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Home , Generating function, Harmonic number, Power series, Riemann zeta function, Telescoping series

A special constant and with zeta values and harmonic numbers

Khristo N. Boyadzhiev Department of and Statistics, Ohio Northern University, Ada, OH 45810, USA [email protected]

Abstract. In this paper we demonstrate the importance of a which is the value of several interesting numerical series involving harmonic numbers, zeta values, and . We also evaluate in closed form a number of numerical and .

Key words and phrases harmonic numbers, skew-harmonic numbers, zeta , ,

2010 Mathematics Subject Classification. 11B34, 11M06, 33B15, 40C15

1. Introduction

The purpose of this paper is to demonstrate the importance of the constant

1 ψγ(t ++ 1) M = ∫ dt ≈1.257746 0 t where ψ ()z=ΓΓ′ ()/() zz is the digamma function and γψ= − (1) is Euler’s constant. 1

in Section 2 shows that M is the numerical value of many interesting and important series

involving logarithms, harmonic numbers Hn and zeta values ζ ()n . The harmonic numbers are defined by

11 Hn=ψγ( ++=+++ 1) 1 ... ,nH ≥ 1, = 0 n 2 n 0

1

with

∞ ln(1− t ) n −=∑ Htn , | t | < 1. 1− t n=1

We also use the skew-harmonic numbers given by

1 1 (− 1) n−1 H−−=−+++1 ... ,nH ≥ 1. = 0 . n 23 n 0

Riemann’s zeta function is defined by

∞ 1 ζ (ss )= ∑ s , Re> 1. n=1 n

− We consider series with terms Hnn (ζ ( )− 1) , Hnn (ζζ ( )−+ ( n 1)) , Hnnn (ζζ( )−+ ( 1) ) and

− others similar. Related power series with coefficients Hnan ζ (+ 1, ) and Hn ζ ( na+ 1, ) , where ζ (,sa ) is the Hurwitz zeta function are also studied. Some series are evaluated in closed form in terms of known constants and . Other interesting series will be presented in form. The main results of this paper are Theorem 1 in Section 2, Theorem 2 in Section 3, and Theorem 3 in Section 4.

2. A mosaic of series

A classical theorem rooted in the works of Christian Goldbach (1690-1764) says that

∞ ∑{ζ (n )−= 1} 1 n=2

(see [1], [9], and [12, p. 142]; a short proof is given in the Appendix). The terms ζ ()n − 1 of the 1 above series decrease like the powers of . The following lemma was proved in [7, p. 51]. It 2 will be needed throughout the paper. (For convenience a proof is given in the Appendix.)

Lemma 1. For every n ≥ 2

2

1 11n + <ζ ()n −< 1 . 2nn 21n −

The lemma shows that the power series with general term {ζ (nx )− 1} n is convergent in the disk ||2x < . The following representation is true [12, p. 173, (139)].

∞ (1) ∑{ζ (nx )− 1}n−1 =−− 1γψ ( 2 −x ) . n=2

Integration of this series yields [12, p. 173, (135)]

∞ ζ ()n − 1 (2) ∑ xxn =−(1γ ) +Γ− ln (2 x ) n=2 n

and in particular, when x =1, this turns into a classical result published by Euler in 1776 (see [11])

∞ ζ ()n − 1 (3) ∑ =1 −γ . n=2 n

With x = −1 we find

∞ζζ()nn−− 1 ∞∞ () (1)n ∑(− 1) nn = ∑∑( − 1) − =γ −+1 ln Γ (3) n=2n nn= 22 nn=

so that (as ln 2= ln Γ (3) )

∞ ζ ()n (4) ∑(−= 1) n γ n=2 n which is another representation of γ often used by Euler.

Cases when the series can be evaluated in explicit closed form are intertwined with similar cases when this is not possible (unless we accept integral form). Series of the form

∞ ζ ()n − 1 ∑ ,p = 1, 2 n=2 np+

3

have been studied and evaluated in a number of papers - see, for instance, [9], [12, pp. 213-219, (469), (474), 517), (518)] and also [13], [14]. At the same time the series

∞∞ζζ(nn+− 1) 1 (+− 1) 1 ∑∑, (− 1) n−1 nn=22nn=

resists evaluation in closed form in terms of recognized constants. However, they can be evaluated in integral form and we want to point out one simple integral which appears in several interesting cases as evidenced by Theorem 1 below. Namely, consider the constant

