BINOMIAL SUMS INVOLVING HARMONIC NUMBERS 1. Introduction

DOI: 10.2478/s12175-011-0006-5 Math. Slovaca 61 (2011), No. 2, 215–226

BINOMIAL SUMS INVOLVING HARMONIC NUMBERS ˇev Marian Genc (Communicated by Georges Grekos ) ABSTRACT. This paper develops the approach to the evaluation of a class of infinite series that involve special products of binomial type, generalized harmonic numbers of order 1 and rational functions. We give new summation results for certain infinite series of non-hypergeometric type. New formulas for the number π are included. c 2011 Mathematical Institute Slovak Academy of Sciences

1. Introduction The number π has been in the centre of view of mathematicians for centuries. Lambert proved the irrationality of π in 1761 and in 1794 Legendre proved that π is not a square root of a rational number. See also [10]. In 1882 Lindemann comes with the proof that π is a transcendental number. Up to date there exist more than 50 different proofs of the transcendence of π. Among them let us mention e.g. [9]. There is a lot of ways how to express the number π. The purpose of this note is to develop a new class ofseries summing up to π. These series involve the central binomial coefficient 2n n , the generalized harmonic n 1 + number of order 1, i.e. jαj , (α, n) ∈ R × N and a special kind of rational j=1

functions. The central binomial coefficient 2n n plays in the theory of the infinite series an important role by their transformations and accelerations. In 1979 Apéry [2] proved the irrationality of ζ(3) — and in the same manner, the irrationality ∞ ∞ (−1)n+1 1 of ζ(2) — by use of the formulae ζ(3) = 52 and ζ(2) = 3 . 2n 3 n (n) n2 (2n n) n=1 n=1 2010 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 33B15; Secondary 40C10. K e y w o r d s: summation methods, infinite series, generalized harmonic numbers.

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ˇ MARIAN GENCEV

Such types of Cantor and factorial series can be also found in [11]–[15] or in [19]. Other important results with evaluations and transformations of infinite series can be seen in [1], [5], [7], [16], [18], [21] or [22]. n 1 Similar importance play harmonic numbers of order 1 defined by Hn := j j=1

for every positive integer n which can be expressed with the help of various definite integrals or infinite series. Harmonic numbers arise by the transformations of the Riemann function ζ(x). Significant results involving infinite series with the term Hn were presented by many authors, c.f. [6] or [8]. An interesting consideration connected with harmonic numbers is presented in [20]. It is not typical to analyse the infinite series with harmonic numbers, binomial coefficients and a rational function. We generalize the construction of such infinite series by considering the generalized harmonic numbers of order 1 and a product of the binomial type introduced in the next definitions. A similar type of this infinite series occurs by the computation of the hyperbolic volume V of the Haleman Ferguson’s sculpture Figure — Eight Complement II, see [4].

1.1

For every ordered pair (α, n) ∈ R+ × N, we define the generalized harmonic number Hn (α) by n 1 . Hn (α) := jαj j=1 Similarly as the usual harmonic numbers Hn (1), there exists a known integral 1 1−(1−x)n dx with L(α) := 1 − α1 . This representation in the form Hn (α) = x L(α)

fact can be easily used to show that ∞ q Hn (α) αq = · ln , n q q−1 αq − 1 n=1 ∞ q αq n · Hn (α) q−1 = · ln + . qn (q − 1)2 αq − 1 αq − 1 n=1

(1) (2)

The identities are valid for all (α, q) ∈ R2 with |q| > 1 and α ≥ 1.

