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Research article Special Issues

The inverse of tails of Riemann zeta function, Hurwitz zeta function and Dirichlet L-function

  • In this paper, we derive the asymptotic formulas Br,s,t(n) such that

    limn{(k=n1kr(k+t)s)1Br,s,t(n)}=0,

    where Re(r+s)>1 and tC. It is evident that the asymptotic formulas for the inverses of the tails of both the Riemann zeta function and the Hurwitz zeta function on the half-plane Re(s)>1 are its corollaries. Subsequently we provide the asymptotic formulas for the Riemann zeta function and the Hurwitz zeta function on the half-plane Re(s)<0. Finally, we study the asymptotic formulas of the inverse of the tails of the Dirichlet L-function for Re(s)>1 and Re(s)<0.

    Citation: Zhenjiang Pan, Zhengang Wu. The inverse of tails of Riemann zeta function, Hurwitz zeta function and Dirichlet L-function[J]. AIMS Mathematics, 2024, 9(6): 16564-16585. doi: 10.3934/math.2024803

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  • In this paper, we derive the asymptotic formulas Br,s,t(n) such that

    limn{(k=n1kr(k+t)s)1Br,s,t(n)}=0,

    where Re(r+s)>1 and tC. It is evident that the asymptotic formulas for the inverses of the tails of both the Riemann zeta function and the Hurwitz zeta function on the half-plane Re(s)>1 are its corollaries. Subsequently we provide the asymptotic formulas for the Riemann zeta function and the Hurwitz zeta function on the half-plane Re(s)<0. Finally, we study the asymptotic formulas of the inverse of the tails of the Dirichlet L-function for Re(s)>1 and Re(s)<0.



    The Riemann zeta function ζ(s) and the Hurwitz zeta function ζ(s,a) are two important functions in the analytic number theory. The Riemann zeta function has an intimate connection with the distribution of primes, and the Hurwitz zeta function plays an important role in studying the properties of the Dirichlet L-function. The properties of ζ(s) and ζ(s,a) have attracted considerable attention from mathematicians.

    In the past ten years, many mathematicians have been working on the tails of various zeta functions. For example, Xin [1] gives the asymptotic formulas of the Riemann zeta function with s=2 and s=3,

    (k=n1k2)1=n1,(k=n1k3)1=2n(n1). (1.1)

    Kim and Song [2] studied the inverses of the tails of the Riemann zeta function and derived results for s in the critical strip 0<s<1,

    (k=n1ks)1{2(121s)(n12)s,if n is even;2(121s)(n12)s,if n is odd. (1.2)

    Lee and Park [3] dealt with the inverses of the tails of the Hurwitz zeta function when s2, sN, and 0a<1, and derived

    (k=n1(k+a)s)1  s1j=0Aj(n+a)j, (1.3)

    where

    As1=s1,Asl=sl1j=1xsjAsl+j,xsj=(s2+jj)Bj,

    and Bj is the Bernoulli numbers.

    Ohtsuka and Nakamura [4] derived the asymptotic formulas for the tails of the Fibonacci zeta function with s = 1 and s = 2,

    (k=n1Fk)1={Fn2, if n is even,n2;Fn21, if n is odd,n1; (1.4)
    (k=n1F2k)1={Fn1Fn1, if n is even,n2;Fn1Fn, if n is odd,n1. (1.5)

    More results associated with the asymptotic formulas can be found in [5,6,7,8,9,10,11,12], where Hwang and Song [5] and Xu [6] study the tails related to the Riemann zeta function, Choi and Choo [7], Komatsu [8], Lee and Park [9,10] and Marques and Trojovsky [11] study the tails of the Fibonacci zeta function, and Xu and Wang [12] studied the tails of Pell zeta function.

    The above results reveal the relations between zeta functions and polynomials. From other perspectives, these sums are the First Newton's identity of an infinite polynomial with roots ks or Fsk for kn; therefore, they also reflect the importance of Newton's identities, which are elaborately introduced in [13].

    In general, the consideration is limited to cases where sN+ or s(0,1). Due to the application of the estimate of the tails of the Riemann zeta function in various other problems, such as the upper of the Dirthlet L-function, the average of many well-known arithmetical functions, and so on, we hope to generalize the results to the complex plane. For this proposal, in this paper, the generalized Riemann zeta function R(r,s,t) and its n-th tail Rn(r,s,t) are defined for Re(r+s)>1, tC by series

    R(r,s,t)=k=11kr(k+t)s   and   Rn(r,s,t)=k=n1kr(k+t)s,

    and then we study the asymptotic formulas of Rn(r,s,t) for r,s in the half-plane Re(r+s)>1 and tC. In addition, inspired by the expressions of ζ(s) with 0<s<1 in [2], we can similarly estimate the inverses of the tails of ζ(s) for Re(s)<0. Similarly, we can derive the asymptotic formulas for the inverses of the tails of the Hurwitz zeta function and the Dirichlet L-function L(1,χ) for Re(s)>1 and Re(s)<0. It is worth noting that the inverses of the Hurwitz zeta function and the Dirichlet L-function possess new characteristics, and we obtain their dominant terms.

