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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 , 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 . In this paper, I have proved that the Beta 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 introduced a properties related to Leibniz’s triangle in quite elegant way. number triangle containing unit which satisfy I have shown that the sum of alternating terms in any row certain 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 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 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 ) xmn11 (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 ) xmn11 (1 x ) we see that m 1 For example, the entry corresponding to m = 0 in Leibniz nm11 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(,) xm1 (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) mm1m ( 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

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4. New Formulation of Leibniz’s concerning Leibniz’s triangle. Triangle 5.1. Symmetric Property 4.1. Let mn,0 be . 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 ) xpq11 (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 LLLm1, 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  xm11(1  x ) n dx  x m (1  x ) n dx m1, n 1 m , n  Proof: From (3.1), (4.1), (4.2) and (4.3), we find that 00 1 11 m n n m11 n m n Lm n, n  x(1 x ) dx x(1  x ) (1  xxdxx )   (1  xdxL )     mn1, 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.  2   2 (5.4) 5. Properties of Leibniz’s Triangle m1 mm( 1) Proof: In Figure 3 of Leibniz’s triangle we consider the In view of (4.2), having identified the entries of Leibniz’s triangle as the Beta integral, in this section I second leading diagonal of the form Lm,1 where prove the following interesting and basic properties m  0,1,2,3,4,...

212 On Some Properties of Leibniz's Triangle

Now from (4.1), we have 6. Centralized Leibniz’s Triangle 1 m 1 1 1 Numbers Lm,1  x(1  x ) dx    (5.5)  m1 m  2 ( m  1)( m  2) We define numbers of the form as centralized 0 Lmm, Triangular numbers are numbers of the form Leibniz’s triangle numbers. Since m + m = 2m is always mm( 1) even, we find from Figure 3 that L occurs as the middle ,m  1,2,3,4,... mm, 2 term of the row beginning with L2m ,0 and ending with Hence the sum of reciprocals of all triangular numbers = L for m  0,1,2,3,4,... 21 0,2m  2 mm10m( m 1) ( m  1)( m  2) The Centralized Leibniz’s Triangle Numbers are given by Now from (5.5), we get Sum of reciprocals of all triangular numbers = m! 1 Lmm,  m (6.1) 2L 2 xm (1 x ) dx 2 (2m 1)!! m,1  mm000 where (2m 1)!!  (2 m  1)(2 m  1)(2 m  3) 5 3 1 11 m 1 is called the of 2m + 1. = 2 x (1 x ) dx  2 (1  x ) dx  2 . m0 1 x 00 Proof: From (4.2) and (4.3), we get This completes the proof. mm!! L B( m  1, m  1)  mm, (2m  1)! 5.5. Sum of Alternating Terms mm!!  The sum of alternating terms in mth row of the Leibniz’s (2m 1)(2)(2 m  m  1) 4321 triangle is zero, if m is odd and a Harmonic number if m is m!  even (5.6) m 2 (2m 1)!! Proof: From Figure 3, we first notice that the sum of This completes the proof. alternating terms in mth row of the Leibniz’s triangle = m If is the centralized Leibniz’s triangle number n ( 1)Lm n, n (5.7)  then LLm, m 4 m 1, m 1 (6.2) as m  n0 Proof: From (6.1) we have We will now try to compute the sum given in (5.7). m! Using (4.1), we get m Lmm, 2 (2m 1)!! 23m  as m . n  24   m m11 m x 1 (m  1)! n n m n n m  Lmmm1, 1 1 ( 1)Lm n, n  (  1) x (1  x ) dx  x dx    x m1 n0 n  000 n  0  2 (2m 3)!! 1 xxmm11( 1) 1 This proves (6.2) and completes the proof. xmm dx 1  (  1)  m  1 2m xm 2 If C  is the mth Catalan number and Lmm, is 0  m  m 1m m n the centralized Leibniz’s triangle number then If m is odd, then ( 1)Lm n, n 0 (5.8) n0 1 m LC (6.3) If m is even, then n 2 . In m, m m ( 1)Lm n, n (5.9) (mm 1)(2 1) n0 m  2 particular if m = 2k, where k = 0,1,2,3,4,… then from (5.9) Proof: m 1 we get ( 1)n Lk  ,  0,1,2,3,4,... (5.10) 1 2m  m n, n Lm, m C m  B( m  1, m  1)   n0 k 1 m 1m Note that 1 for k 0,1,2,3,4,... generates m! m ! 1 (2 m )!    k 1 (2m 1)! m  1 m !  m ! Harmonic numbers. Another occurrence of Harmonic 1 numbers in Leibniz’s triangle.  Hence, equations (5.8) and (5.10) prove (5.6). This (mm 1)(2 1) completes the proof.