1.2 Assume that (a, b, m1, m2) ∈ (R+0 )2 × N2. For (j, n) ∈ N0 × N

define (m1 +m2 )n+1 m m 1n 2n m1 , m2 := ϕn (s + a) (s + b) (s + a + b)−1 , a, b m1 , m2 ϕn,j := a, b

s=1

s=1

s=1

m1 (n+j)

m2 (n+j)

(s + a)

s=m1 n+1

s=m2 n+1

(m1 +m2 )(n+j)+1

(s + b)

(s + a + b)−1 .

s=(m1 +m2 )n+2

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BINOMIAL SUMS INVOLVING HARMONIC NUMBERS

Using the definition of Beta function B(x, y) by means of the Eulerian integral, we obtain the integral representation of the term ϕn in the form 1 1 m1 , m2 = tm1 n+a (1 − t)m2 n+b dt ϕn a, b (a + b + 1) B(a + 1, b + 1)

(3)

0

for every positive integer n. In the later considerations 2n of the infinite series with harmonic numbers, central binomial coefficient n and a rational function, we introduce the notation φ1 (q) and φ2 (q) for the infinite series φ1 (q) :=

∞

Hn (1) n (2n + 1) 2n q n n=1

φ2 (q) :=

and

∞ Hn (1) n 2n q n n=1

which sums are not known in closed form.

2. Main results In this section we give the overview of the main results connected with infinite series involving generalized harmonic numbers Hn (α), products of binomial type and rational functions.

12.1 Assume that (α, a, b, q, m1, m2, k) ∈ (R+0 )4 × N2 × N0 with α ≥ 1

and |q| > 4 . Define

k k m2 j m1 , m2 · ϕn,j , b a, b (−q)j j=0 k k ∞ n · Hn (α) m1 , m2 j m1 , m2 . · ϕn · ϕn,j σ2 := a, b a, b qn (−q)j n=1 j=0

∞ Hn (α) m1 , · ϕn σ1 := n a, q n=1

Then 1 σ1 = (a + b + 1) B(a + 1, b + 1)

1 0

k−1 xm1 (1 − x)m2 xa (1 − x)b 1 − × q × ln

αq dx (4) αq − xm1 (1 − x)m2 217


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and q σ2 = (a + b + 1) B(a + 1, b + 1) × ln

1 x

a+m1

(1 − x)

0

αq αq −

xm1 (1

k−2 xm1 (1 − x)m2 1− q q − xm1 (1 − x)m2 + dx. αq − xm1 (1 − x)m2 (5)

b+m2

− x)m2

Example 2.1. ⎧ ⎪ ⎪ 4− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∞ ⎨ Hn (α) 6n2 + 25n + 25 4− 2n · 3 = 2 + 46n + 15 ⎪ 8n + 36n ⎪ n n=1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩4 −

2 √ π 3 √ 7+4 3 √ π 3(2 + 3) √ 3+2 2 √ π 2(1 + 2)

for α = 1, for α = 2 + for α = 1 +

√

3,

√1 . 2

Example 2.2. ∞ Hn (α) 2n ·

n=1

n

2.2

9n3 + 39n2 + 42n + 36n2 + 46n + 15

8n3

⎧ 40 2 ⎪ ⎪ √ π for α = 1, − ⎪ ⎪ 9 ⎪ 3 ⎪ ⎨ √ √ √ 1 = 100 + 60 3 − 7π(4 + 3 3) for α = 2 + 3, ⎪9 ⎪ ⎪ ⎪ ⎪ √ √ ⎪1 ⎩ 80 + 60 2 − 3π(11 + 4 2) for α = 1 + √12 . 18 Assume that q >

1 4

and define

√ √ 4q − 1 + i 4q − 1 − i √ √ − Li2 i. δ(q) := Li2 2 4q − 1 2 4q − 1

(6)

Then 1 4q − 1 −2q arctan √ · ln + δ(q) , φ1 (q) = √ q 4q − 1 4q − 1 −2q 4q − 1 1 . φ2 (q) = δ(q) − 4 − ln · arctan √ q 4q − 1 (4q − 1)3 218 Unauthenticated Download Date | 9/24/15 11:37 PM

(7) (8)


Example 2.3. The following identities hold

∞

∞ √ 2 π√ Hn (1) 2n = (4 − ln 3) 3 − δ(1) 3, 27 9 n n=1

π Hn (1) 2 2n = − √ ln 3 − √ δ(1). (2n + 1) 3 3 3 n=1 n

(9) (10)