    Theorem 1. Let the function R(r,s,t) and the n-th tail Rn(r,s,t) be as above, for any r,s,tC. If

    Re(r+s)1>0,

    we have

    (Rn(r,s,t))1=Br,s,t(n)+o(1),

    where

    Br,s,t(n)=br+s1nr+s1+br+s2nr+s2++b{r+s}n{r+s}

    and the coefficients

    bjC (j=r+s1,r+s2,,{r+s})

    are constants determined by r,s,t as follows:

    {br+s1=r+s1;br+s1k=  k1+k2=kk10,k21r+s1i=r+s1k1i{r+s1}bi(iirs+2+k1)(sk2)tk2k1+k2=kk1,k21r+s1i=r+s1k1i{r+s1}bi(iirs+1+k1)br+s1k2r+s1+k+r+s1i=r+ski{r+s1}bi[(iirs+2+k)(r+s1)(iirs+1+k)]r+s1+k, for [r+s1]k1; (2.1)

    where [x] denotes the integer part of the real part of x, and {x}=x[x].

    Theorem 2. Let

    ζn(s)=k=n1ks

    be the n-th tail of the Riemann zeta function, then for all Re(s)<0, we have

    (ζn(s))1=πs12Γ(1s2)Γ(s2)B0,1s,0(n)+o(1),

    where

    B0,1s,0(n)=bsns+bs1ns1++b{s}n{s}

    and the coefficients

    bjC (j=s,s1,,{s})

    are constants determined by s as follows:

    {bs=s;bsk=k1+k2=kk10,k21si=sk1i{s}bi(ii+s+1+k1)(sk2)tk2k1+k2=kk1,k21si=sk1i{s}bi(ii+s+k1)bsk2s+k+si=1ski{s}bi[(ii+s+1+k)(s)(ii+s+k)]s+k, for [s]k1; (2.2)

    and [x] denotes the integer part of the real part of x, and {x}=x[x].

    Corollary 1. Let

    ζn(s)=k=n1ks

    be the n-th tail of the Riemann zeta function, then for Re(s)>1, we have

    ζn(s)=k=n1ks=B0,s,0(n)+o(1), (2.3)

    where

    B0,s,0(n)=bs1ns1+bs2ns2++b{s1}n{s1}

    and

    {bs1=s1;bs1k=s1i=ski{s1}bi[(iis+2+k)(s1)(iis+1+k)]s1+kk1+k2=kk1,k21s1i=s1k1i{s1}bi(iis+1+k1)bs1k2s1+k,     for [s1]k1. (2.4)

    Corollary 2. If r=0, 0<t=a1, and |s|>1, we have

    ζn(s,a)=k=n1(k+a)s=B0,s,a(n)+o(1), (2.5)

    where

    B0,s,a(n)=bs1,ans1+bs2,ans2++b{s1},an{s1}

    and

    {bs1=s1;bs1k=k1+k2=kk10,k21s1i=s1k1i{s1}bi(iis+2+k1)(sk2)ak2k1+k2=kk1,k21s1i=s1k1i{s1}bi(iis+1+k1)bs1k2s1+k+s1i=ski{s1}bi[(iis+2+k)(s1)(iis+1+k)]s1+k; for [s1]k1. (2.6)

    Now we start considering the asymptotic formulas for inverses of the tails of the Hurwitz zeta function ζ(s,a) for Re(s)<0 and a=pqQ,p,qN, (p,q)=1.

    In combination with the following functional equation related to the Hurwitz zeta function [14]:

    ζ(1s,pq)=2Γ(s)(2πq)sqm=1cos(πs22πmpq)ζ(s,mq). (2.7)

    For any positive integer n, write

    n=q(l1)+r

    for some qr1 and l,rN+. The n-th tails of the Hurwitz zeta function for Re(s)<0 are defined by

    ζn(s,pq)=2Γ(1s)(2πq)1s(r1m=1amζl+1(1s,mq)+qm=ramζl(1s,mq)),

    where

    am=cos(π(1s)22πmpq)

    for m=1,2,,q.

    When

    n=q(l1)+1,

    rewrite above n-th tails of the Hurwitz zeta function for Re(s)<0 as

    ζn(s,pq)=2Γ(1s)(2πq)1sqm=1amζl(1s,mq).

    Theorem 3. Let function ζ(s,pq) and the n-th tail ζn(s,pq) be as above for Re(s)<0 and p,qN+,(p,q)=1.

    (i) Let n=q(l1)+1 for some lN+, if

    qm=1mam0

    holds, we have

    (ζn(s,pq))1=πq2(2qπ)sΓ(1s)qm=1maml1s+Os,q(ls),

    where

    am=cos(π(1s)22πmpq)

    for m=1,2,,q.

