on the asymptotic behavior of the remainder of a dirichlet series absolutely convergent in a...

Ukrainian Mathematical Journal, Vol. 55, No. 3, 2003

ON THE ASYMPTOTIC BEHAVIOR OF THE REMAINDER OFA DIRICHLET SERIES ABSOLUTELY CONVERGENT IN A HALF-PLANE

L. Ya. Mykytyuk and M. M. Sheremeta UDC 517.537.72

For a Dirichlet series a sn nnexp{ }λ

=

∞∑ 1 with nonnegative exponents and zero abscissa of ab-

solute convergence, we study the asymptotic behavior of the remainder akk n k=

∞∑ exp{ }δλ ,

δ < 0, as n → ∞.

1. Introduction

Let λ = ( )λn n=∞

1 be a sequence of nonnegative numbers increasing to + ∞ and let a Dirichlet series

F ( s ) = n

n na s=

∞

∑ { }1

exp λ , s = σ + i t, (1)

have the abscissa of absolute convergence σ a. = 0. Let Π k ( λ ) denote the class of exponential polynomials of

the form n

kn na s=∑ { }

1exp λ . For δ < 0, we set E n ( F, δ ) = inf { || F – P || δ : P ∈ Π k ( λ ) }, where || F – P || δ =

sup ( ) ( ){ + − +F it P itδ δ : t ∈ }R . Then

| a n + 1 | exp { δ λ n + 1 } ≤ E n ( F, δ ) ≤ k n

k ka= +

∞

∑ { }1

exp δλ . (2)

Inequalities (2) were proved in [1]; for convenience, we present their proof in Sec. 2. On the basis of these in-equalities and the scale of growth, a relationship between the growth of the maximum term of series (1) and thebehavior of E n ( F, δ ) was established in [1, 2]. In the present paper, without using the scale of growth, we de-

scribe a relationship between the asymptotic behavior of the remainder R n ( F, δ ) : = k n k ka=∞∑ { }exp δλ of

the Dirichlet series (1) and the asymptotic behavior of its coefficients a n ; then, using (2), we establish a rela-

tionship between the asymptotic behavior of E n ( F, δ ) and the asymptotic behavior of a n .

In what follows, we assume that ln n = o ( λ n ) as n → ∞ . Note that if lim supn na→∞ = + ∞, then the

coefficients an0 of the Newton majorant of the Dirichlet series (1) satisfy the following condition [3]:

κn F0( ) : = ln lna an n

n n

01

0

1

−−

+

+λ λ ç 0 as n → ∞ .

If the Dirichlet series coincides with its Newton majorant, then

Lviv University, Lviv. Translated from Ukrains’kyi Matematychnyi Zhurnal, Vol. 55, No. 3, pp. 379–388, March, 2003. Original arti-cle submitted June 23, 2001.

456 0041–5995/03/5503–0456 $25.00 © 2003 Plenum Publishing Corporation

ON THE ASYMPTOTIC BEHAVIOR OF THE REMAINDER OF A DIRICHLET SERIES ABSOLUTELY CONVERGENT IN A HALF-PLANE 457

κ n ( F ) : = ln lna an n

n n

−−

+

+

1

1λ λ ç 0 as n → ∞.

We consider a somewhat broader class of Dirichlet series close to their Newton majorants, i.e., such that

| a n | → ∞ as n → ∞ and

κ : = lim infln ln

n

n n

n n

a a

→ ∞+

+

−−

> − ∞1

1λ λ. (3)

It follows from the condition | a n | → ∞ as n → ∞ that κ ≤ 0. The following theorem is true:

Theorem 1. Suppose that the coefficients of a Dirichlet series with zero abscissa of absolute convergence

satisfy condition (3), | a n | → ∞ as n → ∞, and the exponents of this series satisfy the condition

h : = lim infn

n n→ ∞

+ −( ) >λ λ1 0. (4)

Then, for every δ < κ,

1 ≤ lim inf( , )

n

n

n

R F e

a

n

→ ∞

δ δ λ ≤ lim sup

( , )

n

n

n

R F e

a

n

→ ∞

δ δ λ ≤ 1 +

11exp ( )− −{ } −δ κ h. (5)

Theorem 1 yields the following corollary:

Corollary 1. If λ n + 1 – λ n → ∞ and | a n | → ∞ as n → ∞ and

ln lna an n

n n

−−

+

+

1

1λ λ → 0, n → ∞, (6)

then, for every δ < 0,

lim( , )

n

n

n

R F e

a

n

→ ∞=δ δ λ

1. (7)

The following two corollaries are also true:

Corollary 2. Suppose that | a n | ç ∞ and κ n ( F ) → 0 as n → ∞ . In order that equality (7) be true

for every δ < 0, it is necessary and sufficient that λ n + 1 – λ n → ∞ as n → ∞.