Remark 2.1 Using the concept of Clausen’s function Cl2 (x) defined by ∞ sin(nx) Cl2 (x) := n2 n=1

and the relations [17, (4.18), (5.5)] we get π π √ π 4 δ(1) = ln 3 − Cl2 = ln 3 − 3 L−3 (2). 6 3 3 6 Here L−3 (2) denotes the Dirichlet L-series defined as ∞ −3 1 L−3 (2) := · 2, n n n=1

is the Kronecker symbol, see [3]. Thus the formulae in (9) and (10) where −3 n can be written in a more compact form as

∞

∞ 2 2π √ Hn (1) 2n = (2 − ln 3) 3 + L−3 (2), 27 3 n n=1

2π Hn (1) 2n = − √ ln 3 + 2 L−3 (2). (2n + 1) 3 3 n=1 n This identities shows that the irrationality of the values φ1 (1) and φ2 (1) depends essentially on the algebraic character of the value L−3 (2). However, the irrationality of L−3 (2) is still an open problem. Example 2.4. ⎧ 2 ⎪ √ π ⎪ ⎪ ⎪ 3 3 ⎪ ⎪ ⎨π

∞ Hn (1) R(n) 2n · = 2 2n + 1 ⎪ ⎪ n n=1 ⎪ ⎪ ⎪ ⎪ √4 π ⎩ 3

for R(n) = 3n + 1, for R(n) = n · 2n , for R(n) = (n − 1) · 3n . 219


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2.1 Let p(n) ∈ Q[n], deg p(n)1 ≤ 3, q(n) := 8n3 + 36n2 + 46n + 15 and q be a real algebraic number with q > 4 . Then

J1 J2 J3 ∞ H1 (n) p(n) (1) (1) (2) (21) (22) (3) (3) 2n · = Q+ λj ln αj + λj ln αj ln αj + λj Li2 (αj ) n q(n) q n n=1 j=1 j=1 j=1 (11) 3 (k) (k) where (Q, αj , λj ) ∈ Q .

Example 2.5. ∞ Hn (1) 81n3 + 291n2 + 128n − 250 2 2n · = √ π, 3 + 36n2 + 46n + 15 8n 3 n n=1

(12)

∞ 4 Hn (1) 9n3 + 33n2 + 17n − 25 2n · 3 = . 2 8n + 36n + 46n + 15 9 n n=1

(13)

3. Proofs P r o o f o f T h e o r e m 2.1. Using the integral representation in (3) we get ∞ k Hn (α) m1 , m2 −k j k k−j (−1) ϕn+j q σ1 = q a, b qn j n=1 j=0

k ∞ q −k Hn (α) (−1)j k × = (a + b + 1) B(a + 1, b + 1) n=1 q n j=0 q j−k j 1 ×

xm1 (n+j)+a (1 − x)m2 (n+j)+b dx

0

∞ q Hn (α) = × (a + b + 1) B(a + 1, b + 1) n=1 q n −k

1 ×

xm1 n+a (1 − x)m2 n+b (q − xm1 (1 − x)m2 )k dx

0

q −k = (a + b + 1) B(a + 1, b + 1)

1

xa (1 − x)b (q − xm1 (1 − x)m2 )k ×

0

×

∞ n=1

Hn (α) q xm1 (1−x)m2


n dx


q −k = (a + b + 1) B(a + 1, b + 1) ×

1

k

xa (1 − x)b (q − xm1 (1 − x)m2 ) ×

0

q xm1 (1−x)m2 q xm1 (1−x)m2 −

1

· ln

αq xm1 (1−x)m2 αq xm1 (1−x)m2 −

1

dx

where the interchange of the integral and the infinite summation in ∞ 1 an (x) dx with

n=1 0

an (x) :=

Hn (α) m1 n+a x (1−x)m2 n+b (q −xm1 (1−x)m2 )k , qn

|q| >

1 , x ∈ [0, 1], 4

follows easy. It suffices to consider that for M, N ∈ N, M < N , 0≤

N

|al (x)|

l=M

≤ max

0≤x≤1

a

x (1 − x)

b

|q − x

m1

(1 − x)

m2 k

|

N l=M

Hl (α) |q| xm1 (1−x)m2

(14) l .

|q| For |q| > 14 , x ∈ [0, 1] and all positive integers m1 , m2 we have xm1 (1−x) m2 > 1 and thus by (1) there exists a positive integer N0 so that for every M, N ∈ N, N0 < M < N , the finite sum on the left side of (14) can be made arbitrarily small, independent of x ∈ [0, 1]. Finally, the simplification of the last integral gives immediately the formula (4). The proof of (5) uses the same technique and the formula in (2).