    (ii) Let

    n=q(l1)+r

    for some qr1 and l,rN+, if

    qm=1mam0andqm=1mamqqm=ram0

    hold, we have

    (ζn(s,pq))1=πq2(2πq)sΓ(1s)(qm=1mamqqm=ram)l1s+Os,q(ls),

    where

    am=cos(π(1s)22πmpq)

    for m=1,2,,q.

    Remark 1. Since qm=1am=0 (see the proof of Theorem 3 in this paper), we observe that the result (i) of Theorem 3 follows (ii) when r=1.

    Similarly, according to the following functional equation related to the Dirichlet L-function [14]:

    L(s,χ)=kskj=1χ(j)ζ(s,jk), (2.8)

    where χ is a character module k.

    The asymptotic formula of the inverse of the Dirichlet L-function L(s,χ) is trivial when χ=χ1 is the principal character module k. Therefore, we only consider the case for χχ1. Similar to the Hurwitz zeta function, the n-th tail of L(s,χ) could be defined for Re(s)>1 by

    Ln(s,χ)=ks(r1j=1χ(j)ζl+1(s,jk)+kj=rχ(j)ζl(s,jk)),

    where

    n=k(l1)+r

    for some kr1 and l,rN+.

    When r=1, we have

    Ln(s,χ)=ks(kj=1χ(j)ζl(s,jk)).

    In combination with (2.7) and (2.8), for Re(s)<0, we have

    L(s,χ)=kskj=1χ(j)ζ(s,jk)=kskj=1χ(j)2Γ(1s)(2πk)1skm=1cos(π(1s)22πmjk)ζ(1s,mk)=Γ(1s)2sπ1skkj=1km=1χ(j)aj,mζ(1s,mk), (2.9)

    where

    aj,m=cos(π(1s)22πmjk)

    for j=1,2,,k and m=1,2,,k.

    The n-th tail of L(s,χ) could be defined for Re(s)<0 by

    Ln(s,χ)=Γ(1s)2sπ1sk((u1j=1km=1+r1m=1j=u)χ(j)aj,mζl+1(1s,mk)+(km=rj=u+kj=u+1km=1)χ(j)aj,mζl(1s,mk)),

    where

    n=k2(l1)+k(u1)+r

    for some ku1, kr1, and l,u,rN+.

    When r=1, we have

    Lk2+1(s,χ)=Γ(1s)2sπ1skkj=1km=1χ(j)aj,mζl(1s,mk).

    Theorem 4. Let the function L(s,χ) and the n-th tail Ln(s,χ) be as above for χχ1 is the nonprincipal character module k.

    (i) When Re(s)>1, let n=k(l1)+1 for some lN+, if

    kj=1jχ(j)0

    holds, we have

    (Ln(s,χ))1=ks+1kj=1jχ(j)ls+Os,k(ls1).

    (ii) When Re(s)>1, let n=k(l1)+r for some kr1, and l,rN+, if

    kj=1jχ(j)kkj=rχ(j)0andkj=1jχ(j)0

    hold, we have

    (Ln(s,χ))1=k1+skj=1jχ(j)kkj=rχ(j)ls+Os,k(ls1).

    (iii) When Re(s)<0, let n=k2(l1)+1 for some lN+, if

    kj=1km=1mχ(j)aj,m0

    holds, we have

    (Ln(s,χ))1=πk2(2π)sΓ(1s)kj=1km=1mχ(j)aj,ml1s+Os,k(ls),

    where

    aj,m=cos(π(1s)22πmjk)

    for j=1,2,,k and m=1,2,,k.

    (iv) When Re(s)<0, let

    n=k2(l1)+k(u1)+r

    for some ku1, kr1 and l,u,rN+, if

    kj=1km=1mχ(j)aj,mk(km=rj=u+kj=u+1km=1)χ(j)aj,m0andkj=1km=1mχ(j)aj,m0

    hold, we have

    (Ln(s,χ))1=πk2(2π)sΓ(1s)(kj=1km=1mχ(j)aj,mk(km=rj=u+kj=u+1km=1)χ(j)aj,m)l1s+Os,k(ls),

    where

    aj,m=cos(π(1s)22πmjk)

    for j=1,2,,k and m=1,2,,k.

    Before our proof, we first propose several lemmas.

    Lemma 1. For all s,tC and nN, if n>|t|, we have

    (n+t)s=k=0(sk)nsktk.

    Proof. Let

    f(s)=(n+t)s=ns(1+tn)s=nsk=0(sk)(tn)k=k=0(sk)nsktk, (3.1)

    and the third equation holds for |tn|<1.

    Lemma 2. For all r,s,tC and Re(r+s)>1, there exists a unique generalized complex coefficient polynomial

    Br,s,t(n)=br+s1nr+s1+br+s2nr+s2++b{r+s}n{r+s},

    where

    bjC (j=r+s1,r+s2,,{l})

    is determined by r,s, and t, subject to the real part of its order being greater than 0 and

    [Br,s,t(n+1)Br,s,t(n)]nr(n+t)sBr,s,t(n+1)Br,s,t(n)=O(nr+s+{r+s}2).