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This proves (6.3) and completes the proof. 7.1. Inverted Infinite Hockey Stick Property The of centralized Leibniz’s triangle and corresponding centralized binomial coefficients in Pascal’s We first look at the two figures 4 and 5 displayed below. triangle is the reciprocal of the odd positive integers. In Figures 4 and 5 we see that if we add the numbers in the infinite rectangular strip the sum would be a number Proof: Let L be the centralized Leibniz’s triangle mm, indicated in the circle located at North-West direction. numbers. The centralized binomial coefficients in Pascal’s Thus from Figures 4 and 5 we see that triangle are numbers of the form 2m . We P(2 m , m )   1 1 1 1 1 1 1 m       1 (7.1) 2 6 12 20 30 42 56 notice from the definition of Catalan numbers in 6.4 that . Hence the product of centralized and (m 1) Cm P (2 m , m ) 1 1 1 1 1 1 Leibniz’s triangle numbers and corresponding centralized       (7.2) binomial coefficients, using (6.3) is given by 12 30 60 105 168 6 1 . Thus the Lm,, m P(2 m , m )  L m m  ( m  1) C m  Using (5.4), we can easily prove (7.1). 21m  For example, we notice that required product is the reciprocal of odd positive integers 1 1 1 1 1 1 1 1 for each value of m = 0, 1, 2, 3, 4, . . .This completes the  1 proof. 2 6 12 20 30 42 56m1 mm ( 1)  2 To prove (7.2), we use the recurrence relation for entries  Lmm,  (6.4) of Leibniz’s triangle in equation (5.3) repeatedly. For m0 33 example from Figure 5, we find that 1 1 1 1 1 1 1 1 1  ,,     , Proof: From (4.1), we have 12 6 12 30 12 20 60 20 30 1 1 1 1 1 1  11   ,,    Continuing this way L xm(1  x ) m dx  [ x (1  x )] m dx mm,   105 30 42 168 42 56 m0 m  000 m  0 repeatedly we find that 1112dx  dx   1 1 1 1 1 1xx (1 ) 2 2      001333 12 30 60 105 168 x  22 1 1   1 1   1 1              6 12   12 20   20 30  Thus the sum of centralized Leibniz’s triangle numbers 1 1   1 1  1       converges to 2 . This proves (6.4) and completes the     30 42   42 56  6 33 proof. We thus see upon applying the recurrence relation property repeatedly, only the first term in the right hand side remains while all other terms cancel out mutually. This 7. Summations in Leibniz's Triangle kind of summation is often referred as Telescopic Summation. Using the same principle we can select any In Pascal’s triangle there is a wonderful property infinite rectangular strip of numbers in the Leibniz’s concerning summing entries in a particular diagonal called triangle whose sum would be a number just located above Hockey Stick Property. There is a similar version in at the north-west corner of the strip. I refer this property as Leibniz’s triangle producing gallery of infinite series Inverted Infinite Hockey Stick Property. I now provide the summations which I will prove below. general proof of this property.

214 On Some Properties of Leibniz's Triangle

Figure 4. Illustration of Inverted Infinite Hockey Stick Property

Figure 5. Illustration of Inverted Infinite Hockey Stick Property 7.2. Inverted Infinite Hockey Stick Theorem 8. Infinite Triangle Sum Property If m, n, r are whole numbers then In Pascal’s Triangle, we notice a Rhombus Shape as  displayed in Figure 6.  LLm, n r , n 1 (7.3) As shown in Figure 6, if we add all the numbers inside mr the rhombus their sum would be 64 = 84 – 20, which is the Proof: Using (4.1), we have difference of the numbers located just vertically below the rhombus shown in circle and to the right of the number 10.  11  L xm(1  x ) n dx  x m (1  x ) n dx This property of rhombus shape sum holds true for any mn,   rhombus taken in Pascal’s triangle. Is there a similar m r m  r00 m  r property that can be observed in Leibniz’s triangle? Figure  1 7 answers this question. L xrn(1  x ) 1 dx  L . This proves (7.3)  m, n r , n 1 From Figure 7, we notice that if we add all the numbers mr 0 1 and completes the proof. inside the triangle, we get their sum as a number As an illustration, we find that for rn1, 2 from (7.3), 4 . we get located just above the vertex of the infinite triangle. Like Pascal’s triangle, this infinite triangle sum property also  1 1 1 1 1 1 holds true for any triangle considered in Leibniz’s triangle.  LLm,2      1,1  I will now formally prove this fact. m1 12 30 60 105 168 6

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Figure 6. Rhombus Shape Sum Property in Pascal’s Triangle