Remark 3.1 The validity of Theorem 2.1 can be extended by taking mi ∈ N0 , i = 1, 2, with m1 + m2 > 1. P r o o f o f E x a m p√l e 2.1. Take a = b = 0, m1 = m2 = q = k = 1 in (4) with α = 1, α = 2 + 3 and α = 1 + √12 respectively. P r o o f o f E x a m p l e √2.2. Take a = b = 0, m1 = m2 = q = 1 and k = 2 in (5) with α = 1, α = 2 + 3 and α = 1 + √12 respectively. P r o o f o f T h e o r e m 2.2. Using the fact that for every positive integer n we have 1 1 = xn (1 − x)n dx, (2n + 1) 2n n 0


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we obtain for q >

1 4

∞ Hn (1) φ1 (q) = xn (1 − x)n dx n q n=1 1

0

1

1 q · ln dx q − x(1 − x) q − x(1 − x)

=q 0

1 4q 4q ln q arctan √ −√ =√ 4q − 1 4q − 1 4q − 1

√ 1 4q−1

0

2 ln 4q−1 4 + ln(t + 1) dt. t2 + 1

The evaluation of the last integral is based on the evaluation of the integral 2 +1) J(t) := ln(t t2 +1 dt. However, J(t) =

1 − it i + ln(t2 + 1) (ln(t + i) − ln(t − i)) 2 2 i i 2 −i 2 ln (t + i) − ln (t − i) + ln(t − i) · ln − (t + i) 4 2 2 i i − ln(t + i) · ln (t − i) 2 2

i 2

Li2

1 + it 2

− Li2

where ln z, z ∈ C \ {0}, is defined by ln z := ln |z| + i · arg z with −π < arg z ≤ π. d Li2 x = − ln(1−x) . This can be verified by the fact that dx x Hence, for q > √ 1 4q−1

0

1 4

we have

ln 4q−1 + ln(t2 + 1) 4q − 1 1 4 +J dt = ln arctan √ t2 + 1 4 4q − 1

It is easy to see that J(0) = is more complicated. Set

− π2

ln 2. For J

1 −i , η1 := ln √ 4q − 1 −i 1 √ η3 := ln +i , 2 4q − 1

√ 1 4q−1

1 √ − J(0). 4q − 1

(15) 1 , with q > 4 , the situation

1 η2 := ln √ +i , 4q − 1 i 1 √ η4 := ln −i 2 4q − 1



then √ 4q 1 ln − i arctan 4q − 1, 2 4q − 1 1 q 1 , − i arctan √ η3 = ln 2 4q − 1 4q − 1 η1 =

Putting δ(q) :=

√ 4q 1 ln + i arctan 4q − 1, 2 4q − 1 1 q 1 η4 = ln . + i arctan √ 2 4q − 1 4q − 1 η2 =

√ √ 4q − 1 + i 4q − 1 − i − Li2 √ i, Li2 √ 2 4q − 1 2 4q − 1

q>

1 , 4

we obtain δ(q) i(η2 − η1 ) 4q i(η22 − η12 ) iη1 η3 iη2 η4 1 = + · ln − + − J √ 2 2 4q − 1 4 2 2 4q − 1 which can be simplified using the identities for ηi , 1 ≤ i ≤ 4, as follows √ δ(q) 1 π 4q − 1 4q = J √ + ln arctan 4q − 1 + · ln . √ 2 4 q 4 4q − 1 4q − 1 We deduce that 1 − J(0) J √ 4q − 1 1 = − 2 arctan 4q − 1 · ln 2 − arctan 4q − 1 · ln q 2 1 π π δ(q) + arctan 4q − 1 · ln(4q − 1) + π ln 2 + ln q − ln(4q − 1) + 2 4 4 2 π δ(q) 1 16q = − ln · arctan 4q − 1 − + 2 4q − 1 2 2 δ(q) 1 16q 1 + = ln · arctan √ . 2 4q − 1 2 4q − 1 This leads to √ 1 4q−1