Corollary 3. If lim supn na→ ∞ = + ∞ and λ n + 1 – λ n → ∞ as n → ∞, then, for every δ < 0,

lim inf( , )

n

n

n

R F e

a

n

→ ∞=δ δ λ

1. (8)

458 L. YA. MYKYTYUK AND M. M. SHEREMETA

Let us compare ln ( , )R F ennδ δ λ( ) and ln | a n | . Theorem 1 yields the following corollary:

Corollary 4. If | a n | → ∞ as n → ∞ , condition (6) is satisfied, and λ n + 1 – λ n ≥ h > 0, n ≥ 0, then,

for every δ < 0,

limln ( , )

lnn

n

n

R F e

a

n

→ ∞

( )=

δ δ λ

1. (9)

In this corollary, the condition of the positivity of the step of the sequence ( )λn n=∞

1 is not necessary, whichfollows from Theorem 2.

For t ≥ 0, we denote

∆ n ( t ) : = t tn≤ < +

∑λ 1

1.

Theorem 2. Suppose that the coefficients of a Dirichlet series with zero abscissa of absolute convergence

satisfy condition (6) and | a n | → ∞ as n → ∞ . In order that equality (9) be true for every δ < 0, it is nec-

essary and sufficient that

limln ( )

lnn

n

n

n

a→ ∞=∆ λ

0. (10)

Let S0*( )λ denote the class of Dirichlet series (1) absolutely convergent in { s : Re s < 0 }, having a given

sequence of exponents λ = ( )λn n=∞

1, satisfying condition (6), and such that | a n | → ∞ as n → ∞.

Theorem 3. In order that equality (9) be true for every function F ∈ S0*( )λ , it is necessary and suffi-

cient that sup { ∆ n ( t ) : t ≥ 0 } < + ∞.

Let S0**( )λ denote the class of Dirichlet series (1) absolutely convergent in { s : Re s < 0 }, having a given

sequence of exponents λ = ( )λn n=∞

1, and such that lim supn na→ ∞ = ∞.

Theorem 4. In order that, for every function F ∈ S0**( )λ ,

lim infln ( , )

lnn

n

n

R F e

a

n

→ ∞

( )=

δ δ λ

1, (11)

it is necessary and sufficient that sup { ∆ n ( t ) : t ≥ 0 } < + ∞.

The following two statements complement Theorems 2 – 4 in the case where condition (6) may not be satis-fied:

Proposition 1. Suppose that ψ is a positive continuously differentiable function increasing to + ∞ o n

[ 0, + ∞ ) and such that x / ψ ( x ) ç + ∞ a s x → + ∞. If ln | a n | = ( 1 + o ( 1 ) ) λ n / ψ ( λ n ) and ln n =

o ( λ n / ψ ( λ n ) ) as n → + ∞, then equality (9) is satisfied.


Proposition 2. Suppose that ψ is a positive continuously differentiable function increasing to + ∞ o n

[ 0, + ∞ ) and such that x / ψ ( x ) ç + ∞ as x → + ∞. If ln n ( t ) ≤ t / ψ ( t ), t ≥ t0, and λ n / ψ ( λn ) ln | an | → 0

as n → ∞, then equality (11) is satisfied.

2. Proof of Inequalities (2)

Since

lim expT

T

T

Tixt dt

→ + ∞−∫ { }1

2 = 0

for arbitrary x ∈ R \ { 0 }, we conclude that, for every exponential polynomial P ∈ Π k ( λ ) with coefficients bj

and arbitrary β ∈ R and n > k, the following relation is true:

lim ( )expT

T

T

nTP it it dt

→ + ∞−

+∫ + −{ }12 1β λ =

j

k

j jT

T

T

j nbT

it dt= → + ∞

−+∑ ∫{ } −{ }

01

12

exp lim exp ( )βλ λ λ = 0.

