Convergence of a Catalan Series Thomas Koshy and Zhenguang Gao Thomas Koshy ([email protected]) received his Ph.D. in Algebraic Coding Theory from Boston University. His interests include algebra, number theory, and combinatorics. Zhenguang Gao ([email protected]) received his Ph.D. in Applied Mathematics from the University of South Carolina. His interests include information science, signal processing, pattern recognition, and discrete mathematics. He currently teaches computer science at Framingham State University. The well known Catalan numbers Cn are named after Belgian mathematician Eugene Charles Catalan (1814–1894), who found them in his investigation of well-formed sequences of left and right parentheses. As Martin Gardner (1914–2010) wrote in Scientific American [2], they have the propensity to “pop up in numerous and quite unexpected places.” They occur, for example, in the study of triangulations of con- vex polygons, planted trivalent binary trees, and the moves of a rook on a chessboard [1, 2, 3, 4, 6]. D 1 2n The Catalan numbers Cn are often defined by the explicit formula Cn nC1 n , ≥ C j 2n where n 0 [1, 4, 6]. Since .n 1/ n , it follows that every Catalan number is a positive integer. The first five Catalan numbers are 1, 1, 2, 5, and 14. Catalan num- D 4nC2 D bers can also be defined by the recurrence relation CnC1 nC2 Cn, where C0 1. So lim CnC1 D 4. n!1 Cn Here we study the convergence of the series P1 1 and evaluate the sum. Since nD0 Cn n CnC1 P1 x lim D 4, the ratio test implies that the series D converges for jxj < 4. n!1 Cn n 0 Cn Consequently, the series P1 1 converges. We evaluate this infinite sum using gen- nD0 Cn erating functions, plus fundamental tools from the differential and integral calculus. 1 Sum of the series P 1 Cn nD0 To this end, let f .x/ be the generating function of the reciprocals of Catalan numbers, n f .x/ D P1 x . We compute the sum in three steps. First, we find an ordinary dif- nD0 Cn http://dx.doi.org/10.4169/college.math.j.43.2.141 MSC: 05A15, 40A25 VOL. 43, NO. 2, MARCH 2012 THE COLLEGE MATHEMATICS JOURNAL 141 ferential equation satisfied by f .x/, then, after solving the differential equation in the interval .0; 4/, we compute f .1/. n We first rewrite f .x/ as f .x/ D 1 C P1 x . So nD1 Cn 1 n−1 1 0 X nx X n C 1 f .x/ D D x n: Cn CnC nD1 nD0 1 Since nC2 D 4nC2 , by the recurrence relation, this yields Cn CnC1 1 1 X n C 2 X 4n C 2 x n D x n; Cn CnC nD0 nD0 1 1 1 1 1 X n X x n X 4.n C 1/ X x n x n C 2 D x n − 2 ; Cn Cn CnC CnC nD0 nD0 nD0 1 nD0 1 2 x f 0.x/ C 2 f .x/ D 4 f 0.x/ − T f .x/ − 1U; x x.x − 4/ f 0.x/ C .2x C 2/ f .x/ D 2: (1) This is a first-order differential equation for f .x/ with the initial conditions f .0/ D 1 D f 0.0/. 6D D 4−x 3=2 To facilitate solving (1) for x 0, we introduce the function g.x/ x . Then 0 g .x/ D −6 g.x/ x.4−x/ . This implies that [x.x − 4/g.x/]0 D .2x C 2/g.x/: (2) Multiplying (1) by g.x/, we get x.x − 4/ f 0.x/g.x/ C .2x C 2/ f .x/g.x/ D 2g.x/: Using (2), this can be rewritten as [x.x − 4/ f .x/g.x/]0 D 2g.x/: But, again using (2), fx.x − 4/T f .x/ − 1Ug.x/g0 D [x.x − 4/ f .x/g.x/]0 − [x.x − 4/g.x/]0 D 2g.x/ − .2x C 2/g.x/ D −2xg.x/; consequently, Z x.x − 4/T f .x/ − 1Ug.x/ D −2 xg.x/ dx C C1 2 R xg.x/ dx − C f .x/ D 1 C 1 ; x.4 − x/g.x/ where C1 is a constant. 