    Proof. First, we assume

    Br,s,t(n)=blnl+bl1nl1++b{l}+1n{l}+1+b{l}n{l},     Re(l)>0

    satisfies

    [Br,s,t(n+1)Br,s,t(n)]nr(n+t)sBr,s,t(n+1)Br,s,t(n)=O(nr+s+{r+s}2). (3.2)

    Since the absolute value of the order of [Br,s,t(n+1)Br,s,t(n)]nr(n+t)s is greater than |r+s+{r+s}2|, then we have the necessary condition (1): The order of [Br,s,t(n+1)Br,s,t(n)]nr(n+t)s is equal to the order of Br,s,t(n+1)Br,s,t(n).

    In other words, we have l1+r+s=2l, which implies

    l=r+s1. (3.3)

    Then we rewrite

    Br,s,t(n)=br+s1nr+s1+br+s2nr+s2++b{l}+1n+b{r+s1}

    with br+s10, by Lemma 1, we have

    (n+t)s=k=0(sk)nsktk    for  n>|t|.

    Let n>1, we have

    [Br,s,t(n+1)Br,s,t(n)]nr(n+t)s=[r+s1i={r+s1}bi(n+1)ir+s1i={r+s1}bini]nrj=0(sj)nsjtj=r+s1i={r+s1}bik=1(ik)niknrj=0(sj)tjnsj=k=0r+s1i=r+s1ki{r+s1}bi(ii(r+s2k))nr+s2kj=0(sj)tjns+rj=k=0k1+k2=kk1,k20r+s1i=r+s1k1i{r+s1}bi(iirs+2+k1)(sk2)tk2n2r+2s2k (3.4)

    and

    Br,s,t(n+1)Br,s,t(n)=r+s1i={r+s1}bik=0(ik)nikr+s1j={r+s1}bjnj=k=0i=r+s1ki{r+s1}bi(ii(r+s1k))nr+s1k[r+s1]j=0br+s1jnr+s1j=k=0(k1+k2=kk1,k20r+s1i=r+s1k1i{r+s1}bi(ii(r+s1k1))br+s1k2)n2r+2s2k, (3.5)

    where [x] denotes the integer part of the real part of x.

    Now we derive the necessary condition (2): The following system of equations has at least one solution,

    k1+k2=kk1,k20r+s1i=r+s1k1i{r+s1}bi(iirs+2+k1)(sk2)tk2=k1+k2=kk1,k20r+s1i=r+s1k1i{r+s1}bi(iirs+1+k1)br+s1k2, (3.6)

    where k=0,1,2,,[r+s2] and [r+s1].

    By calculation, the system of Eq (3.6) is equivalent with

    (r+s1)br+s1=[br+s1]2   for  k=0,

    and

    (r+s1k)br+s1k+k1+k2=kk10,k21r+s1i=r+s1k1i{r+s1}bi(iirs+2+k1)(sk2)tk2+r+s1i=r+ski{r+s1}bi(iirs+2+k)=(2r+2s2)br+s1k+k1+k2=kk1,k21r+s1i=r+s1k1i{r+s1}bi(iirs+1+k1)br+s1k2+r+s1i=r+ski{r+s1}bi(iirs+1+k)br+s1 (3.7)

    for k=1,2,,[r+s1]. Combining with Re(r+s1)>0, we can then derive the unique solution of the system of Eq (3.6) as follows:

    br+s1=(r+s1)0

    and

    br+s1k=k1+k2=kk10,k21r+s1i=r+s1k1i{r+s1}bi(iirs+2+k1)(sk2)tk2k1+k2=kk1,k21r+s1i=r+s1k1i{r+s1}bi(iirs+1+k1)br+s1k2r+s1+k+r+s1i=r+ski{r+s1}bi[(iirs+2+k)(r+s1)(iirs+1+k)]r+s1+k, (3.8)

    where k = 1, 2, \cdots, [r+s-1].

    So we derive the necessary condition of (3.2):

    \begin{equation} B^*_{r,s,t}(n) = b^*_{r+s-1}n^{r+s-1}+b^*_{r+s-2}n^{r+s-2}+ \cdots + b^*_{\{r+s\}}n^{ \{r+s\} } \end{equation} (3.9)

