#A38 INTEGERS 9 (2009), 491-496 A SHORT PROOF OF A SERIES EVALUATION IN TERMS OF HARMONIC NUMBERS

A SHORT PROOF OF A SERIES EVALUATION IN TERMS OF HARMONIC NUMBERS¹

Alois Panholzer

Institut f¨ur Diskrete Mathematik und Geometrie, Technische Universit¨at Wien, 1040 Wien, Austria

[email protected] Helmut Prodinger

Department of Mathematics, University of Stellenbosch, South Africa [email protected]

Received: 5/21/08, Revised: 2/12/09, Accepted: 3/14/09, Published: 9/30/09

Abstract We give another short and simple proof of

!

n≥1

1 2²ⁿ⁻¹(2n−1)

!

k

"2n−1 k

# 1

k−j−n+¹₂ =−2 j

!j

k=1

1 2k−1.

1. The Main Result

For positive integersj, consider S(j) =!

n≥1

1 2²ⁿ⁻¹(2n−1)

!

k

"2n−1 k

# 1

k−j−n+¹₂.

This quantity arose in [4] and was subsequently evaluated in [3]. Further proofs of the final formula

S(j) =−2 j

!j

k=1

1 2k−1

were given in [2, 1]. Here, we give another short and simple proof.

For our analysis, it is better to consider T(j) =!

n≥1

1 2²ⁿ⁻¹(2n−1)

!

k

"2n−1 k

# 1

k+j−n+¹₂,

1This work was supported by the South African Science Foundation NRF, grant 2053748, and by the Austrian Science Foundation FWF, grant S9608-N23.

so that

S(j) =!

n≥1

1 2²ⁿ⁻¹(2n−1)

!

k

" 2n−1 2n−1−k

# 1

k−j−n+¹₂

=!

n≥1

1 2²ⁿ⁻¹(2n−1)

!

k

"2n−1 k

# 1

(2n−1−k)−j−n+¹₂

=!

n≥1

1 2²ⁿ⁻¹(2n−1)

!

k

"2n−1 k

# 1

n−k−j−¹₂

=−T(j).

It will be advantageous to treat the sum T$j :=!

n≥1

1 2²ⁿ⁻¹(2n−1)

%1 2

2n−1!

k=0

"2n−1 k

# 1

k+j−n+¹₂

−1 2

2n−1!

k=0

"2n−1 k

# 1

k−j−n+¹₂

&

=1 2

'T(j)−S(j)(=T(j).

First we will give a representation of the sum)2n−1 k=0

'_2n−1

k

( ₁

k+m+¹₂,withm∈Z, as a curve integral in the complex plane.

Lemma 1 We have

2n−1!

k=0

"2n−1 k

# 1

k+m+¹₂ =*

Γ

u^2m(1 +u²)²ⁿ⁻¹du,

where the curveΓ is the upper half of the unit circle in the complex plane starting from −1and ending at1, i.e.,Γ={cos(π−t) +isin(π−t) :t∈[0,π]}.

Proof. We have

*

Γ

u^2m(1 +u²)²ⁿ⁻¹du=*

Γ 2n−1!

k=0

"2n−1 k

#

u^2k+2mdu

=²ⁿ⁻¹!

k=0

"2n−1 k

# u^2k+2m+1 2k+ 2m+ 1

++ ++

1

−1

= 2

2n−1!

k=0

"2n−1 k

# 1

2k+ 2m+ 1.

!

Thus we get 1 2

2n−1!

k=0

"2n−1 k

# 1

k+j−n+¹₂ −1 2

2n−1!

k=0

"2n−1 k

# 1

k−j−n+¹₂

=1 2

*

Γ

'u^2j−2n−u^−2j−2n(

(1 +u²)²ⁿ⁻¹du

=1 2

*

Γ

'u^2j−1−u^−2j−1(,1 +u² u

-2n−1

du, and further

T$(j) =!

n≥1

1 2²ⁿ⁻¹(2n−1)

1 2

*

Γ

'u^2j−1−u^−2j−1(,1 +u² u

-2n−1

du

=!

n≥1

*

Γ

1 2(2n−1)

'u^2j−1−u^−2j−1(,1 +u² 2u

-2n−1

du. (1)

Next we consider, forj∈Nandu∈Γ, the series

Qj(u) :=!

n≥1

1 2(2n−1)

'u^2j−1−u^−2j−1(,1 +u² 2u

-2n−1

.

Lemma 2 The seriesQ$_j(ϕ) :=Q_j(e^iϕ)converges uniformly for ϕ∈[0,π], i.e.,

Q$j(ϕ) =ie^−iϕsin(2jϕ)1

2log1 + cosϕ

1−cosϕ. (2)

Proof. Substitutingu=e^iϕ, withϕ∈[0,π], we can write Q$_j(ϕ) =Q_j(e^iϕ) =ie^−iϕ!

n≥1

1 2n−1

e^2jiϕ−e^−2jiϕ 2i

,e^iϕ+e^−iϕ 2

-2n−1

=ie^−iϕ!

n≥1

1

2n−1sin(2jϕ)(cosϕ)²ⁿ⁻¹. Since we have

!

n≥1

z²ⁿ⁻¹ 2n−1 =1

2log1 +z

1−z, for |z|<1, (3) we obtain the pointwise convergence of the series Q$_j(ϕ), for ϕ ∈ (0,π), to the function given in (2).