Therefore, denoting

Sm ( s ) = k

m

k ka s=∑ { }

0

exp λ ,

for any n < m and P ∈ Π n ( λ ) we get

lim ( ) ( ) expT

T

T

m nTS it P it it dt

→ +∞−

+∫ + − +( ) −{ }12 1δ δ λ

= lim exp ( )T

T

T

k n

m

k k k nTa it dt

→ +∞− = +

+∫ ∑ + −{ }12

11δλ λ λ

= lim exp expT

T

T

n n n nTa dt a

→ +∞−

+ + + +∫ { } = { }12 1 1 1 1δλ δλ ,

whence | a n + 1 | exp { δ λ n + 1 } ≤ || S m – P || δ . Since δ < 0, the sequence ( )Sm m=∞

1 uniformly converges to F in

the half-plane { s : Re s ≤ δ }. Hence, || S m – P || δ → || F – P || δ as m → ∞. Thus, by virtue of the arbitrariness

of m, it follows from the last inequality that | a n + 1 | exp { δ λ n + 1 } ≤ || F – P || δ for every exponential polyno-

mial P ∈ Π n ( λ ), which yields the first inequality in (2).

Let Sn be a partial sum of series (1). Then


E n ( F, β ) ≤ || F – S n || δ = max exp :k n

k k ka it t= +

∞

∑ +{ } ∈

1

δλ λ R ≤ k n

k ka= +

∞

∑ { }1

exp δλ ,

i.e., we arrive at the second inequality in (2).

3. Proof of Theorem 1 and Corollaries 2 and 3

Let δ < κ. By virtue of conditions (3) and (4), for any κ* ∈ ( δ, κ ) , h* ∈ ( 0, h ) , and n ≥ n 0 = n 0( κ*, h* ),

we have ln | a n + 1 | – ln | a n | ≤ – κ *

( λ n + 1 – λ n ) and λ n + 1 – λ n ≥ h*. Therefore, for n ≥ n 0, we get

R n ( F, δ ) = a ea

ank n

k

nk n

nδ λ δ λ λ11

+ −{ }

= +

∞

∑ exp ( )

= a e a ank n

k n k nnδ λ δ λ λ1

1

+ − + −{ }

= +

∞

∑ exp ln ln ( )

= a e a ank n j n

k

j j j jnδ λ δ λ λ1

1

1

1 1+ − + −

= +

∞

=

−

+ +∑ ∑exp ln ln ( )

≤ a enk n j n

k

j jnδ λ δ κ λ λ1

1

1

1+ − −

= +

∞

=

−

+∑ ∑exp ( )( )*

≤ a e h k nnk n

nδ λ δ κ11

+ − −{ }

= +

∞

∑ exp ( ) ( )* *

= a e h knk

nδ λ δ κ11

+ −{ }

=

∞

∑ exp ( )* *

= a eh

nnδ λ

δ κ1

1

1+

− −{ } −

exp ( )* *

.

Since a ennδ λ < R n ( F, δ ), one can easily obtain the following inequalities:

1 ≤ lim inf( , )

lim sup( , )

exp ( )* *n

n

n n

n

n

R F e

a

R F e

a h

n n

→ ∞ → ∞≤ ≤ +

− −{ } −δ δ

δ κ

δ λ δ λ1 1

1.

Since the numbers κ* and h* are arbitrary, we obtain inequalities (5). Theorem 1 is proved.

Let us prove Corollary 2. The sufficiency of the condition that λ n + 1 – λ n → ∞ as n → ∞ follows from

Corollary 1. If λn j +1 – λn j ≤ H < + ∞ for an increasing sequence ( )nj j=

∞1 of natural numbers, then, with re-

gard for the inequality | a n | ≤ | a n + 1 | , we get


R F a an n n n n nj j j j j j( , )exp expδ δ λ δ λ λ{ } ≥ + −( ){ }+ +1 1 ≥ a Hnj

1 + { }( )exp δ ,

i.e., relation (7) is not satisfied.

