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]. = 1 2n The Catalan numbers Cn are often defined by the explicit formula Cn n+1 n , ≥ + | 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- = 4n+2 = bers can also be defined by the recurrence relation Cn+1 n+2 Cn, where C0 1. So lim Cn+1 = 4. n→∞ Cn Here we study the convergence of the series P∞ 1 and evaluate the sum. Since n=0 Cn n Cn+1 P∞ x lim = 4, the ratio test implies that the series = converges for |x| < 4. n→∞ Cn n 0 Cn Consequently, the series P∞ 1 converges. We evaluate this infinite sum using gen- n=0 Cn erating functions, plus fundamental tools from the differential and integral calculus.

∞ Sum of the series P 1 Cn n=0 To this end, let f (x) be the generating function of the reciprocals of Catalan numbers, n f (x) = P∞ x . We compute the sum in three steps. First, we ﬁnd an ordinary dif- n=0 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) = 1 + P∞ x . So n=1 Cn

∞ n−1 ∞ 0 X nx X n + 1 f (x) = = x n. Cn Cn+ n=1 n=0 1

Since n+2 = 4n+2 , by the recurrence relation, this yields Cn Cn+1

∞ ∞ X n + 2 X 4n + 2 x n = x n, Cn Cn+ n=0 n=0 1 ∞ ∞ ∞ ∞ X n X x n X 4(n + 1) X x n x n + 2 = x n − 2 , Cn Cn Cn+ Cn+ n=0 n=0 n=0 1 n=0 1 2 x f 0(x) + 2 f (x) = 4 f 0(x) − [ f (x) − 1], x x(x − 4) f 0(x) + (2x + 2) f (x) = 2. (1)

This is a ﬁrst-order differential equation for f (x) with the initial conditions f (0) = 1 = f 0(0). 6= = 4−x 3/2 To facilitate solving (1) for x 0, we introduce the function g(x) x . Then 0 g (x) = −6 g(x) x(4−x) . This implies that

[x(x − 4)g(x)]0 = (2x + 2)g(x). (2)

Multiplying (1) by g(x), we get

x(x − 4) f 0(x)g(x) + (2x + 2) f (x)g(x) = 2g(x).

Using (2), this can be rewritten as

[x(x − 4) f (x)g(x)]0 = 2g(x).

But, again using (2),

{x(x − 4)[ f (x) − 1]g(x)}0 = [x(x − 4) f (x)g(x)]0 − [x(x − 4)g(x)]0 = 2g(x) − (2x + 2)g(x) = −2xg(x), consequently, Z x(x − 4)[ f (x) − 1]g(x) = −2 xg(x) dx + C1

2 R xg(x) dx − C f (x) = 1 + 1 , x(4 − x)g(x) where C1 is a constant.

142 „ THE MATHEMATICAL ASSOCIATION OF AMERICA Suppose 0 < x < 4. Then

Z Z 4 − x 3/2 xg(x) dx = x dx x Z (4 − x)3/2 = dx. x 1/2

Letting x = u2, this implies that

Z Z xg(x) dx = 2 (4 − u2)3/2 du

1 u = u(4 − u2)3/2 + 3u(4 − u2)1/2 + 12 arcsin + C 2 2 2 √ 1√ √ x = x(4 − x)3/2 + 3 x(4 − x)1/2 + 12 arcsin + C , 2 2 2

where C2 is another constant. Therefore, we have

√ √ √ 3/2 1/2 x x(4 − x) + 6 x(4 − x) + 24 arcsin + 2C2 − C1 f (x) = 1 + 2 − 4−x 3/2 x(4 x) x √ √ √ x(4 − x)3/2 + 6x(4 − x)1/2 + 24 x arcsin x + C x = 1 + 2 , (4 − x)5/2

0 where C = 2C2 − C1. Since f (0) = 1 = f (0), C = 0. Thus

√ √ x ∞ x(4 − x)3/2 + 6x(4 − x)1/2 + 24 x arcsin X x n f (x) = 1 + 2 = . 5/2 (4 − x) Cn n=0

When x = 1, this yields

∞ X 1 33/2 + 6 · 31/2 + 24 arcsin 1/2 = 1 + 5/2 Cn 3 n=0 √ 4 3 = 2 + π. (3) 27

√ Thus, the series P∞ 1 converges to the limit 2 + 4 3 π, which is approximately n=0 Cn 27 2.80613305077. Note that already P22 1 ≈ 2.80613305077; so the series converges n=0 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 = 1 in (1), we get

3 f 0(1) = 4 f (1) − 2 √ ! 4 3 = 4 2 + π − 2 27 √ 16 3 f 0(1) = 2 + π. 81

Since f 0(x) = P∞ n+1 x n, it follows that n=0 Cn+1

∞ √ X n + 1 16 3 = 2 + π. Cn+ 81 n=0 1

Since the differential equation is inﬁnitely differentiable, it follows from (1) that

x(x − 4) f 00(x) + (4x − 2) f 0(x) + 2 f (x) = 0. (4)

