Power Series with Binomial Sums and Asymptotic Expansions · Power series with binomial sums 1393 we conclude that the above lemma, among other things, gives a connection between

Int. Journal of Math. Analysis, Vol. 8, 2014, no. 28, 1389 - 1414

HIKARI Ltd, www.m-hikari.com

http://dx.doi.org/10.12988/ijma.2014.45126

Power Series with Binomial Sums

and Asymptotic Expansions

Khristo N. Boyadzhiev

Ohio Northern University

Department of Mathematics and Statistics

Ada, OH 45810, USA

Copyright © 2014 Khristo N. Boyadzhiev. This is an open access article distributed under the Creative Commons

Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the

original work is properly cited.

Abstract

This paper is a study of power series, where the coefficients are binomial expressions (iterated

finite differences). Our results can be used for series summation, for series transformation, or for

asymptotic expansions involving Stirling numbers of the second kind. In certain cases we obtain

asymptotic expansions involving Bernoulli polynomials, poly-Bernoulli polynomials, or Euler

polynomials. We also discuss connections to Euler series transformations and other series

transformation formulas.

Mathematics Subject Classification: Primary 11B83, Secondary 33B15, 40C15, 05A15

Keywords: Binomial transform, Finite differences, Stirling numbers, Bernoulli polynomials,

poly-Bernoulli polynomials, Euler polynomials, Euler series transformation, Lerch transcendent,

Arakawa-Kaneko zeta function, digamma function, asymptotic expansion

1390 Khristo N. Boyadzhiev

1 Introduction

We develop the theory of power series where the coefficients are binomial expressions. In

particular, we extend some known results involving classical special functions and polynomials.

Recently Guillera and Sondow [9] discussed several interesting representations involving

binomial transforms. These include the formulas

0 0

1

1 ( 1)(1 ) ( , )

1 ( )

kn

n ks

ns s a

kn k a

(1.1)

0 0

1( ) ( 1) ( )

1

m nk

n k

mm

nB a k a

kn

(1.2)

1

0 0

1 ( 1)( )

2 ( 1)

kn

nn k

s

ns

k k

(1.3)

0 0

1( ) ( 1) ( )

2

nk

n k

mm n

nE a k a

k

(1.4)

and some others. Here

1

1

( 1)( )

n

ns

sn

and

0

1( , )

( )ns

s an a

(1.5)

are Euler’s eta function and the Hurwitz zeta functions correspondingly; and ( ), ( )m mB a E a are

the Bernoulli and Euler polynomials. Interesting work in this direction was recorded also by

Connon [7]. The representations (1.1) and (1.3) originate from Hasse [11]. The representation

(1.2) results from (1.1), since (1 , ) ( )mm m a B a for 1,2,...m . Independently, (1.2) was

discovered by Todorov in his detailed investigation of Bernoulli polynomials [16]. This

representation is extended in Section 2 to poly-Bernoulli polynomials (see below (2.23)).

In our paper we extend these formulas by including them in a larger theory and connecting them

to the series transformation formulas of Euler [2] and the series transformation formulas

considered in [3]. Our results are organized in one proposition, three theorems, and several

corollaries. The formalism that we develop leads naturally to interesting asymptotic series. In

section 5 we present new proofs of some classical asymptotic expansions. We also obtain the

asymptotic expansion of the recently introduced Arakawa-Kaneko zeta function (Example 7 in

Section 5). Most remarkably, the poly-Bernoulli polynomials and the Bernoulli polynomials

Power series with binomial sums 1391

appear in several important asymptotic series for a large class of functions – see Corollary 2 in

Section 2.

2 Main results

We study power series whose coefficients are binomial sums of the form

0

( 1) ( )k

n

k

nf y zk

k

. (2.1)

Here ,y z are parameters and ( )f t is a formal power series

2

0 1 2( ) ...f t a a t a t (2.2)

Recall now that for any ,m n nonnegative integers, the numbers

0

( 1)( , )

!( 1)

nk m

n

k

nk

kS m n

n

are the Stirling numbers of the second kind. Good references for these numbers are the books of

Comtet [6], Graham et all [10], and Jordan [12].

Lemma1. For any power series ( )f t as in (2.2) and for any integer 0n we have

0 0

( , )

0

( 1) ( ) ( 1) ! m

k n p m pn m

p

p n

mk

nf y zk z y

kn Sa

, (2.3)

where ,y z are parameters. In particular, for 0y ,

00

( 1) ( ) ( 1) ( , )! m

m

k nn

mk

nf zk z S m n

kn a

. (2.4)

Proof. Starting from the binomial expansion

0

( )m

m p p m p

p

m

py zk yz k

we have by changing the order of summation


00 0

( 1) ( 1)( )mn n m

k k p p m p

pk k

n n m

k k py zk yz k

0 00

( 1) ( 1) ! ( , )m p k p n m pm n m

p p

p pk

m n m

p k py k n y S p nz z

.