Finally, we prove Corollary 3. Let an0 be the coefficients of the Newton majorant of the Dirichlet series

(1). Then an0 → + ∞ as n → ∞, an ≤ an

0 for all n, ank = ank

0 for a certain increasing sequence ( )nk k=∞

1

of natural numbers, and κn F0( ) ç 0 as n → ∞ [3]. Therefore, setting Rn0( )δ =

k n k ka=∞∑ { }0 exp δλ and us-

ing Corollary 2, we get R enn0( )δ δ λ = 1 1 0+( )o an( ) as n → ∞. Since R n ( F, δ ) ≤ Rn

0( )δ and ank = ank

0 , we

conclude that R Fn nk k( , )expδ δ λ{ } ≤ 1 1+( )o ank

( ) as k → ∞ , which, in view of the inequality an <

R n ( F, δ )e nδ λ , yields relation (8).

4. Proof of Theorem 2

First, note that, according to condition (6), for every ε > 0 and all n ≥ n 0 = n 0 ( ε ) we have an > e,ln lna an n− +1 < ε ( λ n + 1 – λ n ) , and, hence,

ln lna an p n p− < −ε λ λ , n ≥ n 0 , p ≥ n 0 . (12)

If | λ n – λ p | ≤ 1, then, using inequality (12), we get

| a n | 1

–

ε ≤ | a p | ≤ | a n |

1 +

ε, n ≥ n 0 , p ≥ n 0 . (13)

Therefore,

R n ( F, δ ) ≥ λ λ λ

δ λ ε δ λ λn p n

p na e a e np n n≤ < +

− +∑ ≥1

1 1( ) ( )∆ , n ≥ n 0 ,

whence

ln ( )ln ln

ln ( , )

ln∆n

a a

R F e

an

n n

n

n

nλ ε δ δ δ λ

+ − + ≤( )

1 . (14)

The necessity of condition (10) follows from inequality (14). Taking relations (9) and (14) into account,we get

lim supln ( )

lnn

n

n

n

a→∞≤∆ λ ε,

i.e., by virtue of the arbitrariness of ε, relation (10) is satisfied.

Let us prove sufficiency. We choose a subsequence ( )λn kk =∞

1 of the sequence ( )λn n=∞

1 so that the seg-

ments [ ),λ λn nk k+1 are pairwise disjoint and their union contains the entire set { λ n , n ≥ 1 }.


Let us show that

limln ( , )

lnk

n

n

R F e

ak

nk

k→∞

( )=

δ δ λ

1. (15)

Let n 0 = n 0 ( ε ) be such that condition (13) is satisfied, provided that | λ n – λ p | ≤ 1; in view of (10), we have

∆ n ( λ n ) ≤ | a n | ε. Then

λ λ λ

δλ ε δ λ ε δ λλnk p nk

p

k k

nkk

nka e n a e a ep n n n≤ < +

+ +∑ ≤ ( ) ≤1

1 1 2∆ , k ≥ k 0

,

whence

R F a e anj k

pj k

n nk

n j p n j

p

j j( , ) expδ δλ

λ λ λ

δλ ε= ≤ { }

=

∞

≤ < + =

∞ +∑ ∑ ∑1

1 2.

Since condition (6) implies that [see (12)]

ln lna an n

n n

j j

j j

−

−+

+

1

1λ λ

→ 0, j → ∞,

and λ λn nj j+−

1 ≥ 1, by virtue of Theorem 1 we get

lim supln ( , )exp

lnk

n n

n

R F

ak k

k→∞ +

{ }( )≤

δ δλε1 2 1,

which, by virtue of the arbitrariness of ε, yields relation (15). Now let n ∈ N be an arbitrary number and let λnk

≤ λ n < λnk + 1 for certain k. Then

R F e e a e e a enj n

jj n

jn n j n

k

j( , )δ δ λ δ λ δλ δ λ δλ= ≤=

∞

=

∞

∑ ∑

≤ e e a e e R F enk

k

j

k

nk

j nj n

δ δ λ δλ δ δ λδ=

∞

∑ = ( , ) ,

whence

ln ( , )expln ln

ln ( , )exp

ln

ln

lnR F

a a

R F

a

a

an n

n n

n n

n

n

n

k k

k

kδ δ λ δ δ δ λ{ }( ) ≤ +{ }( )

.


Taking relations (13) and (15) into account and using the last inequality, we get

lim supln ( , )exp

lnn

n n

n

R F

a→∞

{ }( ) ≤ +δ δ λ ε1 ,

which, by virtue of the arbitrariness of ε and the inequality R Fn n( , )expδ δ λ{ } > an , yields relation (9).Theorem 2 is proved.