0

ln 4q−1 + ln(t2 + 1) 4 dt t2 + 1

1 δ(q) 4q − 1 1 16q 1 + ln + arctan √ arctan √ 4 2 4q − 1 2 4q − 1 4q − 1 √ 4 q 1 4q − 1 δ(q) = arctan √ · ln + + ln √ 4 2 4q − 1 4q − 1 1 1 · ln(q(4q − 1)) + δ(q) . = arctan √ 2 4q − 1 = ln


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Finally, this can be used for the evaluation of the infinite series φ1 (q) in the following way 1 4q ln q · arctan √ φ1 (q) = √ 4q − 1 4q − 1 1 1 4q · · ln(q(4q − 1)) + δ(q) arctan √ −√ 4q − 1 2 4q − 1 4q 1 q 2q =√ · arctan √ · ln δ(q) −√ 4q − 1 4q − 1 4q − 1 q(4q − 1) 1 q 4q 2qδ(q) √ √ · arctan · ln = −√ 4q − 1 4q − 1 4q − 1 4q − 1 1 4q − 1 −2q arctan √ · ln + δ(q) . =√ q 4q − 1 4q − 1 This completes the proof of (7). The proof of the identity (8) is based on the same d technique and it uses the obvious differential equation φ1 (q)−φ2 (q) = 2q dq φ1 (q), 1 with q > 4 . P r o o f o f E x a m p l e 2.3. In Theorem 2.2 set q = 1 in the sums φ1 (q) and φ2 (q). P r o o f o f E x a m p l e 2.4. To 2.2 and eval prove the identities use Theorem uate the terms 3φ2 (1)−φ1 (1), φ2 12 −φ1 12 and φ2 13 −3φ1 13 respectively. P r o o f o f C o r o l l a r y 2.1. Using the obvious identity 8n3 + 36n2 + 46n + 15 = (2n + 1)(2n + 3)(2n + 5), we have ∞ 4n2 + 16n + 15 Hn (1) 2n · 3 φ1 (q) = , 8n + 36n2 + 46n + 15 qn n n=1 ∞ Hn (1) 8n3 + 36n2 + 46n + 15 · 3 . φ2 (q) = 8n + 36n2 + 46n + 15 q n 2n n n=1

(16)

Taking a = b = 0 and m1 = m2 = 1 in the series σ1 and σ2 defined in Theorem 2.1 we can write j (n + s)2 ∞ 1 1 Hn (1) s=1 (−1)j q −j σ1 = · 2j+1 n 2n j q n j=0 n=1 (2n + s) s=1

∞ 1 Hn (1) (8q − 2)n2 + (32q − 7)n + 30q − 5 · = 2q n=1 q n 2n 8n3 + 36n2 + 46n + 15 n


(17)


and σ2 =

∞ n=1

j (n + s)2 n · Hn (1) 2 s=1 j −j (−1) q · 2j+1 j q n 2n n j=0 (2n + s) 2

s=1

∞ 1 Hn (1) (16q 2 − 8q + 1)n3 + (64q 2 − 28q + 3)n2 + (60q 2 − 20q + 2)n · = 2 . 4q n=1 q n 2n 8n3 + 36n2 + 46n + 15 n (18)

Next, note

⎞ 16q 2 − 8q + 1 64q 2 − 28q + 3⎟ ⎟. 60q 2 − 20q + 2⎠ 0

⎛

P = (pij ) 1≤i≤4

1≤j≤4

0 8 0 ⎜ 4 36 8q − 2 := ⎜ ⎝16 46 32q − 7 15 15 30q − 5

Then by easy computation we obtain det P ≡ 60 = 0. Thus every polynomial p(n) ∈ R[n], with deg p(n) ≤ 3, can be expressed as a linear combination of 4 4 4 the polynomials Pi (n) := pji nj−1 , 1 ≤ i ≤ 4, say p(n) = ki pji nj−1 . j=1

i=1 j=1

Hence, ∞

4 4

∞

H1 (n) p(n) Hn (1) · · = n 2n n 2n q(n) q q n n n=1 n=1 =

4 i=1

ki

i=1 j=1

ki pji nj−1

q(n)

