Dirichlet-Power Series -.:: Natural Sciences Publishing

J. Ana. Num. Theor. 5, No. 1, 41-48 (2017) 41 Journal of Analysis & Number Theory An International Journal

http://dx.doi.org/10.18576/jant/050107

Dirichlet-Power Series

Hunduma Legesse Geleta∗ Department of Mathematics, Addis Ababa University, Ethiopia

Received: 2 Sep. 2016, Revised: 18 Nov. 2016, Accepted: 23 Nov. 2016 Published online: 1 Jan. 2017

Abstract: In this paper we introduce and investigate a new type of series which we call a ”Dirichlet-power” series. This series includes both Dirichlet series and power series as a special case. It arose during the investigation of series representations of the second order hypergeometric zeta function. The Dirichlet-power series has many properties analogous to Dirichlet series. For a Dirichlet-power series we establish Domain of convergence and show the existence of zero free regions to the right half plane.

Keywords: Multiplicative cone b-metric space, ﬁxed point, contractive mapping.

1 Introduction In this paper we introduce and investigate a new type of series which we call a ”Dirichlet-power series”. This The Riemann Zeta function ζ(s) defined by series includes both Dirichlet series and power series as a ∞ special case. As in the case of Dirichlet series it has a 1 kind of abscissa of absolute convergence is attached to it ∑ s n=1 n with a right half plane convergence. On the other hand as a power series it has a kind of radius of convergence with with an integral representation D-shaped region of convergence. This kind of series actually arose during the investigation of series ∞ s−1 ζ 1 x representations of the second order hypergeometric zeta (s)= x dx Γ (s) Z0 e − 1 function. The hypergeometric zeta function of order N where N is a positive integer, defined as is a special example of a type of series called Dirichlet series. A Dirichlet series is a series of the form ∞ s+N−2 ζ 1 x N (s)= x dx ∞ Γ (s + N − 1) Z0 e − TN−1(x) an ∑ s n=1 n which admits a Dirichlet-type series given by, with a complex number and s complex variable. ∞ n µN (n,s) For a Dirichlet series of this type there is an extended real ζN (s)= ∑ ns+N−1 number σa in the extendedreal line with the propertythat n=1 the series convergesabsolutely for Re(s) > σ , butnotfor a for σ > 1, where Re(s) < σa. Moreover, for any positive number ε, the convergence is uniform for Re(s) > σ + ε, so that the a (N−1)(n−1) a (N,n)Γ (s + N + k − 1) series represents a holomorphic function for all Re(s) > µ (n,s)= ∑ k N kΓ σa. The quantity σa is called the abscissa of absolute k=0 n (s + N − 1) convergence of the Dirichlet series; it is an analogue of the radius of convergence of a power series [8]. In fact, and ak(N,n) is generated by n−1 n−1 x2 x3 xN−1 ∑(N−1)(n−1) k if we fix a prime p, and only allow an to be nonzero (TN−1(x)) = 1 + x + 2! + 3! + ··· + (N−1)! = k=0 ak(N,n)x when n is a power of p, then we get an ordinary power series in p−s. So in some sense, Dirichlet series are a Originally introduced and investigated by Abdul strict generalization of an ordinary power series. Hassen and Heiu D. Nuygen as a generalization of the

∗ Corresponding author e-mail: [email protected]

