The American Mathematical Monthly

A Generalization of Wallis Product --Manuscript Draft--

Manuscript Number: Full Title: A Generalization of Wallis Product Article Type: Note Keywords: Wallis Product Corresponding Author: Mahdi Ahmadinia, Ph.D. University of Qom Qom, Qom IRAN (ISLAMIC REPUBLIC OF) Corresponding Author Secondary Information: Corresponding Author's Institution: University of Qom Corresponding Author's Secondary Institution: First Author: Mahdi Ahmadinia, Ph.D. First Author Secondary Information: Order of Authors: Mahdi Ahmadinia, Ph.D. Hamid Naderi Yeganeh, Undergraduate Student Order of Authors Secondary Information: Abstract: This note presents a generalization of Wallis' product. We prove this generalization by Stirling's formula. Then some corollaries can be obtained by this formula. Corresponding Author E-Mail: [email protected] Additional Information: Question Response 5-Character 2000 AMS/Math Reviews 00-xx Math. Classification Codes for your manuscript. The complete classification is available at http://www.ams.org/msc/.

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1 2 A Generalization of Wallis Product 3 4 Mahdi Ahmadinia and Hamid Naderi Yeganeh 5 6 7 Abstract. This note presents a generalization of Wallis’ product. We prove this generalization 8 by Stirling’s formula. Then some corollaries can be obtained by this formula. 9 10 Wallis’ product (see [1]) is a famous formula, which has been named in honor of 11 English mathematician John Wallis (1616-1703). There are some generalization of this 12 formula, for example see [2]. This note generalizes Wallis product by a simple proof. 13 The generalized formula is as follows: 14 15 ∏∞ mk(mk + m) −(2m+1) 2 16 ∏ = 2πm m−1 (m)! m−1 , m ≥ 2. m−1 2 m−1 17 k=1 l=1 (mk + l) 18 19 Lemma 1. Let m ≥ 2 be an integer number, then 20 2nm m+1 2m ∏n − − − 21 mk(mk + m) 2 m m 1 (n + 1) m 1 n! m 1 ∏ = m! m−1 . , ∀n ∈ N. 22 m−1 2 2 (1) m−1 m−1 23 k=1 l=1 (mk + l) (m(n + 1))! 24 25 Proof. The following identity can be proved by induction on n. 26 ∏n 27 1 (n + 1)m! ∏ − = , 28 m 1(mk + l) (m(n + 1))! 29 k=1 l=0 30 Therefore, 31 32 ∏n mk(mk + m) ∏n 1 33 ∏ = m2nn!2(n + 1) ∏ m−1 2 m−1 2 m−1 m−1 34 k=1 l=1 (mk + l) k=1 l=1 (mk + l) 35 ∏n 1 1 1 36 = m2n(1+ m−1 )n!2(1+ m−1 )(n + 1) ∏ m−1 2 37 m−1 k=1 (mk + l) 38 l=0 2nm m+1 2m − − − 39 2 m m 1 (n + 1) m 1 n! m 1 m−1 40 = m! . 2 . − 41 (m(n + 1))! m 1 42 43 44 Theorem 1. The following formula is true for m = 2, 3, 4,... 45 46 ∏∞ mk(mk + m) −(2m+1) 2 ∏ = 2πm m−1 (m)! m−1 . 47 m−1 2 (2) m−1 48 k=1 l=1 (mk + l) 49 50 Proof. The Stirling’s formula yields 51 √ 1 − 52 2πnn+ 2 e n lim = 1, (3) 53 n→∞ n! 54 55 January 2014] A GENERALIZATION OF WALLIS’ PRODUCT 1 56 57 58 59 60 61 62 63 64 65 Mathematical Assoc. of America American Mathematical Monthly 121:1 October 26, 2014 6:06 p.m. A-Generalization-of-Wallis-Product.tex page 2