That is,

00

( 1) ( ) ( 1) ! ( , )m m pn m

k n p

pk

n m

k py zk n S p n yz

. (2.5)

Multiplying both sides by ma and summing for 0,1,2...m we obtain (2.3).

Remark 1. Note that the summation in the series on the RHS in (2.3) and (2,4) starts, in fact,

from m n , since ( , ) 0S m n when m n . It follows from (2.5) that for any values of the

parameters y and z we have

0

( 1) ( ) 0 ( )mn

k

k

n

ky zk m n

,

and the sum equals ( 1) !n nn z for m n .

When ( )f t is analytic in some disk centered at the origin with radius 0r , the binomial sum on

the LHS of (2.4) is analytic in the disk with radius /r n (when 0n ) and the power series in z

on the RHS is absolutely convergent in that disk, being the Taylor series of the LHS.

Remark 2. We want to point out that the sums (2.1) are iterated finite differences. Namely, if

( ) ( 1) ( )f t f t f t , then

0

( 1) ( ) ( 1) ( )kn

n n

k

nf y k

kf y

.

Recalling that the Newton series of ( )f z has the form

0

( ) ( 1)( 2)...( 1)!

(0)

k

k

f z z z z z kk

f

,


we conclude that the above lemma, among other things, gives a connection between the Newton

series coefficients of ( )f z and its Taylor coefficients. This relationship was observed by Jordan

in [12, pp.189-190]. In our opinion, formula (2.3) is very important and needs further study and

applications.

Lemma 1 is essential for our results. Its first application is in Proposition 1 below, which can be

used for closed form evaluation of series. For this proposition we need also another lemma [12,

p. 189 (17)].

Lemma 2. For every integer 1m we have

1

( , )( 1)! ( 1) 0nm

n

S m n n

, (2.6)

and the value of this sum is 1 when 1m .

Proof. The Stirling numbers of the second kind satisfy the recurrence relation

( , ) ( 1, ) ( 1, 1), 1S m n nS m n S m n n m

and the lemma follows from here by a simple computation. Let 2m . Then

1 1 1

, ( 1)! ( 1) ( 1, ) ! ( 1) ( 1, 1)( 1)! ( 1)( ) n n nm m m

n n n

S m n n S m n n S m n n

11 1

1 0

( 1, ) ! ( 1) ( 1, ) ! ( 1)n km m

n k

S m n n S m k k

(by setting 1n k and noticing that ( 1, ) 0S m m )

1 1

1 1

( 1, ) ! ( 1) ( 1, ) ! ( 1) ( 1,0) 0n km m

n k

S m n n S m k k S m

The case 1m is obvious and the proof is completed.

From the above two lemmas we derive the following proposition.

Proposition 1. Let ( )f t be a formal power series as in (2.2). Then for every z we have


01

( 1) ( )1

(0)kn

kn

nf zk

kf z

n

(2.7)

Proof. We multiply both sides in equation (2.4) by 1

n and sum for 1,2,...n . On the RHS we

change the order of summation. This yields

1

101 0

( 1) ( )1

( , )( 1)!( 1)m

m

k nm

n

n

kn m

nf zk

kz a z

nS m n na

,

which is the desired result. (see also (2.19) below).

When ( )f t is a polynomial of degree p , the series on the LHS in (2.7) truncates, as the

binomial sums become zeros for n p (see Remark 1). A large class of functions for which (2.7)

holds is presented in [4] together with several applications.

We continue with results related to lemma 1.

Example 1. Consider the polynomial

0

1( )

!( , )

pm

m

tg t t

p ps p m

, (2.8)

where 0p is an integer and ( , )s m p are the Stirling numbers of the first kind. From Lemma 1,

00

( 1) ( 1) , ( , )!

!( ) mk n

pn

mk

n zkp m S m n z

k p p

ns

. (2.9)

These sums appeared in Todorov’s paper [17], in the Taylor series expansion

0 0

(1 ) 1 ( 1) ( 1)n

z p n

p

kn

k

n zkt t

k p

.

When p n , both sides in (2.9) are zeros.

In our first theorem we shall use the exponential polynomials

0

( ) ( , )m

m

n

nx S m n x

.


