1 Introduction

The Wallis sequence to which the title refers is

W n = k = 1 n 4 k 2 4 k 2 1 ,nN:={1,2,3,}.
(1.1)

Wallis (1616-1703) discovered that

k = 1 4 k 2 4 k 2 1 = 2 1 2 3 4 3 4 5 6 5 6 7 8 7 = π 2
(1.2)

(see [[1], p.68]). Based on Wallis’ infinite product (1.2), the first infinite continued fraction of π was given by Brouncker (1620-1684):

4 π =1+ 1 2 + 9 2 + 25 2 + 49 2 + .
(1.3)

Euler’s analysis of Wallis’ proof led him to formulas for the gamma and beta functions. Stirling (1692-1770) used (1.2) to determine the constant factor in his asymptotic formula

n! 2 π n ( n e ) n ,n.

Several elementary proofs of (1.2) can be found (see, for example, [24]). An interesting geometric construction produces (1.2) [5]. Many formulas exist for the representation of π, and a collection of these formulas is listed in [6, 7]. For more on the history of π see [1, 810].

Some inequalities and asymptotic formulas associated with the Wallis sequence W n can be found (see, for example, [1123]). In [13], Hirschhorn proved that for nN,

π 2 ( 1 1 4 n + 7 3 ) < W n < π 2 ( 1 1 4 n + 8 3 ) .
(1.4)

Also in [13], Hirschhorn pointed out that if the c j are given by

tanh ( x 4 ) = j = 0 c j x 2 j + 1 ( 2 j ) ! ,
(1.5)

then, as n,

W n π 2 ( 1 + 1 2 n ) 1 j 0 exp ( c j n 2 j + 1 ) = π 2 ( 1 + 1 2 n ) 1 exp ( j = 0 c j n 2 j + 1 ) .
(1.6)

Remark 1 It is well known (see [[24], p.85]) that

tanhz= k = 1 2 2 k ( 2 2 k 1 ) B 2 k ( 2 k ) ! z 2 k 1 ,|z|< π 2 ,

where B n (n N 0 :=N{0}) are the Bernoulli numbers. We then obtain

tanh ( x 4 ) = j = 0 ( 2 2 j + 2 1 ) B 2 j + 2 2 2 j + 1 ( 2 j + 1 ) ( j + 1 ) x 2 j + 1 ( 2 j ) ! ,|x|<2π.
(1.7)

Thus we have

c j = ( 2 2 j + 2 1 ) B 2 j + 2 2 2 j + 1 ( 2 j + 1 ) ( j + 1 ) ,j N 0 :=N{0}.
(1.8)

Let r0 be a given real number and 0 be a given integer. The first aim of this paper is to determine the coefficients c j (,r) (for jN) such that

W(x) π 2 ( 1 + j = 1 c j ( , r ) x j ) x / r as x,

where the function W(x) is defined by

W(x)= π 2 ( 1 + 1 2 x ) 1 1 x [ Γ ( x + 1 ) Γ ( x + 1 2 ) ] 2 .
(1.9)

Clearly, W n =W(n). The second aim of this paper is to establish inequalities for the Wallis sequence W n .

2 A useful lemma

The classical Euler’s gamma function is defined for x>0 by

Γ(x)= 0 t x 1 e t dt.

The logarithmic derivative of Γ(x), denoted by ψ(x)= Γ (x)/Γ(x), is called psi (or digamma) function, and ψ ( k ) (x) (kN) are called polygamma functions.

The following lemma is required in our present investigation.

Lemma 1 ([[25], Corollary 2.1])

Let m,nN. Then for x>0,

j = 1 2 m ( 1 1 2 2 j ) 2 B 2 j ( 2 j ) ! ( 2 j + n 2 ) ! x 2 j + n 1 < ( 1 ) n ( ψ ( n 1 ) ( x + 1 ) ψ ( n 1 ) ( x + 1 2 ) ) + ( n 1 ) ! 2 x n < j = 1 2 m 1 ( 1 1 2 2 j ) 2 B 2 j ( 2 j ) ! ( 2 j + n 2 ) ! x 2 j + n 1 ,
(2.1)

where B n are the Bernoulli numbers.