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∞ Riemann Zeta function, via the integral representation in Suppose there exists an infinite sequence {sk}k=1 of several papers like [2],[4],[5]. Recently we studied complex numbers, with σk = Re(sk) > σ0 for series representation of the second order hypergeometric all k and σk → ∞ such that D(sk)= E(sk) for all k . zeta function in [6] which can be given as: Then D(s)= E(s) for all s with σ = Re(s) > σ0 . ∞ Corollary 23 Let ζ m−1 2(s)= ∑ Dm(s)s ∞ n=1 an D(s)= ∑ s n=1 n where the coefficient Dm(s) is a Dirichlet series for each m = 1,2,3,···. This series representation of the second be a Dirichlet series with abscissa of absolute order hypergeometric zeta function is a source for the title convergence σa . Suppose that for some s with of this paper. σ = Re(s) > σa wehave D(s) 6= 0 . Then there exists a The Dirichlet-power series has many properties half-plane in which the Diriclet series is absolutely analogous to Dirichlet series. For a Dirichlet-power series convergent and never zero. we have a D-shaped region of absolute convergence which can be extended to right half plane. In this From the these properties of Dirichlet series it is known that the Domain of absolute convergence of a extension there is a unique extended real number σa in the extended real line with the property that the series Dirichlet series is one of the following . converges absolutely for Re(s) > σa, but not for 1.empty set, that is for no s does the series absolutely Re(s) < σa just analogous to Dirichlet series. In any converges. compact subset of its absolute convergence the series 2.(−∞,∞), that is the series convergesfor every complex represents an analytic function. Moreover; this series has number s. a zero free regions in the right half-plane. 3.(σ,∞) that is the series converges for every complex number s with Re(s) > σ 4.[σ,∞) that is the series converges for every complex 2 Preliminaries and Auxiliary Results number s with Re(s) ≥ σ Notice that in all cases, there is a unique number σa ∈ In this section we review some main known results on [−∞,∞] such that for all s with Re(s) > σa , the Dirichlet Dirichlet series which we need for our consumption, series is absolutely convergent. And for all s with Re(s) < formally define a Dirichlet-power series and discuss on σa , the Dirichlet series is not absolutely convergent. the domain of absolute convergence.

2.2 Deﬁnition and Examples of Dirichlet-Power 2.1 Review on Dirichlet Series Series

We summarize some of the properties of Dirichlet series Definition 24 Let Dm(s) be a Dirichlet series for each m = as follows: 1,2,.... We define a Dirichlet-Power series to be a series of the form Theorem 21 Suppose a Dirichlet series ∞ m−1 ∞ ∑ Dm(s)s (2.1) −s m=1 D(s)= ∑ amm m=1 Remark 25 is absolutely convergent at a point s0 = σ0 + it0. Then it is also absolutely convergent at all points s with σ = 1.A Dirichlet-power series has a form of a power series Re(s) > σ . but with a Dirichlet series as a coefficient. 0 2.If a Dirichlet-power series converges absolutely Theorem 22 (Uniqueness theorem for Dirichlet series) for Re(s) > σ0, then Dm(s) converges absolutely for each m = 1,2,3,..., for Re(s) > σ0. Let ∞ 3.If Dm(s) is identically zero for m ≥ 2, then the an Dirichlet-power series becomes a Dirichlet series. D(s)= ∑ s 4.If D s D a constant sequence, then the n=1 n m( ) = m, Dirichlet-power series becomes a power series. and ∞ b Proposition 26 Suppose the function F(s) is represented E(s)= ∑ n s by a Dirichlet-power series, for Re(s) > σa, and suppose n=1 n ∞ where D(s) and E(s) are both absolutely convergent m−1 σ σ ∑ Dm(s)s → 0 Dirichlet series in the half-plane = Re(s) > 0 . m=2

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as σ = Re(s) → ∞, independent of t. Then Theorem 28 Suppose the series F(s) → a ∞ 11 m−1 ∑ |Dm(s)s | as σ = Re(s) → ∞ m=1 Proof. By the hypothesis of the proposition we have, does not converge for all t or diverge for all t on a line ∞ with Re(s)= σ0. Then there exists a real number t0, such m−1 ∑ Dm(s)s → 0 that the series ∞ m=2 m−1 ∑ |Dm(s)s | as σ = Re(s) → ∞, independent of t. Then we are left only m=1 with the ﬁrst term of the Dirichlet-power series D1. But D1 converges for all t such that 0 ≤ t ≤ t0 and for any s with is a Dirichlet series and hence by the property of Dirichlet σ σ σ 2 2 σ 2 2 > 0 with the property +t < a +ta , but does not series it converges to its ﬁrst term which is a11 independent converge otherwise. of t as σ = Re(s) → ∞. Proof. Let E = Some examples of a Dirichlet-power series are given here: ∑∞ m−1 {t/t is nonnegative and m=1 |Dm(s)s |converges} 1.the second order Hypergeometric zeta function given where s = σ0 + it. Let t0 be the supremum of t in E such by, that ∞ ∞ m−1 m−1 ζ2(s)= ∑ Dm(s)s ∑ |Dm(s)s |converges} n=1 m=1