The properties of these polynomials together with some applications and notes on their history

can be found in [3] and [5]. The notation BE in (2.11) below stands for Binomial Exponential

Series.

Theorem 1. With ( )f t as in (2.2) and with parameters ,y z , the following representation holds

00 00

( 1) ( )!

( )n

k m pm p

pn m

pn mk

n mxf y zk

k pny xa z

.

(2.10)

When 0y this becomes

BE

0 00

( 1) ( )!

( ) mn

km m

n

n mk

nxf zk z

knxa

(2.11)

Proof. Multiplying both sides of (2.3) by !

nx

n and summing for 0,1...n we come to (2.10) after

changing the order of summation on the RHS. Note that since ( , ) 0S p n for n p , the

summation on the RHS leads exactly to the exponential polynomials ( )p x .

We do not specify convergence in the above formulas and view these series as formal power

series. If ( )f t is an entire function, the partial sums of the series on the LHS in (2.11) are defined

for every z . If the series on the RHS converges for some z , the formula will hold for all z in

some neighborhood of the origin. Depending on the choice of the function ( )f t , the series on the

RHS in the above equations may diverge. As we shall see later, in some cases the RHS

represents an asymptotic expansion as explained in [3].

From the above theorem with 1x we deduce the immediate corollary.

Corollary 1. For any power series ( )f t as in (2.2) we have

0 00

( 1)( 1) ( )

!

m

m

nk

m

n

n mk

nf zk z

knba

, (2.12)

where z is a variable and the numbers


0

(1) ( , )n

k

n n S n kb

are the well-known Bell numbers [5].

Example 2. From (2.9) and Theorem1 we obtain for every integer 0p ,

0 00

1( 1) ( )

! !( , ) m

m

nk

p pn

n mk

n zkxx z

k pn pp ms

. (2.13)

We shall prove now analogous expansions involving another sequence of polynomials. Let

0

( , ) !( )m

n

nm S m n n xx

,

( 0,1,...m ) be the geometric polynomials discussed in [3]. The geometric polynomials m will

replace the exponential polynomials m on the RHS of the new formulas. Multiplying both sides

of (2.3) by nx and summing for 0,1,...n , we obtain (after changing the order of summation)

the following theorem.

Theorem 2. With ( )f t as in (2.2) and with ,y z parameters,

0 0 00

( 1) ( ) ( )k m pm p

pnn m

n m pk

n mf y zk

k pyx a z x

. (2.14)

When 0y ,

BG

0 00

( 1) ( ) ( )k mm m

nn

n mk

nf zk

kz xx a

(2.15)

(BG stands for Binomial Geometric Series).

Example 3. From (2.9) and (2.15) we have for all integers 0p ,

0 00

1( 1)

!( , ) ( ) mk

m

p pnn

n mk

n zkz

k p pp m xsx

. (2.16)


Next we list some variations of these formulas. Integrating for x in (2.14) yields

1

00 0 00

( 1) ( ) ( )1

p

xnk m p

mp

n m

n m pk

n mxf y zk t dt

k pnya z

or, explicitly, after evaluating the integral on the RHS and reducing by x ,

0 0

( 1) ( )1

nk

n

n k

nxf y zk

kn

0 0 01

( )( , ) !m p

m

jp

pm

m p j

m

p j

xy S p j ja z

.

We integrate this equation for x , then reduce by x , and integrate again. Repeating this operation

1r times yields

0 0

( 1) ( )( 1)

nk

r

n

n k

nxf y zk

kn

(2.17)

0 0 0( 1)

( )( , ) !m p

m

jp

r

pm

m p j

m

p j

xy p j ja z S

for any 1r . When 0y this becomes

00 00

( 1) ( )( 1)

( )( , ) !

( 1)

nm

m

jmk

jr r

n

n mk

nxf zk

knz

xS m j j

ja

. (2.18)

We obtain now another formula this way: in (2.15) we remove from both sides the terms without

x (i.e. 0a ), then divide by x and integrate. The results is

101 0

( 1) ( ) ( , )( 1)!( )m

m

n mk p

p

n

kn m

nf zk

k

xz

nS m p p xa

. (2.19)


At this point we want to show some immediate corollaries involving the Bernoulli numbers mB ,

the Bernoulli polynomials,

0

( )m

m jj

m

j

mB y B

jy

,

and also the poly-Bernoulli numbers ( )q

n , and the poly-Bernoulli polynomials ( ) ( )q

n y . The

poly-Bernoulli numbers were introduced by Kaneko [13] by means of the generating function

( )

0

Li (1 )

1 !

t nq

ntn

q e t

e n

, (2.20)

where

1

Li ( )n

q qn

xx

n

is the polylogarithm. Kaneko showed that for 1q we have (1) ( 1)nn nB , i.e. (1)

1B 1/ 2 and (1) ( 1)n n nB . Also proved in [13] was the representation of the poly-Bernoulli numbers in

terms of the Stirling numbers of the second kind,

( )

0

( 1)( 1) ( , ) !

