DePauw University Scholarly and Creative Work from DePauw University Student research—other Student Work Summer 2015 A Short Note On Sums of Powers of Reciprocals of Polygonal Numbers Jihang Wang DePauw University Suman Balasubramanian DePauw University, [email protected] Follow this and additional works at: http://scholarship.depauw.edu/studentresearchother Part of the Number Theory Commons Recommended Citation Wang, Jihang and Balasubramanian, Suman, "A Short Note On Sums of Powers of Reciprocals of Polygonal Numbers" (2015). Student research—other. 1. http://scholarship.depauw.edu/studentresearchother/1 This Article is brought to you for free and open access by the Student Work at Scholarly and Creative Work from DePauw University. It has been accepted for inclusion in Student research—other by an authorized administrator of Scholarly and Creative Work from DePauw University. For more information, please contact [email protected]. A Short Note On Sums of Powers of Reciprocals of Polygonal Numbers Jihang Wang Suman Balasubramanian∗ Department of Mathematics Department of Mathematics DePauw University DePauw University 408 South Locust Street 602 S.College Ave. Greencastle, IN-46135 Greencastle, IN-46135 January 3, 2017 Abstract This paper presents the summation of powers of reciprocals of polygonal numbers. Several summation formulas of the reciprocals of generalized polygonal numbers are presented as examples of specific cases in this paper. Acknowledgement The authors would like to thank the Faculty Development Committee of DePauw University for funding this summer research project during summer 2015. 1 Introduction Polygonal numbers are numbers that represent dots that are arranged in the shape of a geometric regular polygon. Generalized polygonal numbers of rank r are numbers of the form [1][2][16][17] k[(r − 2)k − (r − 4)] P r = (1) k 2 r r r where Pk is k-th r-gonal number. For example, P3 gives the triangular number, while P5 gives the pentagonal number. Cook and Bacon observed in their paper [6], that specific examples of sums of reciprocals of various polygonal numbers are worth further study and thus motivates this paper. Before we present the summation of reciprocal formulas, we note the following interesting lemma about polygonal numbers. Lemma 1 The sum of the first n polygonal numbers is given by [6] n X n(n + 1)[(r − 2)n + 5 − r] P r = (1:2) k 6 k=1 ∗[email protected] 1 Proof. From Equation 1, we can see that, n n n X r − 2 X r − 4 X n(n + 1)[(r − 2)(2n + 1) − 3(r − 4)] n(n + 1)[(r − 2)n + 5 − r] P r = k2 − k = = (2) k 2 2 12 6 k=1 k=1 k=1 Equation 2 gives the results for r = 10; 11; 12; :::; 15 and we state it here, for the sake of completeness, as the following corollary. (For results relating to r = 1;::: 9, refer to [6]). Corollary 2 For r = 10; 11; 12; :::; 15, n X (n + 1)n(8n − 5) P 10 = k 6 k=1 n X (n + 1)n(9n − 6) P 11 = k 6 k=1 n X (n + 1)n(10n − 7) P 12 = k 6 k=1 n X (n + 1)n(11n − 8) P 13 = k 6 k=1 n X (n + 1)n(12n − 9) P 14 = k 6 k=1 n X (n + 1)n(13n − 10) P 15 = k 6 k=1 We now divide the remaining paper into the following sections. The next section will deal with preliminary definitions while sections three and four deal with some results regarding finite and infinite sums of reciprocals of powers of polygonal numbers respectively. 