    with

    \begin{equation} \begin{cases}\;\;\, b^*_{r+s-1} = r+s-1; \\ b^*_{r+s-1-k} = \\ \ \ \frac{\mathop{ \mathop{\sum}\limits_{k_1+k_2 = k} } \limits_{ k_1 \geq0 , k_2 \geq 1 } \mathop{ \mathop{\sum}\limits^{r+s-1}\limits_{i = r+s-1-k_1}}\limits_{i\geq \{ r+s-1 \}} b^*_i\binom{i}{i-r-s+2+k_1} \binom{s}{k_2}t^{k_2} - \mathop{\mathop{\sum}\limits_{k_1+k_2 = k}}\limits_{ k_1,k_2\geq 1 } \mathop{\mathop{\sum}\limits^{r+s-1}\limits_{i = r+s-1-k_1}}\limits_{ i\geq\{ r+s-1\} } b^*_i \binom{i}{i-r-s+1+k_1} b^*_{r+s-1-k_2}} {r+s-1+k} +\frac{\mathop{\mathop{\sum}\limits^{r+s-1}\limits_{i = r+s-k}}\limits_{ i\geq\{ r+s-1\} } b^*_i \left[ \binom{i}{i-r-s+2+k} -(r+s-1) \binom{i}{i-r-s+1+k} \right] } { r+s-1+k }; \\ \quad \text{ for }\; [r+s-1]\geq k\geq 1 , \end{cases} \end{equation} (3.10)

    where [x] denotes the integer part of the real part of x , \{x\} = x-[x].

    Now we show that (3.9) and (3.10) are also sufficient conditions. If Eq (3.10) holds, the functions \left[B^*(n+1, r, s, t)-B^*(n, r, s, t) \right] n^r(n+t)^s and B^*(n+1, r, s, t)B^*(n, r, s, t) are analytic functions of n , which means the series of k in formulas (3.4) and (3.5) are their Taylor expansion.

    Then consider the following function

    [B^*_{r,s,t}(n+1)-B^*_{r,s,t}(n)\,]\,n^r(n+t)^s -B^*_{r,s,t}(n+1)B^*_{r,s,t}(n),

    the coefficients of the first [r+s] terms are equal to zero, and the tail term is convergent, which proves the Eq (3.2).

    Lemma 3. If for all r, s, t \in \mathbb{C} and Re(r+s) > 1 , there exists a generalized complex coefficient polynomial

    B^*_{r,s,t}(n) = b^*_{r+s-1}n^{r+s-1}+b^*_{r+s-2}n^{r+s-2}+ \cdots + b^*_{\{l\}+1}n+b^*_{\{l\}},

    where

    b^*_{j} \ ( j = r+s-1, r+s-2,\cdots, \{l\})\in \mathbb{C}

    subjects to the real part of its order is greater than 0 , and

    \begin{align} &[B^*_{r,s,t}(n+1)-B^*_{r,s,t}(n)\,]\,n^r(n+t)^s -B^*_{r,s,t}(n+1)B^*_{r,s,t}(n) = O(n^{r+s+\{r+s\}-2}). \end{align}

    Then, we have

    \left( \sum\limits^{\infty}_{k = n} \frac{1}{ k^r(k+t)^s } \right)^{-1} = B^*_{r,s,t}(n) +o(1).

    Proof. Using the equation

    \begin{equation} \begin{aligned} &[B^*_{r,s,t}(n+1)-B^*_{r,s,t}(n)\,]\,n^r(n+t)^s -B^*_{r,s,t}(n+1)B^*_{r,s,t}(n) = O(n^{r+s+\{r+s\}-2}). \end{aligned} \end{equation} (3.11)

    Then, we have

    \begin{equation} \begin{aligned} \frac{1}{B^*_{r,s,t}(n)}-\sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s} = &\sum\limits^{\infty}_{k = n} \left[ \frac{1}{B^*_{r,s,t}(k)} -\frac{1}{B^*_{r,s,t}(k+1)}- \frac{1}{k^r(k+t)^s} \right]\\ = &\sum\limits^{\infty}_{k = n} \frac{ [B^*_{r,s,t}(k+1)-B^*_{r,s,t}(k)]k^r(k+t)^s -B^*_{r,s,t}(k+1)B^*_{r,s,t}(k) } { B^*_{r,s,t}(k)B^*_{r,s,t}(k+1)k^r(k+t)^s }\\ = & \sum\limits^{\infty}_{k = n} \frac{ O(k^{{r+s+\{r+s\}-2}}) }{ k^{3r+3s-2}+ O(k^{3r+3s-3}) } \\ = &O\left( \frac{1}{n^{[r+s]}} \sum\limits^{\infty}_{k = n} \frac{ 1 }{ k^{r+s} } \right) \\ = &O( \frac{1}{n^{[r+s]}} ). \end{aligned} \end{equation} (3.12)