1 ψγ(t ++ 1) M = ∫ dt . 0 t

Theorem 1. With M as defined above,

∞ (−+ 1)n−1ζ (n 1) (a) ∑ = M n=1 n

∞ 11 (b) ∑ ln 1+=M n=1 nn

∞ ln (n + 1) (c) ∑ = M n=1 nn(+ 1)

∞

(d) ∑ Hnn (ζ (+−= 1) 1) M n=1

∞

(e) ∑ Hnn (ζγ()− 1) = M +− 1 n=2

∞ 1 (f) ∑ (n−ζζ(2) − (3) −− ... ζ (nM )) = n=1 n

(the first term in this series is just 1)

∞ − (g) ∑ Hnn (ζζ(+− 1) ( n + 2)) = M − ln 2 n=1

4

∞ Ein(x ) (h) dx= M ∫ x 0 e −1

where Ein(x ) is the modified , an

∞ (− 1) nn−1 x Ein(x ) = ∑ n=1 nn!

1 (1−−uu ) ln(1 ) (i) ∫ du= M 0 uuln

∞ 1 n n −k −1ζ += (j) ∑∑n ( 1) (kM 1) . nk=11n2 = k

Before proving the theorem we want to mention that the evaluations (a), (b), (c), and (i) are not new (for instance, they are listed on p. 142 in [3]). The series (c) appears in important applications in – see [5] and [8]. The entire paper [4] is dedicated to that series.

Proof. Starting from the well-known Taylor expansion for ψγ(1++x )

∞ (5) ∑(− 1)nn−1ζ (nx + 1) = ψγ (1 ++ x ) n=1

we divide both sides by x and integrate from 0 to 1 to prove (a). Then (a) implies (b)

n ∞(−+− 1)nn−−11ζ (n 1)∞ ( 1) ∞ 1 ∞∞ 1 (− 1)n−1 1∞ 1  1  = = = + ∑ ∑ ∑n+1 ∑∑  ∑ln  1 . n=1 nn=1 n k= 1 k kn= 11 k = nkk=1 k  k 

Next we prove (b) → (d)

n ∞ ∞ ∞11 ∞∞ 1 ζ +−= = ∑Hnnn( ( 1) 1) ∑ H ∑n+1 ∑∑H n  n=1 n=1 k = 2kkk kn= 21 = 

and using the generating functions for the harmonic numbers we continue this way

5

 ∞11−+ 1 ∞∞ 1 km1  1  =ln 1 −= ln =ln   =M . ∑1 ∑∑−− k=2k1−  k km= 21 k11  k= mm   k

Now starting from the first sum in the last equation we write

 ∞1111111−∞ ∞∞ M =ln 1 −=  −−=ln  1  ∑1 ∑−− ∑∑m k=2k1−  kkk=211  kkmkkm=21= k

∞∞11∞ 1 = = −ζζ − −− ζ ∑∑n ∑(nn(2) (3) ... ( )) nk=12n = kk(− 1) n=1 n by using the evaluation (see [7, p. 51])

∞ 1 =−ζζ − −− ζ ∑ n nn(2) (3) ... ( ) . k =2 kk(− 1)

Thus (f) is also proved. Equation (e) follows from (d) by writing

∞∞11∞ ∞ ∑∑Hnnn(ζ()1−=)  H−1 +( ζζ()1 n −=) ∑ Hm m( ( +−+ 1)1) ∑( ζ ()1n −) nn=22= nnm=1 n= 2 and then applying (3).

Next we prove (h). From the representation

1 ∞ xs−1 ζ ()s = dx, Re()1 s > ∫ x Γ−()se0 1 we find with sn= +1

∞∞(−+ 1)n−−11ζ (n 1)∞∞ ( − 1) nnx dxEin( x ) = = dx ∑∑∫∫xx. nn=11n 00= nn! e−− 11 e

The exponential integral has the representation

6

x 1− e−t Ein(x ) = ∫ dt 0 t

and integrating by parts we find

∞∞ −x ∞ Ein()x Ein()(1 xd− e ) ∞ dx = =Ein(x )ln(1 −− e−−xx ) Ein′ (x )ln(1 − e ) dx ∫∫xx− 0 ∫ 00ee−−11 0

∞ 1− e−x =−−∫ ln(1e−x ) dx . 0 x

∞ Here Ein(xe )ln(1−=−x ) 0 since Ein(x ) grows like ln x when x →∞ and also 0

− limxe ln(1− x )0= . In the last integral we make the substitution ue= −x to find x→0

∞ 1 −e−x 1 (1−−uu ) ln(1 ) −∫∫ln(1 −=e−x ) dx du 00x uuln

and (i) is proved.

