Mathematics and Statistics 9(3): 209-217, 2021 http://www.hrpub.org DOI: 10.13189/ms.2021.090301 On Some Properties of Leibniz's Triangle R. Sivaraman Department of Mathematics, D. G. Vaishnav College, Chennai, India Received February 17, 2021; Revised March 11, 2021; Accepted April 22, 2021 Cite This Paper in the following Citation Styles (a): [1] R. Sivaraman , "On Some Properties of Leibniz's Triangle," Mathematics and Statistics, Vol. 9, No. 3, pp. 209 - 217, 2021. DOI: 10.13189/ms.2021.090301. (b): R. Sivaraman (2021). On Some Properties of Leibniz's Triangle. Mathematics and Statistics, 9(3), 209 - 217. DOI: 10.13189/ms.2021.090301. Copyright©2021 by authors, all rights reserved. Authors agree that this article remains permanently open access under the terms of the Creative Commons Attribution License 4.0 International License Abstract One of the Greatest mathematicians of all numbers, Inverted Infinite Hockey Stick Property, Infinite time, Gotfried Leibniz, introduced amusing triangular Triangle Sum Property array of numbers called Leibniz’s Harmonic triangle similar to that of Pascal’s triangle but with different properties. I had introduced entries of Leibniz’s triangle through Beta Integrals. In this paper, I have proved that the Beta Integral assumption is exactly same as that of entries 1. Introduction obtained through Pascal’s triangle. The Beta Integral The great German polymath Gotfried Wilhelm Leibniz, formulation leads us to establish several significant who was one of the discoverers of Calculus introduced a properties related to Leibniz’s triangle in quite elegant way. number triangle containing unit fractions which satisfy I have shown that the sum of alternating terms in any row certain recurrence relation condition. Leibniz’s triangle of Leibniz’s triangle is either zero or a Harmonic number. described by Leibniz in 1673 contains several interesting A separate section is devoted in this paper to prove properties which were explored by many later interesting results regarding centralized Leibniz’s triangle mathematicians. During the time Leibniz introduced his numbers including obtaining a closed expression, the harmonic triangle, another mathematician Pietro Mengoli asymptotic behavior of successive centralized Leibniz’s discussed the same version of the triangle for finding triangle numbers, connection between centralized quadratures(areas) of a semi-circles with different Leibniz’s triangle numbers and Catalan numbers as well as diameters. But Leibniz defined the elements of the triangle centralized binomial coefficients, convergence of series recursively and so the triangle got the name Leibniz’s whose terms are centralized Leibniz’s triangle numbers. triangle. We may also call Leibniz’s triangle as Harmonic All the results discussed in this section are new and proved Triangle as the outer diagonal of Leibniz’s triangle are for the first time. Finally, I have proved two exceedingly occupied by Harmonic numbers. In this paper, I will important theorems namely Infinite Hockey Stick theorem introduce the elements of Leibniz’s triangle in a different and Infinite Triangle Sum theorem. Though these two perspective through Beta Integrals and discuss several theorems were known in literature, the way of proving properties through this new definition. them using Beta Integral formulation is quite new and makes the proof short and elegant. Thus, by simple re-formulation of entries of Leibniz’s triangle through Beta 2. Definitions Integrals, I have proved existing as well as new theorems in much compact way. These ideas will throw a new light 2.1. The Beta Function denoted by B(,) m n where upon understanding the fabulous Leibniz’s number mn,0 is Defined by the Integral triangle. 