If sup { ∆ n ( t ) : t ≥ 0 } < + ∞, then condition (10) is satisfied and, by virtue of Theorem 2, equality (9) istrue.

We now assume that sup { ∆ n ( t ) : t ≥ 0 } = + ∞ and show that there exists a function F ∈ S0*( )λ for which

relation (9) is not satisfied. Since sup { ∆ n ( t ) : t ≥ 0 } = + ∞, there exists a subsequence λn kk( ) =∞

1 of the

sequence λn n( ) =∞

1 such that ∆ n nkλ( ) ç + ∞ as k → + ∞. Let n 0 = 1.

Assume that bk = exp ln ∆ n nkλ( ){ } , νk = min { 1, λ k–1 – λ k }, and an n

*( ) =∞

1 is a nondecreasing se-

quence tending to ∞ and such that

a1* = 1, an+1

* = ( ) *1 + α νn n na , n ≥ 1, (16)

an* ≤ bk

, n k – 1 < n ≤ n k , (17)

where each term of the sequence αn n( ) =∞

1 is equal to either 0 or 1.

Let us show that the sequence an n*( ) =

∞1 exists. For this purpose, we use the principle of recursive definition

of a sequence (see, e.g., [4, p. 11]). We set α 1 = 0 and assume that α 1 , … , α m are already determined. Thus,

an* are determined for all n = 1, 2, … , m + 1 and an

* ≤ bk for n ∈ ( n k – 1 , n k ] I ( 1 , m + 1 ] . We assume that

n p – 1 ≤ m + 1 < n p for certain p and set

α m + 1 = 0 1

1 1

1 1

1 1

if

if

a b

a b

m m p

m m p

+ +

+ +

+ >

+ ≤

*

*

( ) ,

( ) ,

ν

ν(18)

a am m m m+ + + += +2 1 1 11* *( )α ν .

According to the principle of recursive definition of a sequence, both sequences αn n( ) =∞

1 and an n*( ) =

∞1 are

uniquely defined and relations (16) and (17) are true. It is clear that an n*( ) =

∞1 is a nondecreasing sequence.

Let us show that an* → ∞ as n → ∞. Assume the contrary, i.e., let sup an

* = K < + ∞ and let k 0 be suchthat bk > 2 K for k ≥ k 0 . Then α n = 1 for n ≥ nk0

because if α m + 1 = 0 for certain m ≥ nk0 – 1 and n p – 1 ≤

m + 1 < n p , then relations (16) and (18) yield 2 K < bp < am m+ ++1 11* ( )ν ≤ 2 am+1* ≤ 2 K, which is impossible.

Taking into account that α n = 1 for n ≥ nk0, we obtain


a a Knk n

n

k nk

k

0

0

1

1* *( )=

−

∏ + = ≤ν , n ≥ nk0,

i.e., k k=∞∏ +

11( )ν < + ∞ and, hence,

k k k=∞

+∑ −1 1( )λ λ < + ∞, which is impossible.

Thus, we have constructed a nondecreasing sequence an n*( ) =

∞1 tending to ∞ and satisfying conditions (16)

and (17).

Since an* ç + ∞ as n → ∞, we have

n n n=∞∏ +

11( )α ν = + ∞ and

n n n=∞∑ 1

α ν = + ∞, i.e., there exists a

sequence ηn n( ) =∞

1 such that 1 ≥ η n é 0 as n → ∞ and n n n n=∞∑ 1

η α ν = + ∞.

We set α 1 = 1 and α n + 1 = ( )11

+=∏ η α νk k kk

n for n ≥ 1. Then a n ≤ an

* for n ≥ 1, a n → ∞ as n → ∞ ,

and

0 ≤ ln ln ln( )a an n

n n

n n n

n n

n n n

n nn

+

+ + +

−−

= +−

≤−

≤1

1 1 1

1λ λ

η α νλ λ

η α νλ λ

η → 0, n → ∞,

i.e., condition (6) is satisfied and, by virtue of the Stolz theorem, ln an nλ → 0 as n → ∞ . Since ln n =o ( λ n ) as n → ∞, by virtue of the Valiron theorem the Dirichlet series with the coefficients a n thus definedhas the zero abscissa of absolute convergence.