We shall prove now the implication (a) → (g) by using Abel’s lemma for transformation of series (see, for example, [10, Exercise 10, p.78]):

Lemma 2. Let {}an and{bn }, np≥ , be two of complex numbers and let

Aaan= pp ++1 ++... a n. Then for every np> we have

nn−1 ∑∑abkk=+− bA n n A k() b k b k+1 . kp= kp=

(− 1) k −1 We take here p =1, a = , and bk=ζ ( +− 1) 1 . Then AH= − and we find k k k kk

nnk −1 −1 (− 1) −− ∑∑{ζ (k+−= 1) 1} Hnnk { ζ ( +−+ 1) 1} Hk { ζζ ( +− 1) ( k + 2)} . kk=11k =

7

− Setting n →∞ and using the estimate from lemma1 and also the fact that ||HHnn≤ and

Hnn  ln at infinity we come to the equation

∞∞k −1 (− 1) − ∑∑{ζ (k+−= 1) 1} Hkk { ζζ ( +− 1) ( k + 2)} . kk=11k =

According to (a) we have

∞ (− 1) k −1 ∑ {ζ (kM+−= 1) 1} − ln 2 k =1 k and (g) follows.

The implication (b) → (c) also follows from Abel’s lemma. We take p =1 and

11 akk=ln 1 + = ln(k +− 1) ln( kb ), =. kk

Here Ann =ln ( + 1) and lemma 2 provides the equation

nn1 1 ln(nk++ 1)−1 ln ( 1) ∑∑ln 1+= + kk=11k k n= kk(+ 1) which clearly shows the relation between (b) and (c).

Last we prove (j). The proof is based on Euler’s series transformation [2]. Given a power series f( x )=++ a01 ax ... we have for sufficiently small ||t

∞ n 1 t n n f= ∑∑ tak . 11−−ttnk=00 = k

We take fx( )=ψγ ( x ++ 1) with the expansion (5), where af0 =(0) = 0 . Using the substitution t x = we compute 1− t

1 ψγ(x ++ 1) 1/2 1 t dt ∫∫dx= f  00x 11−−t tt

8

1/2 ∞∞nnnndt 1  =tknk( −+= 1)−−11ζζ ( 1) ( −+ 1)k (k 1) ∫ ∑∑ ∑n ∑ 0 nk=11 = kktnn=12 k = 1

after integrating term by term.

It is easy to see that for n ≥1

n n ∞ 1− Lx () (− 1)k −1ζ (k += 1) n dx ∑ ∫ x k =1 k 0 e −1

where Lxn () are the Laguerre .

The proof of the theorem is completed.

1 Remark 1. Applying Lemma 2 for p =1, a = , and bk=ζ ( +− 1) 1 we find k k k

∞∞ζ (k +− 1) 1 ∑∑=Hkn{ζζ ( +− 1) ( k + 2)} . kk=11k =

At the same time from (1)

∞ ζ(kt+− 1) 11 1 −−γψ (2 − ) ∑ = ∫ dt . k =1 kt0

Remark 2. This is a note on the series in (f). Let

11 1∞ 1 =+++= =−−ζζ −− ζ Sn nn n... ∑ n nn(2) (3) ... ( ) . 2 32 43k =2 kk (− 1)

Obviously,

1 111 1n + 1

That is, we have the same estimate as in Lemma 1

1 11n + <

9

It is interesting to compare equation (g) from the above theorem with the following result.

Proposition 1. We have

∞ 2 − π (6) ∑ Hnnn (ζζ( )− ( + 1)) = −− γ ln 2 . n=2 6

(− 1) k −1 The proof comes from Abel’s lemma again, where we take p = 2 , a = and k k

− bkk =ζ () − 1. Then AHkk= −1 and the lemma yields for every n ≥ 3

nnk −1 −1 (− 1) −− ∑∑{ζ (k )−= 1} ( Hnk − 1){ζ ( n ) −+ 1} ( H − 1){ζζ ( kk ) − ( + 1)} . kk=22k =

From here with n →∞ we find

∞k −1 ∞∞ (− 1) − ∑ζ()k−+= ln21 ∑∑ Hkkk {()ζζ −+− ( 1)} {() ζζ kk −+ ( 1)}. k=2k kk= 22=

The first sum equals −γ and for the last sum we write

∞∞ ∑∑{ζζ (kk )− ( + 1)} = {( ζ ( k ) −− 1) ( ζ ( k +− 1) 1)} kk=22= which is a equal to its first term ζ (2)− 1. Thus

∞ − −γ −ln 2 += 1∑ Hkk {ζζ ( ) − ( k + 1)} − ζ (2) + 1 . k =2 and the proof is finished.

