Abstract
We present the asymptotic expansions of functions involving the ratio of gamma functions and provide formulas for determining the coefficients of the asymptotic expansions. As consequences, we obtain the asymptotic expansions of the Wallis sequence. Also, we establish inequalities for the Wallis sequence.
MSC:40A05, 33B15, 41A60, 26D15.
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1 Introduction
The Wallis sequence to which the title refers is
Wallis (1616-1703) discovered that
(see [[1], p.68]). Based on Wallis’ infinite product (1.2), the first infinite continued fraction of π was given by Brouncker (1620-1684):
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
Several elementary proofs of (1.2) can be found (see, for example, [2–4]). 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, 8–10].
Some inequalities and asymptotic formulas associated with the Wallis sequence can be found (see, for example, [11–23]). In [13], Hirschhorn proved that for ,
Also in [13], Hirschhorn pointed out that if the are given by
then, as ,
Remark 1 It is well known (see [[24], p.85]) that
where () are the Bernoulli numbers. We then obtain
Thus we have
Let be a given real number and be a given integer. The first aim of this paper is to determine the coefficients (for ) such that
where the function is defined by
Clearly, . The second aim of this paper is to establish inequalities for the Wallis sequence .
2 A useful lemma
The classical Euler’s gamma function is defined for by
The logarithmic derivative of , denoted by , is called psi (or digamma) function, and () are called polygamma functions.
The following lemma is required in our present investigation.
Lemma 1 ([[25], Corollary 2.1])
Let . Then for ,
where are the Bernoulli numbers.
It follows from (2.1) that, for ,
where
and
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]):
as , where denotes the Bernoulli polynomials defined by the following generating function:
Note that the Bernoulli numbers (for ) are defined by (3.2) for .
From (3.1), we obtain, as ,
Setting and noting that
(see [[24], p.805]), we obtain from (3.3), as ,
or
We see from (1.6) and (1.9) that
with the coefficients given by (1.5). From (3.5) and (3.6), we retrieve (1.8).
By using the Maclaurin expansion of ,
we obtain
Applying (3.4) and (3.7) yields
with the coefficients (for ) given by
From (3.8), we obtain the following asymptotic expansion for the Wallis sequence :
Using , from (3.10) we deduce that
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 which includes (3.11) as its special case.
Theorem 1 Let be a given real number and be a given integer. Then the function , as defined in (1.9), has the following asymptotic expansion:
with the coefficients (for ) given by
where are given in (3.9), summed over all non-negative integers satisfying the equation
Proof In view of (1.6), we can let
where are real numbers to be determined. Write (3.8) as
where . Further, we have
Equating the coefficients by the equal powers of x in (3.14) and (3.15), we see that
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 be a given real number and be a given integer. Then the function , as defined in (1.9), has the following asymptotic expansion:
with the coefficients (for ) given by the recurrence relation:
where (for ) are given in (3.9).
Proof Taking the logarithm of (3.8)
Write (3.14) as
It follows that
Differentiating each side with respect to x yields
Hence,
and (3.17) follows. The proof of Theorem 2 is complete. □
4 Inequalities
In this section, we establish inequalities for the Wallis sequence .
Theorem 3 For all ,
where
and
That is,
and
Proof In view of the fact that
the inequality (4.1) is equivalent to
To obtain the left-hand inequality, define for by
Using Stirling’s formula, we find that
We now show that is strictly decreasing for , and , so for . By using the second inequality in (2.2), we have
where is a polynomial of degree k with non-negative integer coefficients. In what follows, has the same understanding.
On simplification, using MAPLE, we find that
where is a polynomial of degree 15 with integer coefficients (some positive, some negative). It can be shown further that
so
for and so
for . Direct computation yields
Consequently, the sequence is strictly decreasing. This leads to
which means that the first inequality in (4.1) is valid for .
To obtain the right-hand inequality, define for by
Using Stirling’s formula, we find that
Differentiating and applying the first inequality in (2.2), we obtain
On simplification, using MAPLE, we find that
where is a polynomial of degree 13 with integer coefficients (some positive, some negative). It can be shown further that
so
for and so
for . Direct computation yields
Consequently, the sequence is strictly increasing. This leads to
which means that the second inequality in (4.1) is valid for . The proof of Theorem 3 is complete. □
We propose the following.
Conjecture 1 Let , . The Wallis sequence has the following continued fraction representation:
Theorem 4 The following inequalities hold:
where
and
The first inequality holds for , while the second inequality is valid for all .
Proof Inequality (4.2) can be written as
The lower bound in (4.3) is obtained by considering the function defined by
Using Stirling’s formula, we find that
Differentiating and applying the second inequality in (2.2), we obtain
with
We claim that for . It suffices to show that
Differentiation yields
where is a polynomial of degree 17 with integer coefficients (some positive, some negative). It can be shown further that
so
for and so
for , and we have
This proves the claim.
Hence, is strictly decreasing for . Direct computation yields
Consequently, the sequence is strictly decreasing for . This leads to
which means that the first inequality in (4.2) is valid for .
The upper bound in (4.3) is obtained by considering the function defined by
Using Stirling’s formula, we find that
Differentiating and applying the first inequality in (2.2), we obtain
with
We claim that the function for . It suffices to show that
Differentiation yields
where is a polynomial of degree 15 with integer coefficients (some positive, some negative). It can be shown further that
so
for and so
for , and we have
This proves the claim.
Hence, is strictly increasing for . Direct computation yields
Consequently, the sequence is strictly increasing for . This leads to
which means that the second inequality in (4.2) is valid for . The proof of Theorem 4 is complete. □
In fact, it is proved that
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Lin, L., Deng, JE. & Chen, CP. Inequalities and asymptotic expansions associated with the Wallis sequence. J Inequal Appl 2014, 251 (2014). https://doi.org/10.1186/1029-242X-2014-251
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DOI: https://doi.org/10.1186/1029-242X-2014-251