It follows from (2.1) that, for x>0,

L(x)<ψ(x+1)ψ ( x + 1 2 ) <U(x),
(2.2)

where

L ( x ) = 1 2 x 1 8 x 2 + 1 64 x 4 1 128 x 6 + 17 2 , 048 x 8 31 2 , 048 x 10 + 691 16 , 384 x 12 5 , 461 32 , 768 x 14

and

U ( x ) = 1 2 x 1 8 x 2 + 1 64 x 4 1 128 x 6 + 17 2 , 048 x 8 31 2 , 048 x 10 + 691 16 , 384 x 12 5 , 461 32 , 768 x 14 + 929 , 569 1 , 048 , 576 x 16 .

In Section 4, the proofs of Theorems 3 and 4 make use of inequality (2.2).

3 Asymptotic expansions

The logarithm of gamma function has asymptotic expansion (see [[26], p.32]):

lnΓ(x+t) ( x + t 1 2 ) lnxx+ 1 2 ln(2π)+ n = 1 ( 1 ) n + 1 B n + 1 ( t ) n ( n + 1 ) 1 x n
(3.1)

as x, where B n (t) denotes the Bernoulli polynomials defined by the following generating function:

x e t x e x 1 = n = 0 B n (t) x n n ! .
(3.2)

Note that the Bernoulli numbers B n (for n N 0 ) are defined by (3.2) for t=0.

From (3.1), we obtain, as x,

[ Γ ( x + t ) Γ ( x + s ) ] 1 / ( t s ) xexp ( 1 t s j = 1 ( 1 ) j + 1 ( B j + 1 ( t ) B j + 1 ( s ) ) j ( j + 1 ) 1 x j ) .
(3.3)

Setting (s,t)=( 1 2 ,1) and noting that

B n (0)= ( 1 ) n B n (1)= B n and B n ( 1 2 ) = ( 2 1 n 1 ) B n for n N 0

(see [[24], p.805]), we obtain from (3.3), as x,

[ Γ ( x + 1 ) Γ ( x + 1 2 ) ] 2 xexp ( j = 1 2 ( 1 ( 1 ) j + 1 ( 2 j 1 ) ) B j + 1 j ( j + 1 ) 1 x j ) ,
(3.4)

or

1 x [ Γ ( x + 1 ) Γ ( x + 1 2 ) ] 2 exp ( j = 0 ( 2 2 2 j 1 ) B 2 j + 2 ( 2 j + 1 ) ( j + 1 ) 1 x 2 j + 1 ) .
(3.5)

We see from (1.6) and (1.9) that

1 n [ Γ ( n + 1 ) Γ ( n + 1 2 ) ] 2 exp ( j = 0 c j n 2 j + 1 ) ,n,
(3.6)

with the coefficients c j given by (1.5). From (3.5) and (3.6), we retrieve (1.8).

By using the Maclaurin expansion of ln(1+x),

ln(1+x)= j = 1 ( 1 ) j 1 j x j for 1<x1,

we obtain

( 1 + 1 2 x ) 1 exp ( j = 1 ( 1 ) j j 2 j 1 x j ) as x.
(3.7)

Applying (3.4) and (3.7) yields

W(x) π 2 exp ( j = 1 b j x j ) as x,
(3.8)

with the coefficients b j (for jN) given by

b j = ( 1 ) j 1 ( 1 j 2 j + 2 ( ( 1 ) j + 1 ( 2 j 1 ) ) B j + 1 j ( j + 1 ) ) .
(3.9)

From (3.8), we obtain the following asymptotic expansion for the Wallis sequence W n :