∞ Hn (1) Pi (n) · n 2n q(n) q n n=1

= k1 φ1 (q) + k2 φ2 (q) + 2qk3 σ1 + 4q 2 k4 σ2 . Now use the formulas for σi , φi (q), i = 1, 2, in Theorem 2.1 and Theorem 2.2 to deduce (11). P r o o f o f E x a m p l e 2.5. Let σ1 and σ2 denotes the sums in Example 2.1 and Example 2.2 with α = 1 respectively. Then 9σ2 − 10σ1 = √23 π and σ2 − σ1 = 49 . These results agree with the identities in (12) and (13).

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ˇ MARIAN GENCEV [3] APOSTOL, T. M.: Modular Functions and Dirichlet Series in Number Theory, Springer, New York, 1997. [4] BORWEIN, J.—BAILEY, D.: Mathematics by Experiment (2nd ed.), A. K. Peters, Wellesley, MA, 2008. [5] BORWEIN, M. J.—BROADRHURST, D. J.—KAMNITZER, J.: Central binomial sums, multiple Clausen values, and zeta values, Exp. Math. 10 (2001), 25–34. [6] BORWEIN, D.—BORWEIN, J. M.: On an intriguing integral and some series related to ζ(4), Proc. Amer. Math. Soc. 123 (1995), 1191–1198. [7] BRADLEY, D. M.: A class of series acceleration formulae for Catalan’s constant, Ramanujan J. 3 (1999), 159–173. [8] COFFMAN, S. W.: Problem 1240 and solution: An infinite series with harmonic numbers, Math. Mag. 60 (1987), 118–119. ˇ J.: Two proofs of transcendence of π and e, Czechoslovak Math. J. 35(110) [9] HANCL, (1985), 543–549. ˇ J.: A simple proof of the irrationality of π 4 , Amer. Math. Monthly 93 (1986), [10] HANCL, 374–375. ˇ [11] HANCL, J.: A note to the rationality of infinite series I, Acta Math. Inform. Univ. Ostraviensis 5 (1997), 5–11. ˇ [12] HANCL, J.—TIJDEMAN, R.: On the rationality of Cantor and Ahmes series, Publ. Math. 65 (2004), 371–380. ˇ [13] HANCL, J.—TIJDEMAN, R.: On the irrationality of Cantor series, J. Reine Angew. Math. 571 (2004), 145–158. ˇ J.—TIJDEMAN, R.: On the irrationality of factorial series, Acta Arith. 118 [14] HANCL, (2005), 383–401. ˇ [15] HANCL, J.—TIJDEMAN, R.: On the irrationality of polynomial Cantor series, Acta Arith. 133 (2008), 37–52. [16] LEHMER, D. H.: Interesting series involving the central binomial coefficient, Amer. Math. Monthly 89 (1985), 449–457. [17] LEWIN, L.: Polylogarithms and Associated Functions, North-Holland, Amsterdam, 1981. [18] RIVOAL, T.: Simultaneous generation of Koecher and Almkvist-Granville’s Ap´ ery-like formulae, Exp. Math. 13 (2004), 503–508. [19] SHIDLOVSKI, A. B.: Transcendental Numbers, Nauka, Moskva, 1987 (Russian). [20] SONDOW, J.: Problem 11026: An identity involving harmonic numbers, Amer. Math. Monthly 112 (2005), 367–369. [21] SURY, B.—WANG, T.—ZHAO, F.: Identities involving reciprocals of binomial coefficients, J. Integer Seq. 7 (2004), article 04.2.8. [22] TRIF, T.: Combinatorial sums and series involving inverses of binomial coefficients, Fibonacci Quart. 38 (2000), 79–84. Received 10. 11. 2008 Accepted 23. 9. 2009

Department of Mathematical Methods in Economics Technical University of Ostrava Sokolsk´ a tˇ r. 33 CZ–701 21 Ostrava 1 Czech Republic E-mail : [email protected]