Obviously we also have Q$_j(0) =Q$_j(π) = 0, which shows convergence ofQ$_j(ϕ), for all ϕ ∈ [0,π]. Since (3) converges uniformly for all z, with |z| ≤ q < 1,

we obtain immediately that Q$j(ϕ) converges uniformly for all ϕ ∈ [δ,π−δ], for arbitrary 0<δ<^π₂. But since for allj∈N

ϕ→0limsin(2jϕ) log1 + cosϕ 1−cosϕ = 0,

which can easily be shown, we obtain that for all$>0 there exists aδ >0, such that

++

+ie^−iϕ !

n≥N

1

2n−1sin(2jϕ)(cosϕ)²ⁿ⁻¹+++= !

n≥N

1

2n−1sin(2jϕ)(cosϕ)²ⁿ⁻¹

≤!

n≥1

1

2n−1sin(2jϕ)(cosϕ)²ⁿ⁻¹<$,

for all 0 ≤ ϕ < δ and for all N ∈ N. This, together with the obvious relation Q$_j(π−ϕ) = −Q$_j(ϕ), shows that Q$_j(ϕ) converges even uniformly for all ϕ ∈

[0,π]. !

After back-substitution, we obtain that the seriesQ_j(u) converges uniformly for allu∈Γto the function

'u^2j−1−u^−2j−1(1

4log,1 +^1+u_2u² 1−^1+u2u²

-.

Thus in equation (1) we can interchange summation and integration and obtain the integral representation

T$j =*

Γ

!

n≥1

1 2(2n−1)

'u^2j−1−u^−2j−1(,1 +u² 2u

-2n−1

du

=1 2

*

Γ

'u^2j−1−u^−2j−1(1

2log,1 +^1+u_2u² 1−^1+u_2u²

-du

=1 2

*

Γ

'u^2j−1−u^−2j−1(1 2log,

−(1 +u)² (1−u)²

-du. (4)

Remark Using the substitutionu=e^iϕ one obtains the following representation of the sumT$j as a real integral:

T$_j= 1 2

* π 0

sin(2jϕ) log1 + cosϕ 1−cosϕdϕ, but it seems more involved to evaluate this integral.

We use now that, foru∈Γ:

1 2log,

−(1 +u)² (1−u)²

-= log,

(−i)1 +u 1−u

-,

and the correct determination of the (multi-valued) logarithm function is obtained when considering the real analogue of this equation:

1

2log1 + cosϕ

1−cosϕ= logcos^ϕ₂

sin^ϕ₂, for ϕ∈(0,π).

Then equation (4) gives T$_j =1

2

*

Γ

'u^2j−1−u^−2j−1(log,

(−i)1 +u 1−u

-du

=log(−i) 2

*

Γ

'u^2j−1−u^−2j−1( du+*

Γ

'u^2j−1−u^−2j−1(1

2log,1 +u 1−u

-du

=*

Γ

'u^2j−1−u^−2j−1(1

2log,1 +u 1−u

-du, (5)

since obviously the first integral vanishes.

In order to proceed we consider, forj∈Nandu∈Γ, the series R_j(u) :=!

n≥1

'u^2j−1−u^−2j−1(u²ⁿ⁻¹ 2n−1.

Lemma 3 There is uniform convergence of the series Rj(u), for u ∈ Γ, to the function

f_j(u) ='

u^2j−1−u^−2j−1(1

2log,1 +u 1−u

-. (6)

Proof. It is well-known that equation (3) even holds, with the exception of z = 1 andz =−1, for all complexz with|z|= 1, which proves pointwise convergence of R_j(u) tof_j(u) foru∈Γ\ {−1,1}.

Obviously we also haveRj(−1) =Rj(1) = 0, which shows convergence ofRj(u), for allu∈Γ. Furthermore, since

R_j(u) =−

!j

m=−j+1

u^2m−2

2(m+j)−1+ !

m≥j+1

4j

(2(m−j)−1)(2(m+j)−1)u^2m−2, as can be shown easily, we obtain by simple majorization arguments that R_j(u) converges even uniformly for allu∈Γto the functionfj(u). ! Thus in equation (5) we can replace f_j(u) by the seriesR_j(u) and interchange summation and integration and get

T$_j=!

n≥1

*

Γ

'u^2j−1−u^−2j−1(u²ⁿ⁻¹ 2n−1du,

which can be evaluated easily:

T$j=!

n≥1

*

Γ

,u^2n+2j−2

2n−1 −u^2n−2j−2 2n−1

-du

=!

n≥1

, u^2n+2j−1

(2n−1)(2n+ 2j−1) − u^2n−2j−1 (2n−1)(2n−2j−1)

-++++

1

−1

= 2!

n≥1

, 1

(2n−1)(2n+ 2j−1) − 1

(2n−1)(2n−2j−1) -

= 1 j

!

n≥1

, 1

2n−1− 1 2n+ 2j−1

-−1 j

!

n≥1

, 1

2n−2j−1− 1 2n−1

-

= 1 j

!j

k=1

1 2k−1−1

j

!j

k=1

−1 2k−1= 2

j

!j

k=1

1 2k−1.

AcknowledgmentOne referee’s patience with various versions of this material is gratefully acknowledged.

References

[1] W. Chu and L. De Donno. Transformation on infinite double series and applications to harmonic number identities. Appl. Algebra Engrg. Comm. Comput., 15:339–348, 2005.

[2] C. Krattenthaler. Hypergeometric proof of a curious identity of Lyons, Paule and Riese.South East Asian J. Math. Math. Sci., 3:63–67, 2004.

[3] R. Lyons, P. Paule, and A. Riese. A computer proof of a series evaluation in terms of harmonic numbers. Appl. Algebra Engrg. Comm. Comput., 13:327–333, 2002.

[4] R. Lyons and J. E. Steif. Stationary determinantal processes: phase multiplicity, Bernoullicity, entropy, and domination. Duke Math. J., 120:515–575, 2003.