• 検索結果がありません。

Proof of the best bounds in Wallis’ inequality

N/A
N/A
Protected

Academic year: 2022

シェア "Proof of the best bounds in Wallis’ inequality"

Copied!
4
0
0

読み込み中.... (全文を見る)

全文

(1)

General Mathematics Vol. 13, No. 2 (2005), 117–120

Proof of the best bounds in Wallis’ inequality

Chao-Ping Chen

Dedicated to Professor Dumitru Acu on his 60th anniversary Abstract

Let n≥1 be an integer, then p 1

π(n+ 4π−11) 1·3·5· · ·(2n1)

2·4·6· · ·(2n) < 1 pπ(n+ 1/4).

The constants 4π−11 and 1/4 are the best possible.

2000 Mathematical Subject Classification: Primary 26D20;

Secondary 33B15.

Key words and phrases: Wallis’ inequality, best bounds, gamma function.

The sine has the infinite product representation

(1) sinx=x

Y

n=1

1 x2 π2n2

. 117

(2)

118 Chao-Ping Chen Taking in (1) x=π/2 gives well known the Wallis formula

(2) π

2 = Y

n=1

(2n)2 (2n1)(2n+ 1)

.

Motivated by (2), Kazarinoff [2] proved that

(3) 1

q

π n+ 12 < 1·3·5· · ·(2n1)

2·4·6· · ·(2n) < 1 q

π n+14

for n∈N, the set of positive integers. We here show that, for n∈N,

(4) 1

pπ(n+ 4π−11) 1·3·5· · ·(2n1)

2·4·6· · ·(2n) < 1

pπ(n+ 1/4),

improving the lower bound and confirming the upper in (3), by a very simple argument. We also prove that the bounds in (4) are the best possible.

Proof. It is clear that Γ(n+ 1) =n!, Γ

n+1

2

= (2n1)!!

2n

√π, 2nn! = (2n)!!.

To prove the right hand inequality of (4), it suffices to show that

(5) Rn = Γ n+12q

n+14 Γ(n+ 1) <1.

Using the recurrence relation for the gamma function Γ(x+ 1) =xΓ(x) we conclude that

Rn Rn+1 =

s n+14 n+54

n+ 1

n+ 12 <1 for n 1.

Hence, the sequence {Rn}n=1 is strictly increasing with n∈N.

(3)

Proof of the best bounds in Wallis’ inequality 119 From the asymptotic expansion [1, p. 257]

(6) xb−aΓ(x+a)

Γ(x+b) = 1 + (a−b)(a+b−1)

2x +O x−2

, we conclude that lim

n→∞Rn = 1, thus inequality (5) holds for all n∈N.

The left hand side of inequality (4) is equivalent to

(7) Ln= Γ n+ 12q

n+ 4π 1 Γ(n+ 1) 1.

It is easy to see that Ln Ln+1 =

s

n+ 4π 1 n+π4

n+ 1

n+12 >1 for n≥2.

Hence, the sequence {Ln}n=1 is strictly decreasing for n 2. By (6), we conclude that lim

n→∞Ln = 1, thus inequality (7) holds strictly for all n 2.

Clearly, the sign of equality in (7) holds for n = 1. The proof is complete.

References

[1] M. Abramowitz, I. A. Stegun (Eds), Handbook of Mathematical Func- tions with Formulas, Graphs, and Mathematical Tables, National Bu- reau of Standards, Applied Mathematics Series 55, 4th printing, with corrections, Washington, 1965.

(4)

120 Chao-Ping Chen [2] D. K. Kazarinoff, On Wallis’ formula, Edinburgh. Math. Soc. Notes,

No. 40 (1956), 19–21.

Department of Applied Mathematics and Informatics, Research Institute of Applied Mathematics,

Henan Polytechnic University, Jiaozuo City, Henan 454010, China

E-mail:[email protected], [email protected]

参照

関連したドキュメント