W n π 2 exp ( 1 4 n + 1 8 n 2 5 96 n 3 + 1 64 n 4 1 320 n 5 + 1 384 n 6 25 7 , 168 n 7 + 1 2 , 048 n 8 + 29 9 , 216 n 9 + 1 10 , 240 n 10 695 90 , 112 n 11 + ) , n .
(3.10)

Using e x = j = 0 x j j ! , from (3.10) we deduce that

W n π 2 ( 1 1 4 n + 5 32 n 2 11 128 n 3 + 83 2 , 048 n 4 143 8 , 192 n 5 + 625 65 , 536 n 6 1 , 843 262 , 144 n 7 + 24 , 323 8 , 388 , 608 n 8 + 61 , 477 33 , 554 , 432 n 9 14 , 165 268 , 435 , 456 n 10 8 , 084 , 893 1 , 073 , 741 , 824 n 11 + ) .
(3.11)

Even though as many coefficients as we please on the right-hand side of (3.11) can be obtained by using Mathematica, here we aim at giving a formula for determining these coefficients. In fact, Theorem 1 below presents a general asymptotic expansion for W(x) which includes (3.11) as its special case.

Theorem 1 Let r0 be a given real number and 0 be a given integer. Then the function W(x), as defined in (1.9), has the following asymptotic expansion:

W(x) π 2 ( 1 + j = 1 c j ( , r ) x j ) x / r as x
(3.12)

with the coefficients c j (,r) (for jN) given by

c j (,r)= r k 1 + k 2 + + k j k 1 ! k 2 ! k j ! b 1 k 1 b 2 k 2 b j k j ,
(3.13)

where b j are given in (3.9), summed over all non-negative integers k j satisfying the equation

(1+) k 1 +(2+) k 2 ++(j+) k j =j.

Proof In view of (1.6), we can let

( 2 π W ( x ) ) r / x =1+ j = 1 m c j ( , r ) x j +O ( x m 1 ) as x,
(3.14)

where c 1 (,r),, c m (,r) are real numbers to be determined. Write (3.8) as

ln ( 2 π W ( x ) ) = k = 1 m b k x k + R m (x),

where R m (x)=O( x m 1 ). Further, we have

( 2 π W ( x ) ) r / x = e r R m ( x ) / x e k = 1 m r b k x k + = e r R m ( x ) / x k = 1 m [ 1 + ( r b k x k + ) + 1 2 ! ( r b k x k + ) 2 + ] = e r R m ( x ) / x k 1 = 0 k 2 = 0 k m = 0 1 k 1 ! k 2 ! k m ! × ( r b 1 x 1 + ) k 1 ( r b 2 x 2 + ) k 2 ( r b m x m + ) k m = e r R m ( x ) / x k 1 = 0 k 2 = 0 k m = 0 r k 1 + k 2 + + k m k 1 ! k 2 ! k m ! b 1 k 1 b 2 k 2 b m k m × 1 x ( 1 + ) k 1 + ( 2 + ) k 2 + + ( m + ) k m .
(3.15)

Equating the coefficients by the equal powers of x in (3.14) and (3.15), we see that

c j (,r)= ( 1 + ) k 1 + ( 2 + ) k 2 + + ( j + ) k j = j r k 1 + k 2 + + k j k 1 ! k 2 ! k j ! b 1 k 1 b 2 k 2 b j k j .

The proof of Theorem 1 is complete. □

Theorem 1 gives an explicit formula for determining the coefficients of the asymptotic expansion (3.12). Theorem 2 below provides a recurrence relation for determining the coefficients of the asymptotic expansion (3.12).

Theorem 2 Let r0 be a given real number and 0 be a given integer. Then the function W(x), as defined in (1.9), has the following asymptotic expansion:

W(x) π 2 ( j = 0 c j ( , r ) x j ) x / r as x
(3.16)

with the coefficients c j (,r) (for j N 0 ) given by the recurrence relation:

c 0 (,r)=1and c j (,r)= r j k = 1 j b k (k+) c j k (,r)for jN,
(3.17)

where b j (for jN) are given in (3.9).