( 1)

jnq

n

j

n

qS n j j

j

.

This extends the well known representation of Bernoulli numbers in terms of ( , )S n j , namely,

0

( 1)( , ) !

1

jn

n

j

B S n j jj

.

In analogy to Bernoulli polynomials, one can define the poly-Bernoulli polynomials by

( )( )

0

( ) q

j

q m jm

m

j

m

jy y

.

This definition is equivalent to the definition given by Bayad and Hamahata in [1] by using the

generating function


( )

0

( )Li (1 )

1 !

t nxt q

ntn

qe x

e t

e n

.

The polynomials ( )( )qm x slightly differ from the similar polynomials defined in [8].

When 1q we have (1) ( 1) ( )( ) m

mm B yy , since

(1)

0 0

( 1) ( 1) ( )( ) j m

m

m j m jm j j

m m

j j

m mB B y

j jy y y

.

We shall use equation (2.17) now to obtain representations involving the Bernoulli numbers and

polynomials and also the poly-Bernoulli numbers and polynomials.

Corollary 2. For every power series (2.2), every integer 1r , and with parameters y and z

( ) 1

0 00

1( 1) ( )

( 1)( ) ( )m r

m m

k

r

n

n mk

nf y z k

knz y za

. (2.21)

When 0y ,

( )

0 00

1( 1) ( )

( 1)( )m

m

k rmr

n

n mk

nf zk

knza

. (2.22)

In particular, with 1r we have the following expansions involving the regular Bernoulli

numbers and polynomials,

1

0 00

1( 1) ( )

1( )m

m m

kn

n mk

nf y zk

knz B y za

(2.23)

0 00

1( 1) ( )

1

m

m m

kn

n mk

nf zk

knz Ba

. (2.24)

With the choice ( ) mf t t and 1z in (2.23) we obtain the representation (1.2). An analogous

representation for the poly-Bernoulli polynomials results from (2.21) with 1z


( )

0 0

1( ) ( 1) ( )

( 1)

m nr k

m

n k

m

r

ny y k

kn

(2.25)

(proved by a different method in [1]). Note that the summation here goes to n m , since for

n m we have

0

( 1) ( ) 0n

k

k

mny k

k

.

Now we shall apply (2.24) to the function ( ) tf t e . Obviously,

0

( 1) ( ) (1 )zk nn

k

nf zk

ke

and the LHS of (2.24) becomes

1

0

(1

1

1 ) 1log(1 (1 )

1 1 1

z nz

z z zn

e

n

ze

e e e

(for convergence we need Re 0z ). Therefore, (2.24) reduces to the classical series

01 !

n

n

nz

z zB

e n

,

which converges for | | 2z . This is a good example how we need sometimes to adjust the

range of z or ,x y in the above theorems in order to assure convergence.

Proof of the corollary. With ,y z parameters we first compute

0 0( 1)

( 1)( , ) !m p

jp

r

pm

p j

m

p jy p jS jz

0 0

( 1) (( 1)

( 1)1) ( , ) !

m p

m m pj

r

pm

p j

m y

p z jz p j jS


( ) ( ) 1

0

( 1) ( ) )(m p

m m r m rp m

m

p

m yB z B

p zy zz

.

That is, we proved the equation

( ) 1

0 0

( ) )( 1)

( 1)( , ) (!m p m r

m

jp

r

pm

p j

mz B

p jy p j y zS jz

.

Multiplying this equation by ma and summing for 0,1,...m we obtain

( ) 1

00 0 0

( , ) ! ( ) ( )( 1)

( 1) m r

m m

m pm

jp

r

pm

mm p j

mS p j j z y z

p jy aa z

and (2.21) follows from here and (2.17).

Notice that with the selection ( ) 1 tf t e in (2.22) we arrive at equation (2.20).

It is known that for the geometric polynomials we have

1

1

0

1

2

! 2( 1) 1 2

2 1

mp m

m mpp

pB

m

(2.26)

(see, for instance [10, Problem 6.76, page 317]). Setting 1

2x in (2.15) we find one more

interesting representation.