It was noted by the referee that another proof follows from the fact that the series in (6) has a telescoping property. This property brings to a convenient representation of its partial sum

mmn − ζζ(2) −−(−+ 1) (n 1) ∑∑Hnnn(ζζ( )− ( + 1) ) = − Hmm++11(ζ ( m +−− 1) 1) H + . nn=2221= n +

Setting here m →∞ and using equation (4) we come the desired evaluation.

10

It is also interesting to compare parts (a) and (c) of Theorem 1 to the result in the following proposition.

Proposition 2. For any p >1

∞∞log(n+ 1) ( −+ 1)k −1ζ (pk ) =−+ζ ′ ∑∑p ()p . nk=11nk=

Proof.

∞log(1+nn ) ∞ 1  1∞∞ log 1 1 = += + + ∑p ∑ plogn 1 ∑∑pplog 1 n=1n n= 1 n nnn=11 nn=  n

k ∞∞1 (−− 1)kk−−11 1 ∞( 1) ∞ 1 =−+ζζ′′=−+ ()pp∑∑p () ∑ ∑kp+ nk=11n = kn k=1k n= 1 n

∞ (−+ 1)k −1ζ (kp ) =−+ζ ′()p ∑ . k =1 k

The change of order of is justified by the of the series.

The next proposition parallels some of the results in Theorem 1.

Proposition 3. Let

2 ψγ(1++t ) = ≈ M1 ∫ dt 0.86062 . 1 t

Then

∞ 11 (k) ∑ ln 1+=M1 n=1 nn+1

∞ − (l) ∑ Hnn (ζ (+−= 1) 1) M1 n=1

∞ (− 1) n−1 (m) ∑ (n−ζζ(2) − (3) −− ... ζ (nM )) = 1 n=1 n

11

(the first term in the sum is 1).

Here (k) follows immediately from the well-known representation

ψγ(1++t ) ∞ 1 = ∑ tn=1 nn()+ t

by integration from 1 to 2 (integration between 0 and 1 implies (b) in Theorem 1). Independently (l) follows from Theorem 3 in Section 4 (see the remark at the end of that section). Further details are left to the reader.

3. Variations on a problem of Ovidiu Furdui

The identity (d) of Theorem 1 can be written in the form

∞

∑ Hnn (ζζ(+−= 1) 1) M − (2) + 1 n=2

by starting the summation from n = 2 . Subtracting this from (e) we find

∞ π 2 (7) ∑ Hnn [ζζ( )− ( n += 1)] − γ. n=2 6

This result was displayed in Problem W10 from [6]. We shall extend (7) to power series involving the Hurwitz zeta function

∞ 1 ζ= ζζ= (,sa )∑ s , (,1)s () s , k =0 ()ka+

where Re(s )> 1 and a > 0 .

Theorem 2. For a > 0 and ||xa< we have

∞ n 2 (8) ∑ Hnaxnaxn [ζζ(,)− ( + 1,)] = ζ (2,) ax + ψ () ax + log( Γ−− ax ) log() Γ a . n=2

Proof. The left hand side can be written this way

12

∞∞ ∞ ∞ n11n nn ∑∑Hnnζ(, nax ) =+= H−−11 ζ(, nax ) ∑ H n ζζ(, nax ) + ∑ (, nax ) nn=22= nnn=2 n= 2

∞ n+1 =∑ Hn ζψ( n + 1, ax ) + log Γ−− ( a x ) log Γ ( a ) + ( ax ) n=1

by using the well-known series from [3, p.78], or [12, p. 159]

∞ 1 (9) ∑ ζψ(nax , )n = log Γ−− ( a x ) log Γ ( a ) + ( ax ) n=2 n

for ||xa< . Then we separate the first term in the series

∞∞ nn++1 12 ∑∑Hnnζ( n+= 1,) ax H ζζ ( n ++ 1,) ax (2,) ax nn=12=

and from here

∞∞ nn+12 ∑∑Hnnζ(,) nax− H ζ ( n + 1,) ax = ζψ (2,) ax + () ax + log( Γ−− a x ) log() Γ a. nn=22=

Thus (8) is proved.