Proof Taking the logarithm of (3.8)

ln ( 2 π W ( x ) ) j = 1 b j x j as x.

Write (3.14) as

r x ln ( 2 π W ( x ) ) ln ( j = 0 c j ( , r ) x j ) as x.

It follows that

r k = 1 b k x k ln ( j = 0 c j ( , r ) x j ) as x.

Differentiating each side with respect to x yields

r ( j = 0 c j ( , r ) x j ) ( k = 1 b k ( k + ) x k 1 ) j = 1 c j (,r)j x j 1 .

Hence,

j c j (,r)=r k = 1 j b k (k+) c j k (,r)for jN

and (3.17) follows. The proof of Theorem 2 is complete. □

4 Inequalities

In this section, we establish inequalities for the Wallis sequence W n .

Theorem 3 For all nN,

π 2 α(n)< W n < π 2 β(n),
(4.1)

where

α(n)=1 1 4 n + 5 8 + 3 64 n + 1 2 + 15 64 n + 1 2 + 35 64 n + 1 2 + 63 64 n + 1 2

and

β(n)=1 1 4 n + 5 8 + 3 64 n + 1 2 + 15 64 n + 1 2 + 35 64 n + 1 2 + 63 64 n + 1 2 + 99 64 n + 1 2 .

That is,

α(n)= 32 , 768 n 5 + 77 , 824 n 4 + 133 , 120 n 3 + 116 , 672 n 2 + 55 , 416 n + 10 , 395 32 , 768 n 5 + 86 , 016 n 4 + 149 , 504 n 3 + 143 , 424 n 2 + 73 , 976 n + 16 , 413

and

β(n)= 262 , 144 n 6 + 753 , 664 n 5 + 1 , 781 , 760 n 4 + 2 , 226 , 176 n 3 + 1 , 778 , 048 n 2 + 765 , 768 n + 135 , 135 262 , 144 n 6 + 819 , 200 n 5 + 1 , 945 , 600 n 4 + 2 , 607 , 104 n 3 + 2 , 185 , 600 n 2 + 1 , 043 , 384 n + 211 , 479 .

Proof In view of the fact that

W n = π 2 1 n + 1 2 [ Γ ( n + 1 ) Γ ( n + 1 2 ) ] 2 ,

the inequality (4.1) is equivalent to

1 2 lnα(n)<lnΓ(n+1)lnΓ ( n + 1 2 ) 1 2 ln ( n + 1 2 ) < 1 2 lnβ(n).

To obtain the left-hand inequality, define f(x) for x1 by

f(x)=lnΓ(x+1)lnΓ ( x + 1 2 ) 1 2 ln ( x + 1 2 ) 1 2 lnα(x).

Using Stirling’s formula, we find that

lim x f(x)=0.

We now show that f(x) is strictly decreasing for x3, and f(1)>f(2)>f(3), so f(n)>0 for n1. By using the second inequality in (2.2), we have

f ( x ) = ψ ( x + 1 ) ψ ( x + 1 2 ) 1 2 x + 1 1 2 α ( x ) α ( x ) < U ( x ) 1 2 x + 1 1 2 α ( x ) α ( x ) = U ( x ) 1 2 x + 1 1 2 P 8 ( x ) P 10 ( x ) ,

where P k (x) is a polynomial of degree k with non-negative integer coefficients. In what follows, P k (x) has the same understanding.

On simplification, using MAPLE, we find that

f (x)< N ( x ) P 27 ( x ) ,

where N(x) is a polynomial of degree 15 with integer coefficients (some positive, some negative). It can be shown further that

N(x)=(x3) P 14 (x)+1,752,962,197,350,057,763,671,

so

N(x)>0

for x3 and so

f (x)<0

for x3. Direct computation yields

f(1)=1.83× 10 6 ,f(2)=1.92× 10 8 ,f(3)=7.06× 10 10 .