It follows from (16) that

lim supln ( )

lnlim sup

ln

lnn

n

n k

n

n

n

a

n

ak

k→∞ →∞

≥( )∆ ∆λ λ

≥ lim supln

lnlim sup

ln

ln*k

n

n k

n

k

n

a

n

bk

k

k

→∞ →∞

( )≥

( )∆ ∆λ λ = + ∞.

By virtue of Theorem 2, equality (9) is not satisfied for this Dirichlet series. Theorem 3 is proved.


Let an0 be the coefficients of the Newton majorant F

0 of the Dirichlet series (1). Then an0 → + ∞ as

n → ∞, an ≤ an0 for all n, ank

= ank

0 for a certain increasing sequence nk k( ) =∞

1 of natural numbers, and

κn F0( ) ç 0 as n → ∞, i.e., the coefficients of the Newton majorant F 0 satisfy condition (6) and F

0 ∈ S0*( )λ .

In this case, if sup { ∆ n ( t ) : t ≥ 0 } < + ∞ and Rn0( )δ : =

k n k ka=∞∑ { }0 exp δλ , then, by virtue of Theorem 3,

limln ( , )exp

lnn

n n

n

R F

a→∞

{ }( )=

0

0 1δ δ λ

.

Since R n ( F, δ ) ≤ Rn0( )δ and ank

= ank

0 , we conclude that ln ( , )expR Fn nk kδ δ λ{ }( ) ≤ ( 1 + o ( 1 ) ) ln ank

as k → ∞ , which, with regard for the inequality an < R n ( F, δ )e nδ λ , yields relation (11). Sufficiency isproved.

If sup { ∆ n ( t ) : t ≥ 0 } = + ∞, then there exists a subsequence λn kk( ) =∞

1 of the sequence λn n( ) =

∞1 such that

λnk +1 > λnk

+ 1 and ∆ n nkλ( ) ç ∞ as k → ∞.


We set ank = ∆ n nk

λ( ) and a j = 1 for j ≠ n k . Then limsup j ja→∞ = + ∞ and, taking into account

that

ln lna n n o on n n n nk k k k k= ( ) ≤ +( ) = +( ) = ( )1

212

1 1∆ λ λ λ λ , k → ∞,

where n ( t ) = λn t≤∑ 1 is the counting function of the sequence λn n( ) =∞

1, we conclude that, by virtue of the

Valiron theorem, the Dirichlet series (1) with the coefficients thus chosen has the zero abscissa of absolute con-

vergence and belongs to the class S0**( )λ . It is clear that ln ( , )e R Fj

jδ λ δ( ) ≥ 2 ln a j for j ≠ n k . For j = n k,

we get

ln ( , ) lne R F e a en

k

n

n j n

jk k

k k

n jδ λ δ λ

λ λ λ

δλδ( ) ≥

≤ < +∑

1

≥ ln e e an n

n j n

k k

k k

jδ λ δ λ

λ λ λ

+( )≤ < +

∑

1

1

= δ + ln a nn nk k+ ( ) −( )∆ λ 1

= δ + ln ∆ ∆n nn nk kλ λ( ) + ( ) −( )1

= 1 1+( ) ( )o n nk( ) ln ∆ λ = 2 1 1+( )o ank

( ) ln , k → ∞,

i.e., relation (11) is not satisfied, which proves Theorem 4.

7. Proof of Propositions 1 and 2

We begin with Proposition 1. We have

R n ( F, δ ) ≤ k n

k

kk

n

t

tt dn t

=

∞ ∞

∑ ∫+ +

≤ + +

exp( )

( )exp

( )( )

( )1 1ε λψ λ

δλ εψ

δλ

≤ λ

εψ

δ δψ

n

n tt

tt

t

tdt

∞

∫ + +

−

′

( )exp

( )( ) ( )

1

≤ δ εψ

δ δ εψ

δλ λn n

t

tn t t dt

t

tt dt

∞ ∞

∫ ∫+ + +

≤ + +

exp( )

( )ln ( ) exp

( )( )

1 1 2.