    Hence, we have

    \begin{align*} \frac{1}{B^*_{r,s,t}(n)} = \sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s} = O( \frac{1}{ n^{r+s-1} } ), \end{align*}

    and then

    \begin{equation} \begin{aligned} \lim\limits_{n \rightarrow \infty} \left| \left( \sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s} \right)^{-1} -B^*_{r,s,t}(n) \right| \notag = &\lim\limits_{n \rightarrow \infty} \left| \frac{1}{ \mathop{ \stackrel{\infty} {\sum} } \limits_{k = n} \frac{1}{k^r(k+t)^s} } - \frac{1}{ \frac{1}{ B^*_{r,s,t}( n ) } } \right| \notag \\ = &\lim\limits_{n \rightarrow \infty} \left| \frac{ \frac{1}{B^*_{r,s,t}(n)}- \mathop{ \stackrel{\infty} {\sum} } \limits_{k = n} \frac{1}{k^r(k+t)^s} }{\frac{1}{B^*_{r,s,t}(n)} \mathop{ \stackrel{\infty} {\sum} } \limits_{k = n} \frac{1}{k^r(k+t)^s} } \right|\\ = & \lim\limits_{n \rightarrow \infty} \left| O \left( \frac{\frac{1}{n^{[r+s]}} }{ \frac{1}{n^{r+s-1} } } \right) \right| \notag \\ = &\lim\limits_{n \rightarrow \infty} \left| O \left( \frac{1 }{ n^{1-\{r+s\} } } \right) \right|\notag\\ = &0,\notag \end{aligned} \end{equation}

    which is equivalent with

    \begin{equation} \left( \sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s}\right)^{-1} = B^*_{r,s,t}(n) +o(1). \end{equation} (3.13)

    This completes the proof.

    Proof of Theorem 1. Since Re(r+s) > 1 and series

    \sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s}

    absolutely converges and is analytic in the half-plane Re(r+s) > 1 . Then we have

    \sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s} \rightarrow 0\ \ \ \text{as}\ \ \ n \rightarrow \infty ,

    Let n > |t| , by Lemmas 1 and 2, these are unique generalized complex coefficient polynomials

    B^*_{r,s,t}(n) = b^*_{r+s-1}n^{r+s-1}+b^*_{r+s-2}n^{r+s-2}+ \cdots +b^*_{\{r+s\}}n^{ \{r+s\} } ,

    where

    b^*_{j}\in \mathbb{C}\ ( j = r+s-1, r+s-2,\cdots, \{l\})

    is determined by r, s, t and

    \begin{equation} \begin{aligned} \begin{cases}\;\,\ b^*_{r+s-1} = r+s-1; \\ b^*_{r+s-1-k} = \\ \ \ \frac{\mathop{ \mathop{\sum}\limits_{k_1+k_2 = k} } \limits_{ k_1 \geq0 , k_2 \geq 1 } \mathop{ \mathop{\sum}\limits^{r+s-1}\limits_{i = r+s-1-k_1}}\limits_{i\geq \{ r+s-1 \}} b^*_i\binom{i}{i-r-s+2+k_1} \binom{s}{k_2}t^{k_2} - \mathop{\mathop{\sum}\limits_{k_1+k_2 = k}}\limits_{ k_1,k_2\geq 1 } \mathop{\mathop{\sum}\limits^{r+s-1}\limits_{i = r+s-1-k_1}}\limits_{ i\geq\{ r+s-1\} } b^*_i \binom{i}{i-r-s+1+k_1} b^*_{r+s-1-k_2}} {r+s-1+k} +\frac{\mathop{\mathop{\sum}\limits^{r+s-1}\limits_{i = r+s-k}}\limits_{ i\geq\{ r+s-1\} } b^*_i \left[ \binom{i}{i-r-s+2+k} -(r+s-1) \binom{i}{i-r-s+1+k} \right] } { r+s-1+k }, \\ \quad \mbox{for}\; [r+s-1]\geq k\geq 1; \end{cases}\end{aligned} \end{equation} (3.14)

    subject to the real part of its order being greater than 0 , and

    \begin{align} [B^*_{r,s,t}(n+1)-B^*_{r,s,t}(n)\,]\,n^r(n+t)^s -B^*_{r,s,t}(n+1)B^*_{r,s,t}(n) = O(n^{r+s+\{r+s\}-2}). \end{align}

    Then, in combination with Lemma 3, we have

    \left( \sum\limits^{\infty}_{k = n} \frac{1}{k^r(k+t)^s}\right)^{-1} = B^*_{r,s,t}(n) + o(1).

    This completes thhe proof.

    Proof of Corollary 1. If r = 0 and t = 0 , we have

    R(r, s, t)|_{ r = 0, t = 0} = R(0, s, 0) = \zeta(s)\ \ \ \text{and} \ \ \ R_n(r, s, t)|_{ r = 0, t = 0} = R_n(0, s, 0) = \zeta_n(s).

