir, R. P. Agarwal, N. S. Barnet, Inequality for Beta and Gamma function via some classical and new integral inequalities, J. Inequal. Appl. 2000.

(15)

[3] S.S. Dragomir, M. Raissouli, Iterative refinements of the Hermite-Hadamard inequality, Application to the standard means, J. Inequal. Appl. 2010 (2010), Article ID 107950.

[4] J. E. Pecaric, F. Proschan and Y. L. Tong, Convex functions, partial orderings, and statictical Applications, Academic Press, 1992.

[5] G. Zabandan, A new refinement of Hermite-Hadamard inequality for convex functions, J. Ineq. Pure Appl. Math. 10 (2009), Article ID 45.

[6] G. Zabandan, Bernstein’s polynomials for convex functions and related resutls, Int. J. Nonlinear Anal. Appl. 6 (2015), 23-34.

References

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