√ 00 = 0 + 00 = 8 + 56 3 This yields 3 f (1) 2 f (1) 2 f (1), so f (1) 3 243 π. Since

∞ 00 X (n + 1)(n + 2) f (x) = x n, Cn+ n=0 2 this implies that

∞ √ X (n + 1)(n + 2) 8 56 3 = + π. Cn+ 3 243 n=0 2

Differentiating (4) with respect to x, it follows similarly that f 000(1) = 2 f 0(1), so

∞ √ X (n + 1)(n + 2)(n + 3) 32 3 = 4 + π. Cn+ 81 n=0 3

Clearly, this technique can be employed to evaluate further sums of the form

∞ X (n + 1)(n + 2) ··· (n + k) , Cn+k n=0 where k ≥ 1.

144 „ THE MATHEMATICAL ASSOCIATION OF AMERICA ∞ n Sum of the series P (−1) Cn n=0 We now turn to to solving (1) for −4 < x < 0. We have

Z Z 3/2 4 − x xg(x) dx = x dx x Z (4 − x)3/2 = − √ dx. x

Letting x = −u2, this gives

Z Z (u2 + 4)3/2 xg(x) dx = − (−2u) du u Z = 2 (u2 + 4)3/2 du

u(u2 + 4)3/2 3 p p = 2 + u u2 + 4 + 6 ln u + u2 + 4 + C 4 2 3 1 p p = u(u2 + 4)3/2 + 3u u2 + 4 + 12 ln u + u2 + 4 + C 2 3 1p p p p = |x|(4 − x)3/2 + 3 |x|(4 − x) + 12 ln |x| + |4 − x| + C , 2 3 where C3 is a constant. Consequently, as before, we have √ √ √ √ | | − 3/2 + | | − + |x|+ |4−x| x (4 x) 6 x (4 x) 24 ln C f (x) = 1 + 4 − 4−x 3/2 x(4 x) x √ √ √ √ | | − 3/2 + | | − + | | −x+ 4−x x (4 x) 6 x (4 x) 24 x ln C = 1 − 4 . (4 − x)5/2

0 where C4 is a nonzero constant. Since f (0) = 1 = f (0), C4 = 2. Thus √ √ √ √ | | − 3/2 + | | − + | | −x+ 4−x x (4 x) 6 x (4 x) 24 x ln 2 f (x) = 1 − . (4 − x)5/2

n Since f (−x) generates the alternating series P∞ (−x) for 0 < x < 4, setting x = 1 n=0 Cn we get

∞ X (−1)n 53/2 + 6 · 51/2 + 24 ln φ = 1 − 5/2 Cn 5 n=0 √ 14 24 5 = − ln φ, (5) 25 125 √ 1+ 5 where φ is the well-known golden ratio 2 .

VOL. 43, NO. 2, MARCH 2012 THE COLLEGE MATHEMATICS JOURNAL 145 Formulas (3) and (5) can be employed to compute the sums of the subseries P∞ 1 and P∞ 1 : n=0 C2n n=0 C2n+1

∞ √ √ X 1 32 2 3 12 5 = + π − ln φ, C n 25 27 125 n=0 2 ∞ √ √ X 1 18 2 3 12 5 = + π + ln φ. C n+ 25 27 125 n=0 2 1

Acknowledgment. The authors would like to thank the editor and reviewers for their unwa- vering enthusiasm, and suggestions for the improvement of the exposition.

Summary. This article studies the convergence of the inﬁnite series of the reciprocals of the Catalan numbers. We extract the sum of the series as well as some related ones, illustrating the power of the calculus in the study of the Catalan numbers.

References

1. D. I. A. Cohen, Basic Techniques of Combinatorial Theory, Wiley, New York, 1978. 2. M. Gardner, Catalan Numbers: An Integer Sequence that Materializes in Unexpected Places, Scientiﬁc Ameri- can, 234 (1976), 120–125; available at http://dx.doi.org/10.1038/scientificamerican0676-120. 3. H. W. Gould, Bell and Catalan Numbers: Research Bibliography of Two Special Number Sequences, Revised Edition, Combinatorial Research Institute, Morgantown, West Virginia, 1978. 4. T. Koshy, Catalan Numbers with Applications, Oxford University Press, New York, 2009. 5. , Fibonacci and Lucas Numbers with Applications, Wiley, New York, 2001. 6. R. P. Stanley, Enumerative Combinatorics, Vol. 1, Wadsworth & Brooks/Cole, Monterey, California, 1986.

146 „ THE MATHEMATICAL ASSOCIATION OF AMERICA