In this paper, we using the elementary method to study the convergent property of
one class Dirichlet series involving a special sequences, and give several interesting identities for it

Li, Jing

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Scientia MagnaVol. 3 (2007), No. 3, 52-55An infinity series involving theSmarandache-type functionJing LiDepartment of Mathematics, Northwest UniversityXi’an, Shaanxi, P.R.Chinalj5118@163.comAbstract In this paper, we using the elementary method to study the convergent property ofone class Dirichlet series involving a special sequences, and give several interesting identitiesfor it.Keywords Riemann zeta-function, infinity series, identity.§1. Introduction and ResultsFor any positive integer n and m ≥ 2, the Smarandache-type function Bm(n) is defined asthe largest m-th power dividing n. That is,Bm(n) = max{xm : xm | n}(∀n ∈ N∗).For example, B2(1) = 1, B2(2) = 1, B2(3) = 1, B2(4) = 2, B2(5) = 1, B2(6) = 1, B2(7) = 1,B2(8) = 2, B2(9) = 3, · · · . This function was first introduced by Professor Smarandache. In[1], Henry Bottomley presented that Bm(n) is a multiplicative function. That is,(∀a, b ∈ N)(a, b) = 1⇒ Bm(a · b) = Bm(a) ·Bm(b).It is easily to show that Bm(pα) = pim, α = im + l, (i ≥ 0, 0 ≤ l < m), where p is a prime.So, if n = pα11 pα22 · · · pαrr is the prime powers decomposition of n, then the following identity isobviously:Bm(n) = Bm(pα11 pα22 · · · pαrr ) = pi1m1 pi2m2 · · · pirmr ,where αj = ijm+ lj (ij ≥ 0, 0 ≤ lj < m).Similarly, for any positive integer n and any fixed positive integer m, we define anotherSmarandache function Cm(n) as following:Cm(n) = max{x ∈ N : xm | n}(∀n ∈ N∗).Obviously, Cm(n) is also a multiplicative function.From the definition of Bm(n) and Cm(n), we may immediately getBm(n) = Cm(n)m.Vol. 3 An infinity series involving the Smarandache-type function 53Now let k be a fixed positive integer, for any positive integer n, we define the arithmeticalfunction δk(n) as following:δk(n) = max{d : d | n, (d, k) = 1}.For example, δ2(1) = 1, δ2(2) = 1, δ2(3) = 1, δ2(4) = 1, δ3(6) = 2, · · · . About theelementary properties of this function, many scholars have studied it and got some useful results(see reference [2], [3]). In reference [2], Xu Zhefeng studied the divisibility of δk(n) by ϕ(n), andproved that ϕ(n) | δk(n) if and only if n = 2α3β , where α > 0, β ≥ 0, α, β ∈ N . In reference[3], Liu Yanni and Gao Peng studied the mean value properties of δk(bm(n)), and obtained aninteresting mean value formula for it. That is, they proved the following conclusion:Let k and m are two fixed positive integers. Then for any real number x ≥ 1, we have theasymptotic formula∑n≤xδk(bm(n)) =x22ζ(2m)ζ(m)∏p|kpm + 1pm−1(p+ 1)+O(x32+²),where ² denotes any fixed positive number, ζ(s) is the Riemann zeta-function, and∏p|kdenotesthe product over all different prime divisors of k.Let A denotes the set of all positive integers n satisfying the equation Bm(n) = δk(n).That is, A = {n ∈ N,Bm(n) = δk(n)}. In this paper, we using the elementary method to studythe convergent