where the coefficient Dm(s) is a Dirichlet series for Such supremum exists since the series does not converge each m = 1,2,3,··· , where the Dm(s) is given as in [6]. for all t. 2.a series given by, Observe that from the theorem we have a segment E = ζ s F(s)= (s)e {σ0 + it/ − t0 ≤ t ≤ t0}. Moreover, the series converges absolutely for all t with the property σ = Re(s) > σ0 and 3.a series given by, σ 2 2 σ 2 2 +t < 0 +t0 . If the series convergesfor all t we define G(s)= ζ(s)cosh(s) t0 = ∞ and if it does not converge for any t we define t0 = −∞. Theorem 29 Suppose the series 2.3 Domain of Absolute Convergence ∞ m−1 In this section we establish domain of absolute ∑ |Dm(s)s | convergence for a Dirichlet-power series and characterize m=1 its region of absolute convergence. does not converge for all σ or diverge for all σ. Then Theorem 27 Suppose a Dirichlet-power series there exists a real number σa, called point of absolute ∞ convergence, such that the series m−1 F(s)= ∑ Dm(s)s ∞ m=1 m−1 ∑ |Dm(s)s | is absolutely convergent at a point s0 = σ0 +it0. Then F(s) m=1 converges absolutely at all points s = σ + it with σ > 2 2 2 2 converges if σ > σ with the property σ 2 +t2 < σ 2 +t2 , σ0 and σ +t < σ +t . a a a 0 0 but does not converge otherwise. Proof. From the hypothesis of the theorem, we have: |s|≤ σ σ σ ∑∞ m−1 |s0| and since > 0 we have also |Dm(s)| < |Dm(s0)|. Proof. Let E = { / m=1 |Dm(s)s |diverges} where s = Thus we have, σ +it. Then E is not empty as the series does not converge σ m−1 m−1 for all . E is bounded above because the series does not |Dm(s)s | < |Dm(s0)s0 | diverge for all σ. Therefore, E has a least upper bound call σ σ σ σ σ Hence it a. If < a, then is in E, otherwise would be an ∞ ∞ upper bound for E smaller than the least upper bound. If σ > σ with the property σ 2 +t2 < σ 2 +t2, then σ is not ∑ |D (s)sm−1| < ∑ |D (s )sm−1| < ∞ a a a m m 0 0 in E since σ is an upper bound for E m=1 m=1 , a . Therefore, the Dirichlet-power series converges absolutely Observe that if we know the absolute convergence of 2 2 at all pointss = σ + it,with σ = Re(s) > σ0 and σ +t < the Dirichlet-power series at a point s0 = σ0 + it0, then its σ 2 2 0 +t0 . domain of absolute convergence D is given as follows:

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3 Half-Plane of Absolute Convergence

In case of a Dirichlet series, we know that if a Dirichlet series converges absolutely at a point s0 = σ0 + it0, then it converges absolutely on the whole line with Re(s)= σ0 and σ = Re(s) > σ0 . But this is not generally true for a Dirichlet-power series due to the presence of the power sm−1 in the expression deﬁning it. But if the Dirichlet-power series converges on a line with Re(s) = σ0, then we have a right half plane of convergence analogous to the Dirichlet series. If this is the case then the segment of convergence becomes line of convergence and hence σa which is previously deﬁned becomes abscissa of absolute convergence analogous to the Dirichlet series case. Theorem 31 Suppose a Dirichlet-power series

∞ m−1 F(s)= ∑ Dm(s)s m=1 is absolutely convergent at every point on a line with Re(s)= σ0. Then it is also absolutely convergent at all points s with σ = Re(s) > σ0 .