2 Preliminary Definitions For other notations and terminology not defined here, see Abramowitz and Stegun's Handbook of Mathe- matical Functions with Formulas, Graphs and Mathematical Tables [1] and Gradshteyn and Ryzhik's Table of Integrals, Series, and Products [8]. Definition 1 [1],[8] The Harmonic number Hn is the sum of the reciprocals of the first n natural numbers and is given by, n 1 1 1 X 1 H = 1 + + + ::: + = (3) n 2 3 n k k=1 2 Definition 2 [1],[8] The generalized Harmonic number of order n, m is given by, n X 1 H = (4) n;m km k=1 (s) Definition 3 [1],[8] The generalized Harmonic function Hn (z) is defined by [5][4][11][14] n X 1 H(s)(z) = (5) n (j + z)s j=1 , where n 2 N,s 2 CjZ−; and Z− := {−1; −2; −3; :::g. Definition 4 [1],[8] The error function (denoted by erf(x)) which is typically encountered when integrating the normal distribution, is defined by x 2 Z 2 erf(x) = p e−t dt (6) π 0 3 Finite Sums of Reciprocal of Powers of Polygonal Numbers We now consider the finite sum of the reciprocals of the powers of polygonal numbers. r th Pn 1 2 2 (1) r−4 Lemma 3 Let Pk denote the k polygonal number of rank r. Then r = − Hn;1+ Hn (− ). k=1 Pk β(r−2) β(r−2) r−2 Proof. n n X 1 X 2 = P r k[(r − 2)k − (r − 4)] k=1 k k=1 r−4 Let β = r−2 , then n n 2 X 1 X = r−2 P r k(k − β) k=1 k k=1 n X 2 2 = (− + ) kβ(r − 2) (k − β)β(r − 2) k=1 n n 2 X 1 2 X 1 = − ( ) + ( ) β(r − 2) k β(r − 2) (k − β k=1 k=1 By the definition of harmonic number (Definitions 2 and 3), we know that n X 1 2 2 = − H + H(1)(−β) P r β(r − 2) n;1 β(r − 2) n k=1 k 2 2 r − 4 = − H + H(1)(− ) β(r − 2) n;1 β(r − 2) n r − 2 n X 1 We must mention that our proof above though slightly different, does indeed verify the sum that has P r k=1 k been proved in [10]. 3 n X 1 8 8 r − 4 Lemma 4 Let P r denote the kth polygonal number of rank r. Then ( )2 = H − H(1)(− )+ k P r β3(r − 2)2 n;1 β3(r − 2)2 n r − 2 k=1 k 4 4 r − 4 H + H(2)(− ) for r 6= 4, where β = r−4 β2(r − 2)2 n;2 β2(r − 2)2 n r − 2 r−2 Proof. n n X 1 X 4 ( )2 = P r k2[(r − 2)k − (r − 4)]2 k=1 k k=1 By a partial fraction expansion, the sum becomes, n n 4 X 1 X (r−2)2 ( )2 = P r k2(k − β)2 k=1 k k=1 n X 8 8 4 4 = ( − + + ) kβ3(r − 2)2 (k − β)β3(r − 2)2 k2β2(r − 2)2 (k − β)2β2(r − 2)2 k=1 n n n n 8 X 1 8 X 1 4 X 1 4 X 1 = ( ) − ( ) + ( ) + ( ) β3(r − 2)2 k β3(r − 2)2 (k − β) β2(r − 2)2 k2 β2(r − 2)2 (k − β)2 k=1 k=1 k=1 k=1 By definitions (4) and (5), n X 1 8 8 r − 4 4 4 r − 4 ( )2 = H − H(1)(− ) + H + H(2)(− ) P r β3(r − 2)2 n;1 β3(r − 2)2 n r − 2 β2(r − 2)2 n;2 β2(r − 2)2 n r − 2 k=1 k r th Theorem 5 Let Pk denote the k polygonal number of rank r. Then, for r 6= 4, m ≥ 2, n m−1 1 m 1 m X 1 X ( ) ( ) ( )m = [(−1)mCm r−2 H ] + (−1)mCm r−2 H P r i ( r−4 )m+i−1 n;m−i+1 m ( r−4 )m+i−1 n;m−i+1 k=1 k i=1 r−2 r−2 m−1 1 m 1 m X ( ) r − 4 ( ) r − 4 + [(−1)i−1(−1)mCm r−2 H(m−i+1)(− )] + (−1)m−1Cm r−2 H(m−i+1)(− ) i ( r−4 )m+i−1 n r − 2 m ( r−4 )m+i−1 n r − 2 i=1 r−2 r−2 . r th Proof. Let Pk denote the k polygonal number of rank r. From (1) we can see that 1 2 r = ; Pk k[(r − 2)k − (r − 4)] th r so for the m power of Pk , we have 1 2m ( )m = : r m m r−4 m Pk (r − 2) k (k − r−2 ) By a tedious partial fraction expansion, we can see that n m−1 1 m n 1 m n X 1 X ( ) X 1 ( ) X 1 ( )m = [(−1)mCm r−2 ] + (−1)mCm r−2 P r i ( r−4 )m+i−1 km−i+1 m ( r−4 )m+i−1 k k=1 k i=1 r−2 k=1 r−2 k=1 4 m−1 1 m n 1 m n X ( ) X 1 ( ) X 1 + [(−1)i−1(−1)mCm r−2 ] + (−1)m−1Cm r−2 i ( r−4 )m+i−1 (k − r−4 )m−i+1 m ( r−4 )m+i−1 k − r−4 i=1 r−2 k=1 r−2 r−2 k=1 r−2 m−1 1 m 1 m X ( ) ( ) = [(−1)mCm r−2 H ] + (−1)mCm r−2 H i ( r−4 )m+i−1 n;m−i+1 m ( r−4 )m+i−1 n;m−i+1 i=1 r−2 r−2 m−1 1 m 1 m X ( ) r − 4 ( ) r − 4 + [(−1)i−1(−1)mCm r−2 H(m−i+1)(− )] + (−1)m−1Cm r−2 H(m−i+1)(− ) i ( r−4 )m+i−1 n r − 2 m ( r−4 )m+i−1 n r − 2 i=1 r−2 r−2 m m m−1 Ci = 2Ci−1 + Ci m m−1 Cm = 2Cm (In order to generalize the above formula, we used Mathematica to obtain the partial fraction expansions for m = 2;:::; 10 - see appendix for the mathematica results), 4 Infinite Sums of Reciprocal of Powers of Polygonal Numbers 1 X 1 From results in our previous section, we now validate a formula for , which has already been found P r k=1 k differently in [15].