142 ã THE MATHEMATICAL ASSOCIATION OF AMERICA Suppose 0 < x < 4. Then Z Z 4 − x 3=2 xg.x/ dx D x dx x Z .4 − x/3=2 D dx: x 1=2 Letting x D u2, this implies that Z Z xg.x/ dx D 2 .4 − u2/3=2 du 1 u D u.4 − u2/3=2 C 3u.4 − u2/1=2 C 12 arcsin C C 2 2 2 p 1p p x D x.4 − x/3=2 C 3 x.4 − x/1=2 C 12 arcsin C C ; 2 2 2 where C2 is another constant. Therefore, we have p p p 3=2 1=2 x x.4 − x/ C 6 x.4 − x/ C 24 arcsin C 2C2 − C1 f .x/ D 1 C 2 − 4−x 3=2 x.4 x/ x p p p x.4 − x/3=2 C 6x.4 − x/1=2 C 24 x arcsin x C C x D 1 C 2 ; .4 − x/5=2 0 where C D 2C2 − C1. Since f .0/ D 1 D f .0/, C D 0. Thus p p x 1 x.4 − x/3=2 C 6x.4 − x/1=2 C 24 x arcsin X x n f .x/ D 1 C 2 D : 5=2 .4 − x/ Cn nD0 When x D 1, this yields 1 X 1 33=2 C 6 · 31=2 C 24 arcsin 1=2 D 1 C 5=2 Cn 3 nD0 p 4 3 D 2 C π: (3) 27 p Thus, the series P1 1 converges to the limit 2 C 4 3 π, which is approximately nD0 Cn 27 2.80613305077. Note that already P22 1 ≈ 2:80613305077; so the series converges nD0 Cn to the limit remarkably fast. VOL. 43, NO. 2, MARCH 2012 THE COLLEGE MATHEMATICS JOURNAL 143 Additional consequences of the differential equation Letting x D 1 in (1), we get 3 f 0.1/ D 4 f .1/ − 2 p ! 4 3 D 4 2 C π − 2 27 p 16 3 f 0.1/ D 2 C π: 81 Since f 0.x/ D P1 nC1 x n, it follows that nD0 CnC1 1 p X n C 1 16 3 D 2 C π: CnC 81 nD0 1 Since the differential equation is infinitely differentiable, it follows from (1) that x.x − 4/ f 00.x/ C .4x − 2/ f 0.x/ C 2 f .x/ D 0: (4) p 00 D 0 C 00 D 8 C 56 3 This yields 3 f .1/ 2 f .1/ 2 f .1/, so f .1/ 3 243 π. Since 1 00 X .n C 1/.n C 2/ f .x/ D x n; CnC nD0 2 this implies that 1 p X .n C 1/.n C 2/ 8 56 3 D C π: CnC 3 243 nD0 2 Differentiating (4) with respect to x, it follows similarly that f 000.1/ D 2 f 0.1/, so 1 p X .n C 1/.n C 2/.n C 3/ 32 3 D 4 C π: CnC 81 nD0 3 Clearly, this technique can be employed to evaluate further sums of the form 1 X .n C 1/.n C 2/ ··· .n C k/ ; CnCk nD0 where k ≥ 1. 144 ã THE MATHEMATICAL ASSOCIATION OF AMERICA 1 n Sum of the series P (−1) Cn nD0 We now turn to to solving (1) for −4 < x < 0. We have Z Z 3=2 4 − x xg.x/ dx D x dx x Z .4 − x/3=2 D − p dx: x Letting x D −u2, this gives Z Z .u2 C 4/3=2 xg.x/ dx D − .−2u/ du u Z D 2 .u2 C 4/3=2 du u.u2 C 4/3=2 3 p p D 2 C u u2 C 4 C 6 ln u C u2 C 4 C C 4 2 3 1 p p D u.u2 C 4/3=2 C 3u u2 C 4 C 12 ln u C u2 C 4 C C 2 3 1p p p p D jxj.4 − x/3=2 C 3 jxj.4 − x/ C 12 ln jxj C j4 − xj C C ; 2 3 where C3 is a constant. Consequently, as before, we have p p p p j j − 3=2 C j j − C jxjC j4−xj x .4 x/ 6 x .4 x/ 24 ln C f .x/ D 1 C 4 − 4−x 3=2 x.4 x/ x p p p p j j − 3=2 C j j − C j j −xC 4−x x .4 x/ 6 x .4 x/ 24 x ln C D 1 − 4 : .4 − x/5=2 0 where C4 is a nonzero constant. Since f .0/ D 1 D f .0/, C4 D 2. Thus p p p p j j − 3=2 C j j − C j j −xC 4−x x .4 x/ 6 x .4 x/ 24 x ln 2 f .x/ D 1 − : .4 − x/5=2 n Since f .−x/ generates the alternating series P1 .−x/ for 0 < x < 4, setting x D 1 nD0 Cn we get 1 X .−1/n 53=2 C 6 · 51=2 C 24 ln φ D 1 − 5=2 Cn 5 nD0 p 14 24 5 D − ln φ; (5) 25 125 p 1C 5 where φ is the well-known golden ratio 2 .