    By Theorem 1, we have

    \zeta_n(s) = \sum\limits^{\infty}_{k = n} \frac{1}{k^s} = B^*_{0,s,0}(n) +o(1),

    where

    B^*_{0,s,0}(n) = b^*_{s-1}n^{s-1}+b^*_{s-2}n^{s-2}+ \cdots + b^*_{\{s-1\}}n^{\{s-1\}}

    and

    \begin{equation} \begin{cases}\;\,\ b^*_{s-1} = s-1; \\ b^*_{s-1-k} = \frac{\mathop{\mathop{\sum}\limits^{s-1}\limits_{i = s-k}}\limits_{ i\geq\{ s-1\} } b^*_i \left[ \binom{i}{i-s+2+k} -(s-1) \binom{i}{i-s+1+k} \right] } { s-1+k } - \frac{ \mathop{\mathop{\sum}\limits_{k_1+k_2 = k}}\limits_{ k_1,k_2\geq 1 } \mathop{\mathop{\sum}\limits^{s-1}\limits_{i = s-1-k_1}}\limits_{ i\geq\{ s-1\} } b^*_i \binom{i}{i-s+1+k_1} b^*_{s-1-k_2}} {s-1+k} ,\mbox{for}\; \; [s-1]\geq k\geq 1 . \end{cases} \end{equation} (4.1)

    Proof of Corollary 2. The proof of Corollary 2 follows from Corollary 1.

    Before the proof of Theorem 2, we introduce some basic preliminary properties that could be found in [15].

    Proposition 5. The Riemann zeta function \zeta(s) satisfies

    \begin{equation} \pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})\zeta(s) = \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2})\zeta(1-s). \end{equation} (4.2)

    Proposition 6. (1) The function \frac{1}{\Gamma(s)} is entire and only possesses zeros at s = 0, -1, -2, \cdots .

    (2) The function \Gamma(s) is holomorphic for s \in \mathbb{C} except poles at s = 0, -1, -2, \cdots and \Gamma(s) possesses no zero.

    Proof of Theorem 2. By Proposition 5, we have

    \begin{align*} \zeta(s) & = \pi^{s-\frac{1}{2}} \Gamma(\frac{1-s}{2})\frac{1}{\Gamma( \frac{s}{2} )}\zeta(1-s)\\& = \pi^{s-\frac{1}{2}} \Gamma(\frac{1-s}{2})\frac{1}{\Gamma( \frac{s}{2} )}\sum\limits^{\infty}_{k = 1} \frac{1}{k^{1-s}}. \end{align*}

    By Proposition 6, we have \zeta(s) is entire in half-plane Re(s) < 0 . Hence, for Re(s) < 0, we have

    \begin{align*} \zeta_n(s) & = \pi^{s-\frac{1}{2}} \Gamma(\frac{1-s}{2})\frac{1}{\Gamma( \frac{s}{2} )}\zeta_n(1-s)\\& = \pi^{s-\frac{1}{2}} \Gamma(\frac{1-s}{2})\frac{1}{\Gamma( \frac{s}{2} )}\sum\limits^{\infty}_{k = n} \frac{1}{k^{1-s}}. \end{align*}

    By Theorem 1, we derive

    \zeta_n(s) = \pi^{s-\frac{1}{2} } \frac{\Gamma(\frac{1-s}{2})}{\Gamma( \frac{s}{2} )} B^*_{0,1-s,0}(n) +o(1),

    which completes the proof.

    Proof of Theorem 3. (i) We first prove

    \begin{equation} a_1+ a_2+\cdots +a_q = 0, \end{equation} (5.1)

    where

    a_m = \cos \left( \frac{\pi (1-s)}{2}- \frac{2\pi mp}{q} \right)

    for m = 1, 2, \cdots, q. For any complex variable s = \sigma +it , we have

    \cos s = \frac{ e^{is}+ e^{-is}}{2} ,

    and then

    \begin{equation} \begin{aligned} a_1+ a_2+\cdots +a_q& = \sum\limits^{q}_{m = 1}\cos \left( \frac{\pi (1-s)}{2}- \frac{2\pi mp}{q} \right)\\ & = \frac{1}{2}\sum\limits^{q}_{m = 1}\left( e^{\frac{\pi}{2}( t+i(1-\sigma) ) - \frac{2\pi mpi}{q} } + e^{ -\frac{\pi}{2}( t+i(1-\sigma) ) + \frac{2\pi mpi}{q}} \right)\\ & = \frac{1}{2} e^{\frac{\pi}{2}( t+i(1-\sigma) ) }\sum\limits^{q}_{m = 1} e^{ - \frac{2\pi mpi}{q} }+ \frac{1}{2} e^{-\frac{\pi}{2}( t+i(1-\sigma) ) }\sum\limits^{q}_{m = 1} e^{ \frac{2\pi mpi}{q} }\\ & = 0. \end{aligned} \end{equation} (5.2)

    Since

    \frac{1}{1\pm O(\varepsilon) } = 1\pm O( \varepsilon),

    Lemma 1 and Theorem 1, let n = q(l-1) +1, then for any sufficiently large positive integer l \geq 1 and Re(s) < 0 , if