Proof. For s with σ = Re(s) > σ0 = Re(s0) , we have

|Dm(s)| < |Dm(s0)|

But if a complex number s = σ + it with σ = Re(s) > σ0 = Re(s0) is given,then we can ﬁnd a real number t0, so that |s| ≤ |s0| where s0 = σ0 + it0. Hence

Fig. 1: Caption for D-Shape m−1 m−1 |Dm(s)s | < |Dm(s0)s0 | Hence we have ∞ ∞ ∑ |D (s)sm−1| < ∑ |D (s )sm−1| < ∞ σ σ σ σ 2 2 σ 2 2 m m 0 0 D = {s = + it/ > 0, +t < 0 +t0 } m=1 m=1 Itis a D-shaped region given as in the following ﬁgure: Therefore, the Dirichlet-power series converges absolutely So from this we can conclude that the domain of for Re(s) > σ0 = Re(s0) . absolute convergence of a Dirichlet-power series is one of the following: From the theorem it follows that the Domain of absolute 1.empty set, that is for no s does the series absolutely convergence of a Dirichlet-power series is one of the converge. following similar to that of a Dirichlet series once the 2.a point, that is the series convergesfor s only, actually Dirichlet-power series converges absolutely on the whole 0 σ this is the case where it convergesonly on a point on a line Re(s)= 0. real line. 1.empty set, that is for no s does the series absolutely 3.a D-shaped region, that is the series converges for converges. σ σ σ every complex number s = + it with > 0 and 2.(−∞,∞), that is the series convergesfor every complex σ 2 2 σ 2 2 +t < 0 +t0 . number s. Now one naturally asks, whether this series has a right 3.(σ,∞) that is the series converges for every complex half plane as a region of absolute convergence analogous number s with Re(s) > σ to Dirichlet series. This is the content of the next 4.[σ,∞) that is the series converges for every complex subsection. number s with Re(s) ≥ σ

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Notice that in all cases, there is a unique number 1. σ ∞ ∞ σ ∞ m−1 a ∈ [− , ] such that for all s with Re(s) > a , the ∑ anms Dirichlet-power series is absolutely convergent. And for s n=1 n all s with Re(s) < σa , the Dirichlet-power series is not absolutely convergent. We can also have a right half plane converges absolutely for each m. of absolute convergence if the Dirichlet-power series as a 2. ∞ m−1 power series has radius of convergence infinity. So the anms ∑ s next question would be what are the conditions for which m=1 n the series has such line of convergence. The D-shaped region of absolute convergence is due to the presence of converges absolutely for each n. m−1 3. the power of s as a power series in a Dirichlet-power ∞ ∞ a sm−1 series. If as a power series we have the radius of nm ∑ ( ∑ s ) convergence to be infinity, then we have right half plane m=1 n=1 n of absolute convergence. Here below we first recall a converges absolutely. theorem on Double series and apply for the Dirichlet 4. power series. ∞ ∞ m−1 anms ∑ ( ∑ ) Theorem 32 If the double series, s m=1 n=1 n ∞ ∞ converges absolutely. ∑ ∑ |anm| 5. m=1 n=1 ∞ m−1 anms ∑ s converges, then each of the following holds: n,m=1 n

1. ∞ converges absolutely; moreover, the last three sums in 3,4,5 are all the same. ∑ anm n=1 Therefore, for a Dirichlet-power series converges absolutely for each m. ∞ ∞ m−1 2. ∞ anms ∑ ∑ s ∑ anm m=1 n=1 n m=1 converges absolutely for each n. to have a right half plane of absolute convergence, 3. ∞ ∞ 1.as a Dirichlet series ∑ ( ∑ anm) ∞ m−1 m=1 n=1 anms ∑ s converges absolutely. n=1 n 4. ∞ ∞ converges absolutely for each m with some abscissa of ∑ ( ∑ anm) absolute convergence σa. m=1 n=1 2.as a power series converges absolutely. ∞ m−1 5. ∞ anms ∑ s ∑ anm m=1 n n,m=1 converges absolutely; moreover, the last three sums in converges absolutely for each n with radius 3,4,5 are all the same. convergence inﬁnity.

m−1 Now in particular, if the coefﬁcient anm = a(n,m) as a So if the a ’s are replaced by anms , we have the nm ns function of two variables can be given as anm = bncm for Dirichlet-power series deﬁned earlier, moreover the above each m and n. Then theorem can be restated as, ∞ ∞ m−1 ∞ ∞ Theorem 33 If the double series, anms bn m−1 F(s)= ∑ ∑ s = ∑ s ∑ cms m=1 n=1 n n=1 n m=1 ∞ ∞ a sm−1 ∑ ∑ | nm | s ∑∞ bn m=1 n=1 n Thus if n=1 ns converges absolutely to D(s) with abscissa σ ∑∞ m−1 of absolute convergence a and m=1 cms represents converges, then each of the following holds: an entire function f (s) ( that means as a power series it has