However, according to the l’Hospital rule, we have


lim sup

exp ( )( )

exp ( )( )

x

x

tt

t dt

xx

x→ + ∞

∞

∫ + +

+ +

δ εψ

δ

εψ

δ

1 2

1 2 ≤ lim sup

exp ( )( )

exp ( )( )

( )( )

x

xx

x

xx

x xx

→ + ∞

− + +

+ +

− + +

′

δ εψ

δ

εψ

δ δ εψ

1 2

1 2 1 2

= lim sup

( )( ) ( )

( )

lim sup

( )x x

xx x

x x→ + ∞ → + ∞− + + + ′ ≤

− +δ

δ εψ

ε ψψ

δ

δ εψ

1 2 1 2 1 22

= 1.

Therefore,

R n ( F, δ ) ≤ exp( )

( )exp ( ) ln

1 31 4

+ +

≤ +{ }ε λψ λ

δλ εδλn

nn ne an , n ≥ n 0 ( ε ) .

Using this inequality, one can easily obtain relation (9).

We prove Proposition 2. Let an0 be the coefficients of the Newton majorant of the Dirichlet series (1) and

let Rn0( )δ =

k n k ka=∞∑ { }0 exp δλ . Since, for k > n,

ln ln ln ln` `

a a a ak nj n

k

j jj n

k

j j j n k n0 0

1

10 0

10

10− = −( ) = − −( ) ≤ − −( )

=

−

+=

−

+∑ ∑ κ λ λ κ λ λ ,

for all sufficiently large n we get

Rn0( )δ = a a an

k nk n k

0 0 0

=

∞

∑ − +{ }exp ln ln δλ ≤ ank n

n k n k0 0

=

∞∑ − −( ) +{ }exp κ λ λ δλ

= an n nk n

n k0 0 0exp expκ λ δ κ λ{ } −( ){ }

=

∞

∑ = a t dn tn n n n

n

0 0 0exp exp ( )κ λ δ κλ

{ } −( ){ }∞

∫

≤ a n t t dtn n n n n

n

0 0 0 0exp ( )expκ λ κ δ δ κλ

{ } −( ) −( ){ }∞

∫

≤ at

tt dtn n n n n

n

0 0 0 0exp exp( )

κ λ κ δψ

δ κλ

{ } −( ) + −( )

∞

∫ .

As in the proof of Proposition 1, using the l’Hospital rule we get

κ δψ

δ κλ

n n

n

t

tt dt0 0−( ) + −( )

∞

∫ exp( )

≤ 1 1 0+( ) + −( )

o n

nn n( ) exp

( )λ

ψ λδ κ λ , n → ∞,


whence

Rn0( )δ ≤ 1 1 0+( )

o a enn

n

n( ) exp( )

δλ λψ λ

, n → ∞.

Taking the inequality an ≤ an0 into account, we obtain

ln ( , ) ln ( )R F e R en nn nδ δδ λ δ λ( ) ≤ ( )0 ≤ ln

( )( ) ( ) lna o o an

n

nn

0 01 1 1+ + = +( )λψ λ

, n → ∞.

Since ank = ank

0 for an increasing sequence nk k( ) =∞

1 of natural numbers, we get

ln ( , ) ( ) lnR F e o an nk

nkk

δ δ λ( ) ≤ +( )1 1 0 , k → ∞.

Using the inequality an ≤ R F enn( , )δ δ λ , we obtain relation (11).

8. Remark

By virtue of (2), we have an ≤ E F enn

−1( , )δ δ λ ≤ R F enn( , )δ δ λ . Therefore, in relations (5), (7) – (9),

and (11), one can replace R Fn( , )δ by E Fn−1( , )δ .

REFERENCES

1. L. Ya. Mykytyuk and M. M. Sheremeta, “On the approximation of Dirichlet series by exponential polynomials,” Visn. L’viv.

Univ., Ser. Mekh.-Mat., Issue 53, 35–39 (1999).

2. L. Ya. Mykytyuk, “A remark on the approximation of Dirichlet series by exponential polynomials,” Visn. L’viv. Univ., Ser.

Mekh.-Mat., Issue 57, 25–28 (2000).

3. F. I. Geche and S. V. Onipchuk, “On the abscissas of convergence of a Dirichlet series and its Newton majorant,” Ukr. Mat.

Zh., 26, No. 2, 161–168 (1974).

4. H. L. Royden, Real Analysis, Macmillan (1989).

on the asymptotic behavior of the remainder of a dirichlet series absolutely convergent in a...

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