Consequently, the sequence ( f ( n ) ) n N is strictly decreasing. This leads to

f(n)> lim n f(n)=0,nN,

which means that the first inequality in (4.1) is valid for nN.

To obtain the right-hand inequality, define g(x) for x1 by

g(x)=lnΓ(x+1)lnΓ ( x + 1 2 ) 1 2 ln ( x + 1 2 ) 1 2 lnβ(x).

Using Stirling’s formula, we find that

lim x g(x)=0.

Differentiating g(x) and applying the first inequality in (2.2), we obtain

g ( x ) = ψ ( x + 1 ) ψ ( x + 1 2 ) 1 2 x + 1 1 2 β ( x ) β ( x ) > L ( x ) 1 2 x + 1 1 2 β ( x ) β ( x ) = L ( x ) 1 2 x + 1 1 2 P 10 ( x ) P 12 ( x ) .

On simplification, using MAPLE, we find that

g (x)> 1 32 , 768 M ( x ) P 26 ( x ) ,

where M(x) is a polynomial of degree 13 with integer coefficients (some positive, some negative). It can be shown further that

M(x)=(x9) P 12 (x)+83,067,781,256,008,661,351,549,403,

so

M(x)>0

for x9 and so

g (x)>0

for x9. Direct computation yields

g ( 1 ) = 7.03 × 10 7 , g ( 2 ) = 4.217 × 10 9 , g ( 3 ) = 9.61 × 10 11 , g ( 4 ) = 4.88 × 10 12 , g ( 5 ) = 4.22 × 10 13 , g ( 6 ) = 5.3 × 10 14 , g ( 7 ) = 8.78 × 10 15 , g ( 8 ) = 1.8 × 10 15 , g ( 9 ) = 4.37 × 10 16 .

Consequently, the sequence ( g ( n ) ) n N is strictly increasing. This leads to

g(n)< lim n g(n)=0,nN,

which means that the second inequality in (4.1) is valid for nN. The proof of Theorem 3 is complete. □

We propose the following.

Conjecture 1 Let a k = ( 2 k 1 ) ( 2 k + 1 ) 64 , kN. The Wallis sequence W n has the following continued fraction representation:

W n = π 2 ( 1 1 4 n + 5 8 + a 1 n + 1 2 + a 2 n + 1 2 + a 3 n + 1 2 + ) .

Theorem 4 The following inequalities hold:

π 2 ( 1 1 4 n + 5 2 ) λ ( n ) < W n < π 2 ( 1 1 4 n + 5 2 ) μ ( n ) ,
(4.2)

where

λ(n)=1 3 64 n 2 + 3 64 n 3 23 1 , 024 n 4

and

μ(n)=1 3 64 n 2 + 3 64 n 3 23 1 , 024 n 4 1 512 n 5 .

The first inequality holds for n5, while the second inequality is valid for all n1.

Proof Inequality (4.2) can be written as

( 1 1 4 n + 5 2 ) λ ( n ) / 2 < 1 n + 1 2 Γ ( n + 1 ) Γ ( n + 1 2 ) < ( 1 1 4 n + 5 2 ) μ ( n ) / 2 .
(4.3)

The lower bound in (4.3) is obtained by considering the function F(x) defined by

F(x)=lnΓ(x+1)lnΓ ( x + 1 2 ) 1 2 ln ( x + 1 2 ) λ ( x ) 2 ln ( 1 1 4 x + 5 2 ) .

Using Stirling’s formula, we find that

lim x F(x)=0.

Differentiating F(x) and applying the second inequality in (2.2), we obtain

F ( x ) = ψ ( x + 1 ) ψ ( x + 1 2 ) V ( x ) 24 x 2 36 x + 23 512 x 5 ln ( 8 x + 3 8 x + 5 ) < U ( x ) V ( x ) 24 x 2 36 x + 23 512 x 5 ln ( 8 x + 3 8 x + 5 ) ,

with

V(x)= 8 , 192 x 6 + 10 , 240 x 5 + 2 , 944 x 4 96 x 3 + 48 x 2 + 2 x 23 128 ( 8 x + 3 ) ( 8 x + 5 ) x 4 ( 2 x + 1 ) .