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radius of convergence infinity). Then the Dirichlet-power Theorem 41 Let series ∞ ∞ ∞ m−1 bn m−1 F(s)= ∑ Dm(s)s F(s)= ∑ s ∑ cms m=1 n=1 n m=1 be a Dirichlet-power series with σ σ converges absolutely to D(s) f (s) for s = + it with > ∞ σa. −s Dm(s)= ∑ anmn n=1 Theorem 34 Let σ0 > 0 and suppose a Dirichlet-power series a Dirichlet series for each m = 1,2,3,··· . Ifa11 is not zero ∞ and , limσ ∞ |F(s)| = a uniformly for all t, then F(s) m−1 → 11 F(s)= ∑ Dm(s)s has a zero free region in the right half plane. m=1 Proof. Assume F(s) has no zero free region in the right- σ σ is absolutely convergent for some > 0. The Dirichlet- half plane. Then there is a sequence of complex numbers power series ∞ {sk}k=1 with the property ∞ σ σ σ m−1 1 < 2 < 3 < ··· F(s)= ∑ Dm(s)s m=1 such that, F(sk)= 0 for all k. But this contradicts the fact that as σk → ∞ , wehave |F(sk)| tends to a11 which is converges absolutely to the right of σ if the power series not equal to zero by hypothesis, since as k tends to infinity σ ∞ k is also tends to infinity. m−1 ∑ anms Theorem 42(Uniqueness theorem) m=1 Given two Dirichlet-power series, ∞ represents an entire function. m−1 F(s)= ∑ Dm(s)s m=1 Proof. Suppose ∞ and ∞ ∑ a sm−1 nm G(s)= ∑ E (s)sm−1 m=1 m m=1 represents an entire function. Then its radius of both absolutely convergent for σ = Re(s) > σa. IfF(sk)= convergence is infinity. But the domain of F(s) is the ∞ G(sk) for all k in an infinite sequence {sk}k=1 of complex intersection of the domain of the power series and the numbers, with σk = Re(sk) > σa such that σk → ∞, as Dirichlet series attached to it, we have right half plane as k → ∞ with the property that F(sk)= G(sk) implies that a domain of absolute convergence for the Dirichlet-power Dm(sk)= Em(sk) for each m = 1,2,3,··· . Then F(s)= series. G(s) for all s with σ = Re(s) > σa. ∞ Observe that On any compact subset in the domain of Proof. Suppose {sk}k=1 be an infinite sequence with the σ ∞ ∞ absolute convergence, the function represented by a property that k → , as k → . Assume Dirichlet-power series is analytic. Hence if we have a F(sk)= G(sk right half-plane in which it converges absolutely, then the function represented by a Dirichlet-power series is for each k = 1,2,3,··· , then analytic in the same right half-plane. Dm(sk)= Em(sk) m = 1,2,3,··· and k = 1,2,3,··· . Hence by identity 4 Zero free Region for a Dirichlet-Power theorem for Dirichlet series we have Series Dm(s)= Em(s) for all s and each m = 1,2,3,··· Thus we have, In this section we analyze whether such a series has a m−1 m−1 zero free region in the right half plane where it converges Dm(s)s = Em(s)s absolutely in the whole right half plane. Actually in Therefore, Dirichlet series case the existence of zero free region ∞ ∞ depends on the unique representation of the Dirichlet m−1 m−1 series. In this case we show by using the limit of the ∑ Dm(s)s = ∑ Em(s)s modulus of F(s) as σ tends to infinity. This is the content m=1 m=1 of the following theorem: that is F(s)= G(s) for all s in the domain.

