Volume 10 (2009), Issue 3, Article 69, 8 pp.
DERIVATIVES OF CATALAN RELATED SUMS
ANTHONY SOFO
S CHOOL OF E NGINEERING AND S CIENCE
V ICTORIA U NIVERSITY, PO B OX 14428 M ELBOURNE
VIC 8001, AUSTRALIA .
anthony.sofo@vu.edu.au
Received 19 March, 2009; accepted 29 July, 2009
Communicated by J. Sándor
A BSTRACT. We investigate the representation of sums of the derivative of the reciprocal of
Catalan type numbers in integral form. We show that for various parameter values the sums
maybe expressed in closed form. Finally we give bounds for the sums under investigation, in
terms of the parameters.
Key words and phrases: Catalan numbers, Binomial coefficients, Estimates, Integral identities.
2000 Mathematics Subject Classification. Primary 05A10. Secondary 26A09, 26D20, 33C20.
1. I NTRODUCTION
We define Cn (j) , the Catalan related numbers, for j ∈ N ∪ {0} , N = {1, 2, 3, ...} as
 1 2n+j 2n+j+1
1
, for n = 1, 2, 3, ...
 n+1 n = 2n+j+1
n+1
(1.1)
Cn (j) =

1,
for n = 0,
and in particular, from (1.1), the Catalan numbers Cn = Cn (0) are defined by
 1 2n 2n+1
1
, for n = 1, 2, 3, ...
 n+1 n = 2n+1
n
Cn =
.

1,
for n = 0.
Catalan numbers have many representations, see Adamchik [1], in particular Penson and Sixdeniers [3], by employing some ideas of the Mellin transform gave the integral representation,
(1.2)
1
Cn =
2π
Z
4
tn−1/2
√
4 − t dt.
0
By the change of variable tw2 = 4 − t, in (1.2) we find that
Z
22n+2 ∞
w2
Cn =
dw,
π
(1 + w2 )n+2
0
079-09
2
A NTHONY S OFO
which, see Bailey et.al. [2], is closely related to cbn , the central binomial coefficient
Z
2n
22n+1 ∞
1
=
cbn =
dw.
n
π
(1 + w2 )n+1
0
Moreover, it can be seen that cbn = (n + 1) Cn .
Lemma 1.1. For j ≥ 0 and n ∈ N, let the Catalan related numbers
 1 2n+j 2n+j+1
1
, for n = 1, 2, 3, ...
 n+1 n = 2n+j+1
n+1
Cn (j) =

1,
for n = 0
be an analytic function in j then
0
d
1
1
(1.3)
=
dj Cn (j)
Cn (j)
n
n+1 X
= − 2n+j
n
r=1
1
r+j+n
n+1
= − 2n+j [Ψ (1 + j + 2n) − Ψ (1 + j + n)] ,
n
where the Psi(or digamma function)
Ψ (z) =
(Γ (z))0
d
ln (Γ (z)) =
dz
Γ (z)
and the Gamma function
Z
Γ (w) =
∞
tw−1 e−t dt,
0
for < (w) > 0.
Proof. For the first derivative of
1
Cn (j)
with respect to j, let for integer n,
1
n+1
= 2n+j Cn (j)
n
(n + 1) Γ (n + 1) Γ (n + j + 1)
Γ (2n + j + 1)
(n + 1) Γ (n + 1)
= Q
.
n
(n + r + j)
=
r=1
Taking the logs on both sides we have
" n
#
X
1
ln
= ln (n + 1) + ln [Γ (n + 1)] − ln
(n + r + j)
Cn (j)
r=1
and differentiating with respect to j we obtain the result (1.3).
It may be seen that for j = 0, we obtain
0 n
i
1
n+1X 1
n + 1 h (1)
(1)
= − 2n = − 2n H2n+1 − Hn+1 ,
Cn (j)
r+n
j=0
n
n
r=1
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
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C ATALAN R ELATED S UMS
3
where the nth Harmonic number
Hn(1)
Z
1
= Hn =
t=0
n
X1
1 − tn
dt =
= γ + Ψ (n + 1) ,
1−t
r
r=1
and γ denotes the Euler-Mascheroni constant, defined by
!
n
X
1
γ = lim
− log (n) = −Ψ (1) ≈ 0.577215664901532860606512.........
n→∞
r
r=1
An extension of the nth harmonic numbers is introduced and studied by Sandor [4]. As an aside
it is possible to consider higher derivatives of (1.3).
Some numbers of (1.3), without the negative sign are
0


1
n
Cn (j)




2
1

(j+2)2




3!(2j+7)
2

2
2


(j+3)
(j+4)




4!(3j 2 +30j+74)


3

2
2
2
(j+4)
(j+5)
(j+6)




5!(2j 3 +39j 2 +251j+533)


4

2
2
2
2
(j+5)
(j+6)
(j+7)
(j+8)




4
3
2


6!(5j +160j +1905j +10000j+19524)
5

2
2
2
2
2
(j+6) (j+7) (j+8) (j+9) (j+10)