We claim that F (x)<0 for x7. It suffices to show that

G(x):= 512 x 5 ( U ( x ) V ( x ) ) 24 x 2 36 x + 23 ln ( 8 x + 3 8 x + 5 ) <0for x7.

Differentiation yields

G (x)= R ( x ) 2 , 048 x 12 ( 8 x + 3 ) 2 ( 8 x + 5 ) 2 ( 2 x + 1 ) 2 ( 24 x 2 36 x + 23 ) 2 ,

where R(x) is a polynomial of degree 17 with integer coefficients (some positive, some negative). It can be shown further that

R(x)=(x7) P 16 (x)+89,314,236,262,237,854,773,083,

so

R(x)>0

for x7 and so

G (x)>0

for x7, and we have

G(x)< lim x G(x)=0,x7.

This proves the claim.

Hence, F(x) is strictly decreasing for x7. Direct computation yields

F(5)=2.399× 10 9 ,F(6)=1.494× 10 9 ,F(7)=7.947× 10 10 .

Consequently, the sequence (F(n)) is strictly decreasing for n5. This leads to

F(n)> lim n F(n)=0,n5,

which means that the first inequality in (4.2) is valid for n5.

The upper bound in (4.3) is obtained by considering the function H(x) defined by

H(x)=lnΓ(x+1)lnΓ ( x + 1 2 ) 1 2 ln ( x + 1 2 ) μ ( x ) 2 ln ( 1 1 4 x + 5 2 ) .

Using Stirling’s formula, we find that

lim x H(x)=0.

Differentiating H(x) and applying the first inequality in (2.2), we obtain

H ( x ) = ψ ( x + 1 ) ψ ( x + 1 2 ) J ( x ) 48 x 3 72 x 2 + 46 x + 5 1 , 024 x 6 ln ( 8 x + 3 8 x + 5 ) > L ( x ) J ( x ) 48 x 3 72 x 2 + 46 x + 5 1 , 024 x 6 ln ( 8 x + 3 8 x + 5 ) ,

with

J(x)= 8 , 192 x 7 + 10 , 240 x 6 + 2 , 944 x 5 96 x 4 + 48 x 3 + 2 x 2 27 x 2 128 ( 8 x + 3 ) ( 8 x + 5 ) x 5 ( 2 x + 1 ) .

We claim that the function H (x)>0 for x2. It suffices to show that

I(x):= 1 , 024 x 6 ( L ( x ) J ( x ) ) 48 x 3 72 x 2 + 46 x + 5 ln ( 8 x + 3 8 x + 5 ) >0for x2.

Differentiation yields

I (x)= S ( x ) 16 x 9 ( 8 x + 3 ) 2 ( 8 x + 5 ) 2 ( 2 x + 1 ) 2 ( 48 x 3 72 x 2 + 46 x + 5 ) 2 ,

where S(x) is a polynomial of degree 15 with integer coefficients (some positive, some negative). It can be shown further that

S(x)=(x2) P 14 (x)+5,637,072,199,918,

so

S(x)>0

for x2 and so

I (x)<0

for x2, and we have

I(x)> lim x I(x)=0,x2.

This proves the claim.

Hence, H(x) is strictly increasing for x2. Direct computation yields

H(1)=0.000462508,H(2)=0.000005843.

Consequently, the sequence (H(n)) is strictly increasing for n1. This leads to

H(n)< lim n H(n)=0,n1,

which means that the second inequality in (4.2) is valid for n1. The proof of Theorem 4 is complete. □

In fact, it is proved that

W n = π 2 ( 1 1 4 n + 5 2 ) 1 3 64 n 2 + 3 64 n 3 23 1 , 024 n 4 + O ( n 5 ) ,n.
(4.4)