Corollary 3. With ( )f t as in (2.2) and z a complex variable

1

1

1

0 00

1( 1) ( )

2

1 2

1m

m

mk

mn

n

n mk

nf zk B z

ka

m

. (2.27)

Again, we do not specify convergence in the above series. When ( )f t is a polynomial, all

formulas in Theorems 1 and 2, as well as the formulas in Corollaries 2 and 3 provide closed form

evaluations of the series on the LHS. By choosing different functions ( )f t in Corollaries 2 and 3

we can generate various asymptotic expansions. It is really amazing that the poly-Bernoulli and

the Bernoulli polynomials are present in all these situations.


3 Series with Euler polynomials

We shall record here another corollary involving Euler’s numbers mE and Euler’s polynomials

( )mE x . First, we show a special connection between the geometric polynomials and Euler’s

polynomials. The geometric polynomials ( )m x are related to the geometric series in the

following way: For every | | 1x and every 0,1,2...m , we have

0

1

1 1

m nm

n

x

x xn x

(3.1)

(see [3]). We shall use this property to find the Taylor series of the function

1

( )1t

h te

,

where , are parameters. For this purpose we need to evaluate the higher derivatives of h at

zero. Assuming | | 1te we expand as geometric series

0

1 1

1 1 ( )( )

n

n tnt te e

e

.

From this

0

1

1( )m m

n

m

n tnt

d

dt en e

and in view of (3.1) ,

1 1

1 1 1m

m t

t t t

d e

dt e e e

(3.2)

so that, for 0t ,

( )

0

1(0)

1 11

mm

m

m

tt

dh

dt e

, (3.4)

which provides the Taylor series representation


0

1 1

1 !1 1

mm

mm

t me

t

. (3.5)

In particular, with 1 ,

0

2

1 !

1

2

m

mm

te m

t

. (3.6)

Multiplying both sides by xte and using Cauchy’s rule for multiplication of power series (see

also (4.4) below) we can write

00

1

2

2

1 !m k

xt m m

kkm

t

mex

ke

t

m

. (3.7)

At the same time, the function on the LHS here is the generating function for Euler’s

polynomials ( )mE x , i.e.

0

2

1 !( )

m

m

xt

mt

e

e m

tE x

. (3.8)

Comparing both series we find

0

1

2( ) m k

m

m

kk

mE x x

k

. (3.9)

In particular, when 0x ,

1

1

1

2

2(0) 1 2

1

m

m mmE Bm

(3.10)

Euler’s numbers mE are defined by

1

22

m

m mE E

. (3.11)

Thus we arrive at the following corollary.

Corollary 4. With ( )f t as in (2.2),


0 00

1( 1) ( ) ( )

2m

kmn

n

n mk

nf y k E y

ka

, (3.12)

where ( )mE y are Euler’s polynomial. In particular, with 1

2y ,

0 00

1 1( 1)

2 2 2

1m mm

k

n

n

n mk

nf k E

ka

. (3.13)

Proof. In (2,14) we set 1

2x and 1z . Then we use (3.9) for (3.12) and (3.11) for (3.13).

Note that applying the above corollary to the function ( ) mf t t yields the representation (1.4).

4 Connection to other series transformation formulas

In [3] the present author considered two series transformation formulas (STF). Namely, the

Exponential STF,

ES 0 0

( ) ( )!

nx m

m m

n m

xf zk e a z x

n

(4.1)

and the Geometric STF (for | | 1x ).

GS 0 0

1( )

1 1

n m

m m

n m

xx f zk a z

x x

, (4.2)

where ( )m x and ( )m x are the exponential polynomials and the geometric polynomials

correspondingly. It was shown that in some cases the ES and GS can be used to evaluate power

series in a closed form. In other cases these representations provide asymptotic expansions. To

show the connection of ES and GS to the expansions in the preceding section we shall use

Euler’s series transformations as described in [2]. Let ( )f t be defined by (2.2). The Exponential

Euler Series Transformation (EE) says that


EE 0 0 0

( )( ) ( 1) ( )

! !

n n nx k

n n k

nx xe f zk f zk

kn n

(4.3)

or, more generally,

0 0 0

( )( ) ( 1) ( )

! !

n n nx k n k

n n k

nx xe f zk f zk

kn n

, (4.4)

( with ,z parameters). The geometric version is ([1], [3])

EG 0 0 0

1( 1) ( ) ( 1) ( )

1 1

n nn n k

n n k

ntf zk t f zk

kt t

. (4.5)

Theorem 3. Consider the three formulas BE, ES, and EE (that is, (2.11), (4.1), and (4.3)). Then

any two of them together imply the third one. The same is true for BG, GS, and EG.