5
4
3
2

+5380j +50445j +234908j+434568) 
 6 7!(6j +285j



(j+7)2 (j+8)2 (j+9)2 (j+10)2 (j+11)2 (j+12)2
...
.............................
In the next theorems we give integral representations for various summation expressions of
Catalan type numbers with parameters. In some particular cases we give closed form values of
the sums and then determine upper and lower bounds in terms of the given parameters.
The following theorem is proved.
2. C ATALAN R ELATED S UMS
Theorem 2.1. Let the Catalan related numbers, with parameter j = 0, 1, 2, 3, 4, ..., Cn (j) =
2n+j
1
and |t| < 4, then
n+1
n
Sj (t) =
∞
X
n=1
n
tn X
1
Cn (j) r=1 r + j + n
∞ n
n
X
t (n + 1) X
=
2n+j
=
n
n=1
∞ n
X
t (n + 1)
[Ψ (1 + j + 2n) − Ψ (1 + j + n)]
2n+j
n
n=1
Z
(2.1)
r=1
1
r+j+n
= −2t
0
1
(1 − x)j+1 log(1 − x)
dx.
(1 − tx (1 − x))3
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
http://jipam.vu.edu.au/
4
A NTHONY S OFO
Proof. Consider
∞
X
n=1
∞
X tn (n + 1)
tn
=
2n+j
Cn (j) n=1
n
∞ n
X
t (n + 1) Γ (n + 1) Γ (n + j + 1)
=
Γ (2n + j + 1)
n=1
.
Now by the use of the Gamma property Γ (n + 1) = nΓ (n) we have
∞
∞ n
X
t n (n + 1) Γ (n) Γ (n + j + 1) X n
=
t n (n + 1) B (n, n + j + 1) ,
Γ
(2n
+
j
+
1)
n=1
n=1
where
Γ (α) Γ (β)
=
B (α, β) =
Γ (α + β + 1)
1
Z
α−1
(1 − y)
y
β−1
Z
dy =
0
1
(1 − y)β−1 y α−1 dy
0
for α > 0 and β > 0 is the classical Beta function, therefore
∞
X
n
t n (n + 1) B (n, n + j + 1) =
n=1
∞
X
Z
n
t n (n + 1)
=
0
1
(1 − x)n+j xn−1 dx
0
n=1
Z
1
j
(1 − x)
x
∞
X
n (n + 1) (tx (1 − x))n dx
n=1
by interchanging sum and integral. Now applying Lemma 1.1 we obtain
1
Z
Sj (t) = −2t
0
(1 − x)j+1 log(1 − x)
dx,
(1 − tx (1 − x))3
which is the result (2.1).
Other integral representations involving binomial coefficients and Harmonic numbers can be
seen in [5], [6], [7] and [8].
Remark 1. For j ∈ N we obtain for t = 2
∞
n
X
2n (n + 1) X
Sj (2) =
2n+j
n
n=1
1
r=1
j+1
1
r+j+n
(1 − x) log(1 − x)
dx
(1 − 2x (1 − x))3
0
= α1 G + α2 ζ (2) + α3 π ln (2) + α4 π + α5 ,
Z
= −4
where the Catalan constant
Z
G=
π
4
ln (cot (x)) dx = .0.91596559...,
0
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
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C ATALAN R ELATED S UMS
5
and ζ (z) is the Zeta function. Some specific cases of Sj (2) are


j
α1
α2
α3
α4
α5




1
3
3
0
0 
 0
4
4




3
7
3
27
7
 6

− 32
− 32
8
32
4




35
35
27
3
6920563

 13
−
−
−

32
256
128
32
6350400 


....... ....... ....... ....... .......
....... 
Remark 2. For j ∈ N we obtain for t = − 12
n
X
∞
n
− 12 (n + 1) X
1
1
=
Sj −
2n+j
2
r+j+n
n
n=1
r=1
Z 1
(1 − x)j+1 log(1 − x)
=
3 dx
0
1 + 12 x (1 − x)
= β1 ζ (2) + β2 (ln (2))2 + β3 ln (2) + β4 .
Some specific cases of Sj − 21 are

j
β1


8
− 81
 0


8
 2
81


 8 − 31412

81

. . .
...
β2
β3
4
81
2
− 27
8
− 81
4
− 27
31744
81
6596
27
...
...
β4



0 

2
−9 
.

