A new asymptotic expansion and sharp inequality for the volume of the unit ball in R n

For n ∈ { 1 , 2 , . . . } , let Ω n = π n/ 2 / Γ( n 2 + 1) be the volume of the unit ball in R n . In this paper, we give a new asymptotic expansion for Ω n . Based on the obtained result, we also establish a sharp double inequality for Ω n .


Introduction
A considerable number of properties of the volume Ω n of the unit ball in R n have been reported by several researchers in the recent past, where Ω n = π n/2 Γ( n 2 + 1) , n ∈ N := {1, 2, . . .}.
From the right-hand side of (1), it follows that Chen and Paris [10,Equation (3.18)] developed (2) to produce a complete asymptotic expansion: where the coefficients b j are given by and a j are given by Here, B n (n ∈ N 0 := N ∪ {0}) are the Bernoulli numbers defined via the generating function: (n + 633922198537106197381 1270643701163933024775 ) 6 + . . . , as n → ∞.
The first aim of the present paper is to determine the constants λ and µ such that In view of (6), it is natural to ask: what is the smallest value of α and what is the largest value of β such that the inequality is valid for every n ∈ N? Answering this question is the second aim of the present paper.
The gamma function Γ(x) is one of the most important functions in mathematical analysis and has applications in many diverse areas. The logarithmic derivative of Γ(x), denoted by ψ(x) = Γ (x)/Γ(x), is called the psi (or digamma) function. It is known that Γ(x + 1) = xΓ(x) and ψ(x + 1) = ψ(x) + 1 x .
The following inequalities are needed in the present study: and where x > 0. The inequalities (7) and (8) can be found in [10]. We remark that the inequalities (7) and (8) follow from Theorem 8 of [1]. We end this introductory section with the remark that the numerical values reported in this paper are calculated by using MAPLE 11 (a software).

Main results
where the constants λ and µ are given by a pair of recurrence relations as follows: and with λ 1 = − 1 36 and µ 1 = 23 45 .
Here b j are given in (4).
Proof. In view of (6), we can assume that as n → ∞, where λ and µ are real numbers to be determined. This can be written as follows: We then obtain as n → ∞. On the other hand, it follows from (3) that as n → ∞, where b j are given in (4). By equating coefficients of the term n −j on the right-hand sides of (12) and (13), we obtain By setting j = 2 and j = 2 + 1 in (14), respectively, we find and From (15) and (16) we obtain for = 1, Also, for ≥ 2, we have Consequently, we arrive at the recurrence relations (10) and (11).
Here, we give explicit numerical values of some first terms of λ and µ by using the formulas (10) and (11). This shows how easily we can determine the constants λ and µ in Theorem 2.1. We see from (3) We obtain from (10) and (11) that We note that the values of λ and µ (for = 1, 2, 3), given above, are equal to the constants appearing in (6). From a computational viewpoint, the formula (6) improves the formula (3). is valid for n = 1, 2 and 3. For n = 1, the equality sign on the right-hand side of (17) holds. We now prove that the double inequality (17) with α = 23 45 and is valid for n ≥ 4. It suffices to show that for x ≥ 2, where .
The double inequality (18) can be written as In order to prove the double inequality (19) for x ≥ 2, it suffices to show that By (7), we obtain lim x→∞ f (x) = lim x→∞ g(x) = 0.
Hence, f (x) < 0 for x ≥ 2. So, f (x) is strictly decreasing for x ≥ 2 and we have Therefore, the left-hand side of (17) is valid for n ∈ N. Now, by differentiating g(x) and applying the left-hand side of (8), we obtain for x ≥ 2, 144x 2 + 144 · 7 10 · x + 36 · 7 10 2 − 1 (2x + 7 10 ) Hence, g(x) is strictly increasing for x ≥ 2, and we have Therefore, the right-hand side of (17) holds for n ∈ N. If we write the double inequality (17) as we find that