Proof. The proof of the first part is straightforward. In fact, we can unite the formulas together.

0 0 0 0

( )( 1) ( ) ( ) ( )

! !

n nnk x m

m m

n k n m

nx xf zk e f zk a z x

kn n

. (4.6)

For the second part of the theorem, we first use the substitution ,1 1

t xx t

t x

, in GS to put it

in the form

0 0

1( )

1 1

n

m

m m

n m

tf zk a z t

t t

, (4.7)

and then we write (using EG for the first equality and (3.5) for the second)

0 0 0

1( 1) ( ) ( 1) ( )

1 1

nnn k

n k n

nn tt f zk f zk

k t t

0

m

m m

m

a z t

. (4.8)

The proof is complete.

Example 4. For any complex number s consider the function


0

1( ) , | | 1

(1 ) m

ms s

sf t t t

mt

.

According to (4.6) we have

0 0 0 0

( 1) ( )( )

! (1 ) !(1 )

n k nnx m

m

n k n ms s

n sx xe z x

k mn k z n nz

. (4.9)

Replacing here z by 1

and x by x we come to the representation

0 0

( 1)( , ) ( )

!( )

n m

s mm sn m

s

sxe x x

mn n

, (4.10)

which is the asymptotic expansion of the polyexponential function ( , )se x in terms of . This

expansion was obtained in [3].

5 Asymptotic expansions

In this section we give new proofs of some classical asymptotic expansions and also provide

some new results. In particular, in Example 7 we find the asymptotic expansion of the recently

introduced Arakawa-Kaneko zeta function.

Example 5. Expansions related to the Lerch Transcendent.

The Lerch Transcendent is defined by the series

0

( , , )( )

k

sk

zz s a

k a

( 0a ). When 1z and Re 1s this is the Hurwitz zeta function

0

( , ) (1, , )( )

1s

k

s a s ak a

.

We have the representations


1 1

0 0

1 1( , , ) , ( , )

( ) ( )1 1

s a t s a t

t tx s a s a

s s

t e t edt dt

xe e

, (5.1)

0

1

0

( 1) 1

( ))

( )(1

k kn

s

s a t tn

k

n x

k st e dt

k axe

(5.2)

1

1

0

1

( )(1 )

1

st n

a t

t

t

s

exe dt

xe

,

and for 0 1x also,

0 0

1( 1)

1( , 1, ) log ( , , )

( )k

k

sn

n

k

n

k

x

ns x s a x x s a

a k

. (5.3)

With 1x this becomes

0 0

1

1

( 1)( 1, )

( )

k

sn

n

k

n

kns s a

a k

, (5.4)

also,

1

0 0

1( 1)

2( , , )

( )k

k

n sn

n

k

n

k

xx s a

k a

, (5.5)

and in particular, when 1x ,

1

0 0

1

2

( 1)( , ) ( 1, , )

( )

k

n sn

n

k

n

ks a s a

k a

. (5.6)

Proof. Starting from the well-known representation

0

11

( )

1

( )s

s k t a t

st e e dt

k a

(5.7)


we multiply both sides by kx and sum for 0,1,...,k to get (5.1). In order to prove (5.2) we

multiply (5.7) by ( 1)k knk

x and sum for 0,1,...,k n . From (5.2) and with 0 1x we have

11

0 000

1 1 (1 )( 1)

1 ( ) 1( ) 1

t sk

k n

sn n

n a t

tk

n

k

x xe t

n s n

edt

k a xe

1 1

0 0

1 1log( ) log

( ) ( )1 1

s st

a t a t

t t

t txe x t

s s

e edt dt

xe xe

1

0 0

log 1

( ) ( )1 1

s a t s a t

t t

x

s s

t e t edt dt

xe xe

log ( , , ) ( , 1, )x x s a s x s a ,

which proves (5.3). Next, from (5.2) again,

1

10 000

1 1 1 1( 1)

2 ( ) 2 2( )k s

nk t

n sn n

na t

k

n

k

xt

s

xee dt

k a

1 1

0 0

1 1 1 1 1

( ) 2 ( ) 111

2

s s

tt

a t a tt ts s

e dt e dtxexe

( , , )x s a ,

which is (5.5).

For some similar results see [7].