1 Keywords Leibniz’s Triangle, Pascal’s Triangle, B( m , n ) xmn11 (1 x ) dx (2.1) . We may also call Harmonic Numbers, Centralized Leibniz’s Triangle 0 210 On Some Properties of Leibniz's Triangle B(,) m n as Beta Integral. row m, and multiply them by 1 . By considering f( x ) xmn11 (1 x ) we see that m 1 For example, the entry corresponding to m = 0 in Leibniz nm11 f(1 x ) x (1 x ) 11 triangle would be 1. By the property of the definite integrals 11 11 Similarly the entries corresponding to m = 1 in Leibniz f( x ) dx f (1 x ) dx (2.2) we see that 1 1 1 1 1 1 00 triangle are , 11 2 1 2 2 1 2 Bmn(,) xm1 (1) xdx n 1 x n 1 (1) x m 1 dxBnm (,) The entries corresponding to m = 2 in Leibniz triangle 1 1 1 1 1 1 1 1 1 00 are ,, Hence, B( m ,) n B (, n m )(2.3) . Thus the Beta 3 1 3 3 2 6 3 1 3 Integral defined in (2.1) is symmetric with respect to the The entries corresponding to m = 3 in Leibniz triangle indices m and n. are 11111 111 1111 ,,, 3. Construction of Leibniz’s Triangle 41 443 1243 1241 4 The Pascal’s triangle consists of entries which are Proceeding in this way for each value of m, we get Leibniz’s triangle as shown in Figure 2, from the Pascal’s binomial coefficients of the form m where P(,) m n triangle of Figure 1. n Sometimes, we call the Leibniz’s triangle in Figure 2 as 0 nm. The Pascal’s triangle is displayed in Figure 1. Harmonic triangle, since the outer diagonal entries are 1 1 1 1 1 Harmonic numbers 1, , , , , , Further, we note 23456 that the nth entry in row m of Leibniz’s triangle where shown in Figure 2 is given by 1 1 (m n )! n ! where PL( m , n ) (3.1) mm1m ( 1)! n PL(,) m n is the Leibniz’s triangle entry obtained from corresponding Pascal’s triangle entry. For knowing more about Leibniz and his fantastic number triangle refer [1 – 10]. Leibniz’s triangle exhibit many exciting properties Figure 1. The first 11 rows of Pascal’s Triangle similar to that of Pascal’s triangle. The upcoming sections Pascal’s Triangle possesses so many exciting properties focus on exhibiting such properties. For doing this, I will that a whole book can be devoted to listing them. Now to introduce entries of Leibniz’s triangle in a completely construct Leibniz’s triangle from Pascal’s triangle in different way. Figure 1, we need to take the reciprocal of each number in Figure 2. The first 8 rows of Leibniz’s Triangle Mathematics and Statistics 9(3): 209-217, 2021 211 4. New Formulation of Leibniz’s concerning Leibniz’s triangle. Triangle 5.1. Symmetric Property 4.1. Let mn,0 be integers. We denote L(,) m n to be The entries of Leibniz’s triangle are symmetric with the entries of Leibniz’s triangle as shown in Figure respect to vertical line through the centre i.e. LLm,, n n m 3 (5.1) Proof: By (2.3) we know that Beta Integral is symmetric with respect to its indices m and n. Hence, from (4.2), we get Lm,, n B( m 1, n 1) B ( n 1, m 1) L n m proving (5.1) 5.2. Harmonic Numbers Property The outermost diagonal entries in the Leibniz’s triangle 1 are Harmonic numbers of the form ,k 0,1,2,3,4,... k 1 Proof: The outermost diagonals in the Leibniz’s triangle (by Figure 3) are numbers of the form or where Lk,0 L0,k Figure 3. Arrangement of entries in Leibniz’s Triangle But by (5.1), we find that for all k 0,1,2,3,4,... LLkk,0 0, k. 4.2. The Entries of Leibniz’s Triangle in Figure 3 is 1 Lmn, From (4.1), we have L xmn(1 x ) dx . Hence, 1 mn, Defined by the Integral mn 0 Lmn, x(1 x ) dx (4.1) 1 0 1 L xk dx (5.2) . But we know that . k,0 In view of (2.1), we can write (4.1) as 0 k 1 Lmn, B( m 1, n 1) (4.2) Hence from (5.2), we see that the outmost diagonal entries Using the relationship between Beta and Gamma of the Leibniz’s triangle are Harmonic numbers of the form Integrals and noting that (kk 1) ! for all whole numbers k, we have 1 ()p () q ( p 1)! ( q 1)! B( p , q ) xpq11 (1 x ) dx (4.3) 5.3. Recurrence Relation (p q ) ( p q 1)! 0 The entries of the Leibniz’s triangle satisfy the recurrence relation given by 4.3. Lemma 1 LLLm1, n 1 m 1, n m , n (5.3) If be integers are such that 0 nmthen Proof: Using (4.1), we have 11 PL( m , n ) Lm n, n (4.4) L L xm11(1 x ) n dx x m (1 x ) n dx m1, n 1 m , n Proof: From (3.1), (4.1), (4.2) and (4.3), we find that 00 1 11 m n n m11 n m n Lm n, n x(1 x ) dx x(1 x ) (1 xxdxx ) (1 xdxL ) mn1, 0 00 B( m n 1, n 1) as required. This provides (5.3) and completes the proof. (m n )! n ! Many sources about Leibniz’s triangle use the PL(,) m n (m 1)! recurrence relation property in (5.3) to generate subsequent Lemma 1, thus establishes the fact that the entries of entries of Leibniz’s triangle. Leibniz’s triangle obtained through Pascal’s triangle are exactly the same as entries of the reformulation as defined 5.4. Infinite Sum Property in (4.1). The sum of reciprocals of all triangular numbers is 2, i.e.