property of the Dirichlet series involving the set A, and give several interestingidentities for it. That is, we shall prove the following conclusions:Theorem 1. Let m ≥ 2 be a fixed positive integer. Then for any real number s > 1, wehave the identity:∞∑n=1n∈A1ns= ζ(ms)∏p|k(1− 1pms )21− 1ps.Theorem 2. For any complex number s with Re(s) > 2, we have the identity:∞∑n=1n∈ABm(n)ns= ζ(ms− 1)∏p|k(1− 1pms )21− 1ps,where ζ(s) is the Riemann zeta-function, and∏pdenotes the product over all primes.§2. Proof of the theoremsNow we complete the proof of our Theorems. First we define the arithmetical functiona(n) as follows:a(n) = 1, if n ∈ A,0, if otherwise.54 Jing Li No. 3For any real number s > 0, it is clear that∞∑n=1n∈A1ns=∞∑n=1a(n)ns<∞∑n=11ns,and∞∑n=11nsis convergent if s > 1, thus∞∑n=1n∈A1nsis also convergent if s > 1. Now we find the set A.From the definition of Bm(n) and δk(n) we know that Bm(n) and δk(n) both are multiplicativefunctions. So in order to find all solutions of the equation Bm(n) = δk(n), we only discuss thecase n = pα. Let α = im+ l, where i ≥ 0, 0 ≤ l < m, then Bm(n) = pim. If n = pα, (p, k) = 1,now δk(n) = pα, then the equation Bm(n) = δk(n) has solution if and only if α = im, i ≥ 0.If n = pα, p | k, now δk(n) = 1, then the equation Bm(n) = δk(n) has solution if and only ifα = l, 0 ≤ l < m.Thus, by the Euler product formula, we have∞∑n=1n∈A1ns=∏p(1 +a(p)ps+a(p2)p2s+ · · ·+ a(pm−1)p(m−1)s+ · · ·)=∏p|k(1 +1ps+1p2s+ · · ·+ 1p(m−1)s)∏p†k(1 +1pms+1p2ms+ · · ·)=∏p|k1− 1pms1− 1ps∏p†k11− 1pms=∏p11− 1pms∏p|k(1− 1pms )21− 1ps= ζ(ms)∏p|k(1− 1pms )21− 1ps.This completes the proof of Theorem 1.Now we come to prove Theorem 2. Let s = σ + it be a complex number. Note thatBm(n) ¿ n, so it is clear that∞∑n=1Bm(n)nsis an absolutely convergent series for Re(s) > 2, byVol. 3 An infinity series involving the Smarandache-type function 55the Euler product formula and the definition of Bm(n) we get∞∑n=1n∈ABm(n)ns=∏p(1 +Bm(p)ps+Bm(p2)p2s+ · · ·)=∏p|k(1 +Bm(p)ps+Bm(p2)p2s+ · · ·+ Bm(pm−1)p(m−1)s)∏p†k(1 +Bm(pm)pms+Bm(p2m)p2ms+ · · ·)=∏p|k(1 +1ps+1p2s+ · · ·+ 1p(m−1)s)∏p†k(1 +pmpms+p2mp2ms+ · · ·)=∏p|k1− 1pms1− 1ps∏p†k11− 1pms−1=∏p11− 1pms−1∏p|k(1− 1pms )21− 1ps= ζ(ms− 1)∏p|k(1− 1pms )21− 1ps.This completes the proof of Theorem 2.References[1] Henry Bottomley, Some Smarandache-type Multiplicative Functions, Smarandache No-tions, 13(2002), 134-139.[2] Zhefeng Xu, On the function ϕ(n) and δk(n), Research On Smarandache Problems InNumber Theory, Hexis, 2005, 9-11.[3] Liu Yanni and Gao Peng, On the m-power free part of an integer, Scientia Magna,1(2005), No. 1, 203-206.[4] Zhang Pei, An Equation involving the function δk(n), Scientia Magna, 2(2006) No. 4.[5] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York,1976.[6] Pan Chengdong and Pan Chengbiao, Foundation of Analytic Number Theory, Beijing,Science Press, 1997.[7] Yang Hai and Fu Ruiqing, On the K-power Part Residue Function, Scientia MagnaJournal, 1(2005), No. 1.

An infinity series involving the Smarandache-type function

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