Now let 0 1x and consider the function

( )( )

t

s

xf t

a t

(5.8)


We write logt t xx e and expand this in power series for t . Then using Cauchy’s rule for

multiplication of power series and setting m k n we write

00 0

1 log 1 log1

( ) ! !

t k m k

s s mm

s

k k

sk k

sx t x t xt t

ma t a a k a a k

(5.9)

0 0

log

!

n kn

n kn s k

s xt

n k k a

When 1x this becomes

0

1( )

( )

m

m sm

s

s tf t

ma t a

. (5.10)

We apply now formula (5.3) to obtain (in view of (5.9) and (2.24)) the asymptotic expansion

0 0

log

!( , 1, ) log ( , , )

m

m

m k

k

m s k

s xB

m k k as x s a x x s a

.

In particular, when 1x only the 0k term in the second sum is nonzero and we come to the

well-known asymptotic series for the Hurwitz zeta function (when a )

0

( 1, ) m

mm s

s B

m as s a

. (5.11)

(See [14, p. 25] or [15, p. 610]. The expansion is written there in a somewhat different, but

equivalent form.)

From (3.12) we also derive the asymptotic expansion

0

10 0

( )1 1( , )

2 2

( 1)

( )m

m sm

k

n sn

n

k

n

k

s E ys y a

m ak y a

. (5.12)

In particular, when 0y ,

1

1

0 0

1 2(0)1( , )

2 ( 1)

m

mm

m s m sm m

Bs sEs a

m ma m a

. (5.13)


Example 6. The log-gamma and digamma functions.

Here we show how the results in Corollary 2 can be used to derive the classical asymptotic

expansions of the log-gamma function log ( )z and the digamma function ( ) log ( )d

z zdz

(see [18]). We shall use the integral representation

0

1 1( ) log

1 x

zxz z e d xx e

. (5.14)

Let Re 0z and consider the function

( ) log 1t

f tz

(5.15)

for 0t . Using the representation

0

1log 1

t xz xt e

e dxz x

,

we find

0 00 0

log 1( 1) ( 1) ( 1)k k k

zxnk x

k

n n

k k

n n n

k k k

k ee dx

z x

. (5.16)

Note that when 0n both sides are zeros. For 1n we have

0 00 0

log 1 1( 1) ( 1)k k xzx zx

nk xn n

k k

n n

k k

k e ee dx e dx

z x x

0

11

1

x

x

zxn e dx

exe

From here,

1

1 00 1

log 111

( 1)1 1 1

k

nx

xn

zxn

k n

n

k

k

z

e e dx

n n xe


0 0

log 1 (1 ) (1 ) (1 )1 1

x x x

x x

zx zxe dx e dxe e x e

x xe e

0

1 1( ) log

1 x

zxe d x z zx e

.

The summation in the very first sum can be started from 0n , as the first term there will be

zero anyway. Thus

0 0

log 11

( ) log ( 1)1

k

n

n

k

n

k

k

zz z

n

. (5.17)

Writing log 1 log( ) logk

zz k z

and noticing that for 1n we have

0 0

log log 0( 1) ( 1)k k

n n

k k

n n

k kz z

,

we conclude that

0 00 0

log 1 log( )1 1

( 1) log ( 1)1 1

k k

n n

n n

k k

n n

k k

kz k

zz

n n

and therefore,

0 0

log( )1

( ) ( 1)1

k

n

n

k

n

kz kz

n

. (5.18)

This representation can also be found in [9]. When | | | |t z the function ( )f t in (5.15) has the

representation

1

1

( 1)( ) log 1

mm

mm

tf t t

z m z

,

and according to formula (2.24) we obtain from (5.17) the asymptotic expansion ( )z

1

1

( 1) 1( ) log

m

m

mm

Bz z

m z

.


For the same function ( )f t from (5.15) and with 0y we compute

0 00 0

( ) log1 1

( 1) ( 1) 11 1

k k

n n

n nk k

n n

k kf y k

y k

n n z

0 0

log1

( 1) log ( ) log1

k

n

n k

n

ky z k z y z z

n

,

where for the last equality we used (5.18). This leads, in view of (2.23), to the more general and

interesting asymptotic representation ( z )

1

1

( 1) ( ) 1( ) log

m

m

mm

B yy z z

m z

. (5.19)

As 1

1( )

2B y y , we can write this in the form

1

2

( 1) ( )1 1( ) log

2

m

m

mm

B yyy z z

z z m z

. (5.20)

Integrating for z yields

1

12

( 1) ( )1 1log ( ) log

2 ( 1)

m

m

mm

B yy z z y z z C

m m z

.

The constant of integration is log 2C , as follows from Stirling’s formula

1

2( ) 2z

zz z e

( )z .