5045 
18 

... 
Next we shall give upper and lower bounds for the series Sj (t) given by (2.1).
Theorem 2.2. For j = 0, 1, 2, 3, ..., and 0 < t < 4
(2.2)
2t
< Sj (t)
(j + 2)2

2t


(j+2)2


≤
q



 2t
4 3
4−t
2
(2j+3)3
,
2t4 −41t3 +342t2 −1490t+3860
5(4−t)
Proof. Consider the integral inequality
Z x1
|f (x) g (x)| dx ≤
Z
+
0
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
t
2
t(4−t)11
2
.
|g (x)| dx,
x0
and from the integral (2.1) we can identify
Z x1
Z 1
(2.3)
|g (x)| dx =
(1 − x)j+1 ln (1 − x) dx =
x0
√
√ 1
x1
sup |f (x)|
x∈[x0 ,x1 ]
x0
5
1008 arcsin
1
.
(j + 2)2
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6
A NTHONY S OFO
Similarly
f (x) =
(2.4)
2t
(1 − tx (1 − x))3
is monotonic on x ∈ [0, 1] with lim f (x) = 2t, lim f (x) = 0, hence
x→0
x→1
3
4
sup f (x) = 2t
.
4−t
x∈[0,1]
The series (2.1) is one of positive terms and its lower bound is given by the first term, hence
combining these results we obtain the first part of the inequality (2.2). For the second part of
the inequality (2.2), consider the Euclidean norm where for α = β = 2, α1 + β1 = 1, and
for f (x) and g (x) defined by (2.4) and (2.3) respectively we have that |f (x)|2 and |g (x)|2
are integrable functions defined on x ∈ [0, 1] . From (2.1) and by Hölder’s integral inequality,
which is a special case of the Cauchy-Buniakowsky-Schwarz inequality,
Z 1
12 Z 1
12
2
2
Sj (t) ≤
|f (x)| dx
|g (x)| dx ,
0
0
where
Z
1
|g (x)|2 dx
12
Z
=
0
0
1
2 12
j+1
=
(1 − x) ln (1 − x) dx
s
2
(2j + 3)3
and
Z
1
|f (x)|2 dx
21
0
√  12
t
1008
arcsin
2
2t
−
41t
+
342t
−
1490t
+
3860
 ,
= 2t 
+ q
11
5 (4 − t)5
t (4 − t)

4
3
2
so that
s
Sj (t) ≤ 2t
√  12
t
1008
arcsin
2
2
2t
−
41t
+
342t
−
1490t
+
3860


+ q
11
(2j + 3)3
5 (4 − t)5
t (4 − t)

4
3
2
and the second part of (2.2) follows.
Remark 3. For a particular value of t and j ≥ 0 we can plot the exact value of Sj (t) , (2.1)
against the upper bounds from (2.2)
3
4
2t
(2.5)
Aj (t) =
(j + 2)2 4 − t
and
s
(2.6) Bj (t) = 2t
√  12
t
1008
arcsin
2
2
2t
−
41t
+
342t
−
1490t
+
3860
 .

+ q
11
(2j + 3)3
5 (4 − t)5
t (4 − t)

4
3
2
Remark 4. For t = 1.5 we have the graph, Figure 2.1, showing that Bj (1.5) is a better estimator
of Sj (1.5), (the lower curve in Figure 2.1), than Aj (1.5) up to about j = 4.5. The exact value
of j for a given value of t can be exactly calculated from (2.5) and (2.6).
The graph in Figure 2.1 suggests that the sum in (2.1) may be convex. We prove the convexity
of (2.1) in the following theorem.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
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C ATALAN R ELATED S UMS
7
Figure 2.1: A plot of Sj (1.5), Aj (1.5) and Bj (1.5) showing the crossover point at about j = 4.5.
Theorem 2.3. For j ≥ 0 and 0 < t < 4 the function j 7→ Sj (t), as given in Theorem 2.1 is
strictly decreasing and convex with respect to the parameter j ∈ [0, ∞) for every x ∈ [0, 1] .
Proof. Let
(1 − x)j+1 ln (1 − x)
gj (x, t) =
(1 − tx (1 − x))3
be an integrable function for x ∈ [0, 1] and put
Z 1
Sj (t) = −2t
gj (x, t) dx,
0
so that
1
1
(1 − x) ln (1 − x)
3 dx.
0
0 (1 − tx (1 − x))
Applying the Leibniz rule for differentiation under the integral sign, we have that
Z 1
∂
0
Sj (t) =
gj (x, t) dx
0 ∂j
Z 1
(1 − x)j+1 (ln (1 − x))2
dx.
= −2t
(1 − tx (1 − x))3
0
Z
S0 (t) = −2t
Z
g0 (x, t) dx = −2t
Since j ≥ 0 and 0 < t < 4
(1 − x)j+1 (ln (1 − x))2
> 0 for x ∈ (0, 1) ,
(1 − tx (1 − x))3
then Sj0 (t) < 0, so that the sum in (2.1), is a strictly decreasing sum with respect to the parameter j for x ∈ [0, 1] .
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
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8
A NTHONY S OFO
Now
Sj00
1
∂2
gj (x, t) dx
2
0 ∂ j
Z 1
(1 − x)j+1 (ln (1 − x))3
= −2t
dx,
(1 − tx (1 − x))3
0
Z
(t) =
and since
(1 − x)j+1 (ln (1 − x))3
< 0,
(1 − tx (1 − x))3
then Sj00 (t) > 0 so that (2.1) is a convex function for x ∈ [0, 1] .
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J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 69, 8 pp.
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