and 1 √ 1 − 2 2π(m(n + 1))m(n+1)+ 2 e m(n+1) lim = 1. (4) 3 n→∞ (m(n + 1))! 4 5 Then (3) and (4) imply that 6 (√ ) 2m 7 ( ) 2nm m+1 1 − m−1 2nm m+1 2m m−1 m−1 n+ n 8 m m−1 (n + 1) m−1 n! m−1 m (n + 1) 2πn 2 e lim = lim ( ) 9 n→∞ 2 n→∞ √ 2 (m(n + 1))! m−1 1 − m−1 10 2π(m(n + 1))m(n+1)+ 2 e m(n+1) 11   ( ) −m(2n+1) 12 − 2m −(2m+1) n + 1 m 1 13 = lim 2πe m−1 m m−1  . 14 n→∞ n 15 16 The following relation is obvious, 17 ( ) −m(2n+1) 18 − n + 1 m 1 −2m 19 lim = e m−1 . 20 n→∞ n 21 22 Hence, 23 ( ) 2nm m+1 2m 24 m m−1 (n + 1) m−1 n! m−1 −(2m+1) lim = 2πm m−1 . (5) 25 →∞ 2 n (m(n + 1))! m−1 26 27 Proof of the theorem will be completed by Lemma and (5). 28 29 Note that (2) is Wallis product for m = 2, 30 ∞ 31 ∏ 2k(2k + 2) π 32 = . (6) (2k + 1)2 4 33 k=1 34 35 It shows that (2) is a generalization of Wallis formula. Also as a result of (2), we can 36 prove 37 ∏∞ 38 4k(4k + 4) 3π = √ . (7) 39 (4k + 1)(4k + 3) 8 2 40 k=1 41 The theorem implies the following equality 42 43 ∏∞ 4k(4k + 4) −9 2 44 3 3 2 = 2π.4 .4! , (8) 3 45 k=1 ((4k + 1)(4k + 2)(4k + 3)) 46 47 when m = 4. Wallis product (6) immediately yields 48 ( ) ∏∞ 2 1 49 (4k + 2) 3 4 3 50 1 = . (9) 3 π 51 k=1 (4k(4k + 4)) 52 53 Finally, equality (7) will be obtained by (8) and (9). 54 55 2 ⃝c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 56 57 58 59 60 61 62 63 64 65 Mathematical Assoc. of America American Mathematical Monthly 121:1 October 26, 2014 6:06 p.m. A-Generalization-of-Wallis-Product.tex page 3

REFERENCES 1 1. M. Abramowitz, and I. A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and 2 Mathematical Tables, 9th printing, Dover, New York, 1972. 3 2. T. J. Osler, The general Vieta-Wallis product for π, The Mathematical Gazette, 89 No. 516. (2005) 371– 4 377. 5 6 7 MAHDI AHMADINIA received his PhD in mathematics from Kerman University under the guidance of professor Mehdi Radjabalipour and was supported by Mahani Mathematical Research Center. 8 Department of Mathematics, University of Qom, Qom, Iran. P. O. Box. 37185-3766. 9 [email protected] & [email protected] 10 11 12 HAMID NADERI YEGANEH is a Bachelor student of mathematics in University of Qom. He won gold 13 medal at the 38th Iranian Mathematical Society’s Competition (2014). [email protected] 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 January 2014] A GENERALIZATION OF WALLIS’ PRODUCT 3 56 57 58 59 60 61 62 63 64 65 Cover Letter

Mahdi Ahmadinia Department of Mathematics, University of Qom, Qom, Iran. P. O. Box. 37185-3766.

Scott T. Chapman Professor and Scholar in Residence Sam Houston State University Department of Mathematics and Statistics Box 2206 Huntsville, Texas 77341-2206.

Oct 26, 2014

Dear Professor Scott T. Chapman

I am pleased to submit an original research article entitled: ”A Generalization of Wallis Product” by M. Ahmadinia and H. Naderi Yeganeh for consideration for possible publication in the American Mathematical Monthly.

This manuscript has not been published and is not under consideration for publication else- where.

Best Regards,

Mahdi Ahmadinia Assistant Professor of Department of Mathematics University of Qom Qom, Iran.