The convergence in these series is supported by the estimate given in the Appendix

11 11na+ +<ζ (,na ) ≤+  (a+ 1) nn a aan(+− 1) n n 1

( n ≥ 2 ) and by the slow growth of the harmonic numbers at infinity, Hnn  ln .

Corollary 1. For ||1x <

∞ n 2 (10) ∑ Hn [ζζ( n )− xn ( + 1)] x = ζ (2) x − γ x + log Γ− (1 x ) , n=2

and for ||2x <

∞ n 22 (11) ∑ Hn [ζζ( n )− xn ( ++− 1) x 1] x = ζ (2) x − x +− (1 γ ) x + log Γ− (2 x ) . n=2

13

Proof. Setting a =1 in (8) we have ψγ(1) = − and (10) follows. With a = 2 in (8) we have

ζζ(nn ,2)= ( ) − 1, ζζ(1,2)(1)1nn+ = +−, and ψγ(2)= 1 − . Thus (8) turns into (11).

With x =1 (11) becomes (7). Setting x = −1 in (11) we find also

∞ n (12) ∑(− 1)Hnn [ζζ ( ) + ( n + 1) − 2] = ζγ (2) +−+ 2 log 2 . n=2

The series in (8) can be regularized, modified to extend the interval of convergence.

Corollary 2. For every a > 0 and every ||xa<+ 1

∞ ax− ζζ− +− n (13) ∑ Hn (,) na x ( n 1,) an+1 x n=2 a

()a− xx =ζψ(2,ax )2 + ( ax ) + +log Γ+−− (a 1 x ) log Γ+ ( a 1) . a2

Proof. For ||xa< we write using the generating function for the harmonic numbers

log(ax− ) 1   x log(ax−=− ) ( ax ) =−( ax )  loga+ log  1 − ax−− ax  a

−1 logax 1  x =−(ax ) +− 1 log 1 − axa−  a a

∞ Hxn x∞ () ax− = −− n = −− − n loga ( ax ) ∑ n+1 loga ( ax ) 21∑ Hn n+ x. n=1 a aan=2

This result can be put in the form

∞ ()ax− x − n = −− +− ∑ Hn n+12 xlog( ax ) log a ( ax ) . n=2 aa

Now we add this equation to equation (8). On the left hand side we unite the two sums into one. On the right hand side we write log(ax−+ ) log Γ− ( ax ) = log Γ+− ( a 1 x ) and

14 logaa+ log Γ ( ) = log Γ+ ( a 1) . This way from (8) we obtain (13). The right side in (13) is obviously defined for ||xa<+ 1. To see that the series on the left side converges in this interval we write it in the form

∞ ax− ζζ− +− n ∑ Hn (,) na x ( n 1,) an+1 x n=2 a

∞∞11   =ζζ −nn − +− +1 ∑∑Hnaxnn(,) nn  Hna( 1,) +1  x. nn=22aa = 

Here both series converge for ||xa<+ 1 in view of the estimate in the Appendix

1 11ma+ <ζ (,)ma −≤  (a+ 1) maa mm(+− 1) m 1 true for every m ≥ 2. The proof is completed.

4. Two interesting generating functions

The series in (10) is a power series, the difference of the two powers series with terms

n n+1 Hn ζ () nx and Hnxn ζ (+ 1) .

Although the difference of these two series can be evaluated in closed form, evaluating each one separately brings to difficult . We shall demonstrate this in the following theorem. For

− comparison we also include the series with skew-harmonic numbers Hn .

Theorem 3. With a > 0 and ||xa<

∞∞(− 1)k −1ζ (k + 1, ax − )x ψψ ( axt −+ ) − ( ax − ) ζ +=nk= (14) ∑∑Hn ( n 1, a ) x x ∫ dt nk=11= kt0

∞∞(− 1)kk−1 (2 − 1)ζ (k + 1, ax − )x ψψ ( ax −+ 2 t ) − ( axt −+ ) − ζ +=nk= (15) ∑∑Hn ( n 1, a ) x x ∫ dt . nk=11= kt0

15

The proof is based on the lemma:

2 n Lemma 3. Let f() t= at12 + at ++... atn + ... be a power series. Then the following representations hold

∞∞k −1 n (− 1) kk() (16) ∑∑Haxnn = x f() x nk=11= kk!

∞∞kk−1 − n (−− 1) (2 1) kk() (17) ∑∑Hnn ax = x f() x. nk=11= kk!