Figure 7. Infinite triangle sum property in Leibniz’s triangle This proves (8.1) and completes the proof. Infinite Triangle Sum Theorem I provide two illustrations explaining (8.1). First, considering Figure 7, we find that according to Figure 3, If m,,, n r k are whole numbers then we have k = 1, r = 4. Note that the topmost entry of the  (8.1.) infinite triangle of Figure 7 is 1 . Also note that the LLm, n k 1, r 1 L1,4  m k n r 30 number in the circle above the vertex is 1 . L0,3  Proof: First, we note that Lkr, is the topmost number in 4 our infinite triangle. Also, we note from Figure 3 that Hence the sum of all the numbers in the infinite triangle  1 Lkr1, 1 is the number located just vertically above the according to (8.1) would be LLas  mn, 0,3 4 vertex of the infinite triangle. We now use Inverted Infinite mn14 expected. Hockey Stick Identity obtained in (7.3) twice to prove this theorem. First using (5.1), we get Similarly, the sum of all the numbers in the infinite triangle shown in Figure 8 below can be obtained by taking     k = 1, r = 1. LLm,, n   n m  m k n  r m  k n  r Using (8.1), we have , a number LLmn, 0,0 1  mn11 LLLr, m 1 m  1, r k  1, r  1 located above the vertex of the infinite triangle shown in m k m k the circle.

216 On Some Properties of Leibniz's Triangle

Figure 8. Infinite triangle sum property in Leibniz’s triangle 9. Conclusions as well as unknown properties of the not so well known Leibniz’s triangle. Most importantly by observing that the Usually, the entries of Leibniz’s triangle would either be entries are Beta Integral values, I could prove almost every identified through or obtained from Pascal’s result in this paper quite beautifully and in short. This triangle as explained in section 3. But in this paper, I proves that by thinking little differently we can see deeper identified the entries of Leibniz’s triangle as Beta Integral truths in a more elegant way. values which is not done before in any paper. I proved that the Beta Integral assumption is exactly same as that of entries obtained through Pascal’s triangle in section 4 through equation (4.4). This equivalence made me establish so many identities in subsequent sections, that too REFERENCES very elegantly. [1] Ivan D. Stones, “The harmonic triangle: Opportunities for In section 5, I have proved five results in which the first pattern identification and generalization”, The Mathematics four are commonly known, but the proofs are given with Teacher, vol. 76, no. 5, pp. 350 – 354, May 1983, https://www.jstor.org/stable/27963528 respect to integrals, which were not done before. The fifth result of section 5, namely sum of alternating terms of [2] Peter Hilton and Jean Pedersen,“Looking into Pascal's entries of Leibniz’s triangle is not so well known, again triangle: , arithmetic, and geometry”, Mathematics Magazine, vol. 60, no. 5, pp. 305 – 316, proved easily by using integrals. The six results proved in December 1987, https://doi.org/10.1080/0025570X.1987.11 section 6 are all relatively new and not much known in the 977330 literature about them. Hence, they are completely new results of this paper. In particular the proof of sum of all [3] Thomas Koshy,“Leibniz’s Barmonic Triangle”, Triangular Arrays with Applications, Oxford University Press, New centralized Leibniz’s triangle numbers is convergent York, 2011. provides a new dimension to existing properties. In section 7, I had proved Infinite Hockey Stick Theorem, [4] MarjorieBickNell-Johnson,“Diagonal sums in the harmonic triangle”, Fibonacci Quarterly, vol. 19, no. 3, pp. 196 – 199, which is done so easily. As far as I knew there is no such August 1981, https://www.fq.math.ca/Scanned/19-3/bicknel easy way of proving as I did for that theorem. Further the l.pdf Infinite Hockey Stick Theorem provides us a way of [5] ANTOGNAZZA, Maria Rosa, “Leibniz: An Intellectual constructing gallery of several infinite series with unit Biography, Cambridge”, Cambridge University Press, 2009. fractions whose sum can be readily obtained by the equation (7.3). Similarly, in section 8, I had proved Infinite [6] GeorgePolya, “Mathematical Discovery – On Understanding, triangle sum property whose proof like previous one was Learning and Teaching Problem Solving”, Wiley, 1981. done in a single line. Illustrations and Figures were [7] Belaval, Yvon and Look, Brandon C.. "Gottfried Wilhelm provided for better understanding wherever necessary. Leibniz", Encyclopedia Britannica, https://www.britannica. Figures 4,5,6,7,8 were incorporated from [11]. com/biography/Gottfried-Wilhelm-Leibniz. This paper, thus contain almost all the known properties [8] Jerphagnon, Lucien and Orcibal, Jean. "Blaise Pascal",

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Encyclopedia Britannica, https://www.britannica.com/biogr [10] Alan Tucker, “Applied Combinatorics”, Wiley (6th Edition, aphy/Blaise-Pascal. USA, 2012 [9] S.C. Althoen and C.B. Lacampagne,“Tetrahedral numbers as [11] James, Lacey Taylor, "Analogues between Leibniz's sums of square numbers”, Mathematics Magazine, pp. 104 – Harmonic Triangle and Pascal's Arithmetic Triangle" (2019). 108, 1991,https://doi.org/10.1080/0025570X.1991.1197758 Electronic Theses, Projects, and Dissertations, 835, 6 https://scholarworks.lib.csusb.edu/etd/835