Finally,

1

1

( 1) ( )1 1log ( ) log log 2

2 ( 1)

m

m

mm

B yy z z y z z

m m z

. (5.21)

This is the classical asymptotic representation of the log-gamma function [14], [15], [18].

Example 7 The Arakawa-Kaneko zeta function.


Recently in [1] and [8] the following function was introduced and studied

1

0

1( , ) Li ( )

( )1

1

s a tt

tr rs as

t ee dt

e

, (5.22)

where 0a . In particular, it was proved in [1] and [8] that ( , )r s a has the representation

0 0

1

( 1)

( 1)

( )( , )

k

r r s

n

n k

n

kn k as a

, (5.23)

and when 1r , it is related to the Hurwitz zeta function by the formula

1( , ) ( 1, )s a s s a . (5.24)

Applying equation (2.22) from Corollary 2 (with 1z ) to the function ( ) ( ) sf t t a defined in

(5.10) we find the representation

( ) ( )

0 0

1( 1)( , )

r r

m m

m

r m s m sm m

s m s

m ma as a

, (5.25)

which is the asymptotic expansion of ( , )r s a for a large.

References

[1] Bayad, Abdelmejid and Yoshinori Hamahata, Polylogarithms and poly-Bernoulli

polynomials, Kyushu J. Math., 65 (2011) 15-24

[2] Boyadzhiev, Khristo N. Series Transformation Formulas of Euler Type, Hadamard

Product of Functions, and Harmonic Number Identities, Indian Journal of pure and

Applied mathematics 42 (2011) 371-387

[3] Boyadzhiev, Khristo N. A Series transformation formula and related polynomials,

Internat. J. Math. Math. Sci.Vol. 2005 (2005), Issue 23, Pages 3849-3866

[4] Boyadzhiev, Khristo N. Evaluation of series with binomial sums, Analysis Mathematica,

40 (1), (2014), 13-23.

http://www.hindawi.com/GetArticle.aspx?doi=10.1155/IJMMS.2005.3849


[5] Boyadzhiev, Khristo N. Exponential polynomials, Stirling numbers, and evaluation of

some Gamma integrals, Abstract and Applied Analysis, Volume 2009, Article ID 168672

(electronic, open source).

[6] Comtet, L. Advanced Combinatorics (D. Reidel Publ. Co. Boston) (1974)

[7] Connon, Donal F. Some series and integrals involving the Riemann zeta function,

binomial coefficients and the harmonic numbers. Volumes I and IV;

http://arxiv.org/abs/0710.4022 and http://arxiv.org/abs/0710.4028

[8] Coppo, M.-A. and B. Candelpergher, The Arakawa-Kaneko zeta function,

Ramanujan J.,22 (2010) 153-162

[9] Guillera, Jesus, Jonathan Sondow, Double integrals and infinite products for some

classical constants via analytic continuations of Lerch's transcendent, Ramanujan J. 16

(2008) 247-270

[10] Graham, Ronald L., Donald E. Knuth, and Oren Patashnik, Concrete Mathematics,

(Addison-Wesley Publ. Co., New York) (1994)

[11] Hasse, H., Ein Summierungsverfahren fur die Riemannische -Reihe, Math. Z, 32

(1930) 458-464

[12] Jordan, Charles, Calculus of finite differences (Chelsea, New York) (1950) (First

edition: Budapest 1939).

[13] Kaneko, M. Poly-Bernoulli numbers. J. Théorie de Nombres , 9 (1997) 221–228

[14] Magnus, W. Oberhettingeer, F. and R.P. Soni, Formulas and Theorems for the

Special Functions in Mathematical Physics (Springer) (1966)

[15] Olver, Frank W.J. et al. NIST Handbook of Mathematical Functions (NIST) (2010)

[16] Todorov, Pavel G. On the theory of the Bernoulli polynomials and numbers, J. Math.

Anal. Appl. 104 (1984) 309-350

[17] Todorov, Pavel G. Taylor expansions of analytic functions related to (1 ) 1xz , J.

Math.Anal. Appl. 132 (1988) 264-280

[18] Whittaker, E.T. and G.N. Watson, A Course of Modern Analysis (Cambridge) (1992)

Received: May 9, 2014

http://www.hindawi.com/journals/aaa/2009/168672.html

http://www.hindawi.com/journals/aaa/2009/168672.html

Documents

Power Series with Binomial Sums and Asymptotic Expansions · Power series with binomial sums 1393 we conclude that the above lemma, among other things, gives a connection between