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This theorem implies the existence of a right half plane 5 Conclusion in which a Dirichlet-power series does not vanish, unless of-course it is identically zero. As we have seen a new type of series called Dirichlet-Power series is introduced. It has many Theorem 43 Suppose properties analogous to an ordinary Dirichlet series and ∞ power series. The series has a D-shaped region of m−1 F(s)= ∑ Dm(s)s absolute convergence which can be extended to to right m=1 half plane. There is a unique extended real number σa with the property that the series converges absolutely and with the property that F(s )= 0 implies that D (s )= 0 k m k uniformly to the right of it and diverges to the left of it. In foreach m = 1,2,3,··· and assume that F(s) 6= 0 for some any compact subset of its absolute convergence the series s with σ = Re(s) > σ . Then there is a half plane in which a represents an analytic function. This series has zero free F(s) is never zero. region in the right half-plane. We have also forwarded Proof. Assume no such half plane exists. Then for each some open problems for discussions. k = 1,2,3,··· , there is a point sk with σk = ℜ(sk) > k such that F(sk)= 0. Since σk → ∞, as k → ∞, the uniqueness theorem shows that F(s)= 0 for all s with σ = Re(s) > σa. Acknowledgments This contradicts the hypothesis that F(s) 6= 0 for some s with σ = Re(s) > σa. I would like to thank Professor Abdulkadir Hassen for introducing me such an interesting topics. I also thank the referees for their insightful comments on the drafts of this 4.1 Open problems paper.

1.can we remove the condition F(sk)) = G(sk) implies that Dm(sk)= Em(sk) for each m = 1,2,3,··· in the References Uniqueness Theorem? 2.can we improve the limit condition on the theorem [1] H. M. Edwards, Riemann’s Zeta Function, Pure which guarantees the existence of zero free regions? and Applied Mathematics Series, Academic Press, 1974. The above problem is concerned to the prove or [2] A. Hassen and H. D. Nguyen, Hypergeometric Zeta disprove of the following: Function, Intern. J. Number Theory, 6 (2010), No. 6, 99-126. Theorem 44 (Uniqueness theorem for Dirichlet-power series) [3] A. Hassen and H. D. Nguyen, The Error Zeta Function, Intern. J. Number Theory, 3 (2007), No. Given two Dirichlet-power series, 3, 439-453. ∞ [4] A. Hassen and H. D. Nguyen, Hypergeometric m−1 F(s)= ∑ Dm(s)s Bernoulli Polynomials and Appell Sequences , m=1 Intern. J. Number Theory, 4 (2008) No. 5, 767-774 [5] A. Hassen and H. D. Nguyen, A Zero Free and ∞ Region for Hypergeometric Zeta Function, Intern. m−1 J. Number Theory, 7 (2011), No. 4,1033-1043. G(s)= ∑ Em(s)s m=1 [6] H. Geleta, A. Hassen, and S.Mohammed Series Representation of Second Order Hypergeometric both absolutely convergent for σ = Re(s) > σa. If F(sk)= ∞ Zeta Function Preprint G(sk) for all k in an inﬁnite sequence {sk}k=1 of complex [7] A.S.B. Holland,Introduction to the Theory of Entire numbers, with σk = Re(sk) > σa such that σk → ∞, as Functions, Academic Press, 1973. k → ∞. Then F(s)= G(s) for all s with σ = Re(s) > σa. [8] Tom M. Apostol, Introduction to Analytic Number Theory, Spring-Verlag, New York Inc, 1976. Corollary 45 [9] B. E. Peterson, Riemann Zeta Function, Lecture Let Notes, 1996. ∞ [10] B. Riemann, Ueber die Anzahl der Primzahlen m−1 F(s)= ∑ Dm(s)s unter einer gegebenen Grosse (On the Number of m=1 Prime Numbers less than a Given Quantity), 1859, Translated by D. R. Wilkins (1998). be a Dirichlet-power series with abscissa of absolute [11] E. C. Titchmarsh, The Theory of the Riemann Zeta- σ convergence a . Suppose that for some s with Function, Oxford University Press, 1967. σ = Re(s) > σa wehave F(s) 6= 0. Then there exists a half-plane in which the Diriclet-power series is absolutely convergent and never zero.

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Hunduma Legesse Geleta is an assistant professor of Mathematics at Department of Mathematics, College of Natural and Computational Sciences, Addis Ababa University, Ethiopia.

c 2017 NSP Natural Sciences Publishing Cor.