Proof. The harmonic numbers have the well-known integral representation

1 t n −1 = Hn ∫ dt . 0 t −1

n Multiplying both sides here by axn and summing for n =1,2,... we find for x small enough

∞ 1 fxt( )− fx () n = ∑ Hnn a x ∫ dt n=1 0 t −1

(cf. [10, Exercise 20, p.79]). Now Taylor’s formula applied to the function g() t= f ( xt ) and centered at t =1 gives ∞ xfkk()() x fxt()()=+− fx ∑ (t 1) k . k =1 k! Substituting this into the integral and integrating term by term we arrive at (16).

The proof of (17) is done the same way by using the representations ∞ 1 f() x−− f ( xt ) − n = ∑ Hnn a x ∫ dt n=1 0 t +1 and ∞ (− 1)kkxf() k ( x ) f()()−= xt f x +∑ (t + 1) k . k =1 k! Proof of the theorem. We apply the lemma to the function ([12, p.159])

16

∞ (18) ft( )=∑ζ ( n + 1, at )n =−−+ ψψ ( a t ) ( a ), | t | < | a | . n=1 Here f()k( x )=− ( 1) kk+1ψζ () (ax −= ) k ! ( k + 1, ax − ) and from the lemma

∞∞k −1 nk(− 1)ζ (k +− 1, ax ) (19) ∑∑Hn ζ ( n+= 1, ax ) x. nk=11= k

Now we write (18) in the form

∞ ∑(− 1)kk−1ζ (k + 1, at ) = ψψ ( a +− t ) ( a ), | t | < | a | . k =1

Dividing by t and integrating between 0 and x we find

∞ (− 1)k −1ζ (k + 1, a )x ψψ ( at+− ) ( a ) ∑ xk = ∫ dt . k =1 kt0

Replacing here a by ax− gives

∞ (− 1)k −1ζ (k + 1, ax − )x ψψ ( axt −+ ) − ( ax − ) (20) ∑ xk = ∫ dt . k =1 kt0

Now (19) and (20) bring to (14). The proof of (15) is left to the reader.

Remark 3. By setting a = 2 and x =1 in (14) we obtain

∞∞(− 1)k −1ζ (kt + 1)1 ψγ (1 ++ ) ζ +−= = ∑∑Hn [ ( n 1) 1] ∫ dt nk=11= kt0 that is, a new proof of equations (a) and (d) in Theorem 1, as ζζ(1,2)(1)1nn+ = +−. With ax=2, = 1 in (15) we find

∞ 12ψ(1+ 2tt ) − ψ (1 + ) ψγ (1 ++ t ) − ζ +−= = (21) ∑ Hn ( ( n 1) 1) ∫∫dt dt n=1 01tt which is (l) in Proposition 3. To show the equality of the above two integrals we write

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1ψ(1+− 2tt ) ψ (1 + ) 11 ψ (1 ++ 2 t ) γψ (1 ++ t ) γ ∫dt = ∫∫dt − dt 0t 00 tt and then in the first integral on the right hand side we make the substitution ut= 2 .

Appendix

Proof of Goldbach’s theorem. The theorem has a simple two-line proof

 n ∞ ∞∞1 ∞∞ 1∞ 11 {ζ (n )−= 1}   =  =  ∑ ∑∑n ∑∑ ∑ 2 1 n=2 nk=22 = k kn= 22 =  kkk=21− k

∞∞1 11 =∑∑  = −= 1 kk=22kk(−− 1)=  k 1 k

where the last sum is a well-known telescoping series. Proof of Lemma 1 (extended version). Let as>>0, 1 . Then

1 11sa+ <ζ (,sa ) −≤ . (a+ 1) s aa ss(+− 1) s 1 Furdui’s proof from [7, p.51] is modified here for the Hurwitz zeta function. The left hand side inequality is obvious and for the right hand side inequality we write using the reminder estimate from the integral test for series

∞ 11∞ 1 ≤=dt ∑ ss∫ s−1 . k =2 (ka+ )1 ( ta + ) ( s −+ 1)( a 1) Therefore, 11∞ 1 1 1 1sa+ ζ −= + ≤ + = (,sa ) ss∑ ss s−1 s. a( a+ 1)k =2 ( ka + ) ( a + 1) ( s −+ 1)( a 1) ( a + 1) s − 1 The proof is completed.

References

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[11] Mits Kobayashi, A dissection proof of Euler's series for 1-γ , Math. Mag., 90 (2017), 360-364.

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