    \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m\neq 0

    holds, we have

    \begin{align} \begin{aligned} \left( \zeta_{n}\left( s, \frac{p}{q} \right)\,\right)^{-1} & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left( \sum\limits^{q}_{m = 1} a_m \zeta_l \left( 1-s, \frac{m}{q} \right) \right) ^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left( \sum\limits^{q}_{m = 1}a_m \sum\limits^{\infty}_{k = l} \frac{ 1 } { ( k+\frac{m}{q} )^{1-s} } \right)^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left[\ \sum\limits^{\infty}_{k = l} \left( \frac{ a_1 } { ( k+\frac{1}{q} )^{1-s} } + \frac{ a_2 } { ( k+\frac{2}{q} )^{1-s}} +\cdots + \frac{ a_q } { ( k+1 )^{1-s} } \right)\ \right]^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left[ \ \sum\limits^{\infty}_{k = l} \frac{1}{k^{1-s}} \left( \frac{a_1}{ (1+\frac{1}{qk} )^{1-s}} +\frac{a_2}{ (1+\frac{2}{qk} )^{1-s} } +\cdots + \frac{a_q}{ (1+\frac{1}{k} )^{1-s} } \right)\ \right]^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left[ \sum\limits^{\infty}_{k = l} \frac{1}{k^{1-s}} \left( \mathop{\sum}\limits^{q}_{m = 1} a_m \mathop{\sum}\limits^{\infty}\limits_{l_1 = 0} \binom{s-1}{l_1}\left( \frac{m}{qk} \right)^{l_1} \right) \ \right]^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left[ \sum\limits^{\infty}_{k = l} \frac{1}{k^{1-s}} \left(a_1 +a_2 +\cdots+ a_q +\frac{s-1}{qk}\mathop{\sum}\limits^{q}\limits_{ m = 1 }ma_m + O_{q, s}\left( \frac{1}{k^2}\right)\ \right)\ \right]^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \left[ \sum\limits^{\infty}_{k = l} \frac{1}{k^{2-s}} \left( \frac{s-1}{q}\mathop{\sum}\limits^{q}\limits_{ m = 1 }ma_m+ O_{q, s}\left( \frac{1}{k}\right)\ \right)\ \right]^{-1} \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \times \left( \sum\limits^{\infty}_{k = l} \frac{1}{k^{2-s}} \right)^{-1} \left( \frac{s-1}{q}\mathop{\sum}\limits^{q}\limits_{ m = 1 }ma_m \right)^{-1} \left( 1 + O_{s, q}\left( \frac{1}{n} \right) \right) \\ & = \frac{(2\pi q)^{1-s}}{2\Gamma(1-s)} \frac{ q }{ (s-1)\mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } \times (1-s) l^{1-s} + O( l^{-s} ) \\ & = \frac{ - \pi q^{2}} { (2q\pi)^s \,\Gamma(1-s)\mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m }l^{1-s} + O_{s,q} \left( l^{-s} \right) .\end{aligned} \end{align} (5.3)

    (ii) Let n = q(l-1)+r for some q \geq r\geq 1 and l, r \in \mathbb{N_+}, if

    \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m\neq 0

    and

    \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m -q \mathop{\sum}\limits^{q} \limits_{ m = r }a_m \neq 0

    hold, we have

    \begin{align} \begin{aligned}\begin{aligned} \left( \zeta_{n}\left( s, \frac{p}{q} \right)\,\right)^{-1} & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) }\left(\, \sum\limits^{r-1}_{m = 1} a_m \zeta_{l+1} \left( 1-s, \frac{m}{q}\right) + \sum\limits^{q}_{m = r} a_m \zeta_{l} \left( 1-s, \frac{m}{q}\right)\, \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left(\ \sum\limits^{q}_{m = 1} a_m \zeta_{l+1} \left( 1-s, \frac{m}{q}\right) + \frac{a_r} { \left( l+ \frac{ r }{q} \right)^{1-s} } + \frac{a_{r+1}} { \left( l+ \frac{ r+1 }{q} \right)^{1-s} } + \cdots +\frac{a_q} { \left( l+ 1 \right)^{1-s} } \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left(\ \sum\limits^{\infty}_{k = l+1} \sum\limits^{q}_{m = 1}\frac{ a_m } { ( k+\frac{m}{q} )^{1-s} } + \frac{a_r} { \left( l+ \frac{ r }{q} \right)^{1-s} } + \frac{a_{r+1}} { \left( l+ \frac{ r+1 }{q} \right)^{1-s} } + \cdots +\frac{a_q} { \left( l+ 1 \right)^{1-s} } \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left(\ \sum\limits^{\infty}_{k = l+1} \sum\limits^{q}_{m = 1}\frac{ a_m } { ( k+\frac{m}{q} )^{1-s} }\right)^{-1} \left(1 + \frac{ \mathop{\sum}\limits^{ q } \limits_{m = r} \frac{a_m} { \left( l+ \frac{ m }{q} \right)^{1-s} } }{ \mathop{\sum}\limits^{\infty}\limits_{k = l+1} \mathop{\sum}\limits^{q}\limits_{m = 1}\frac{ a_m } { ( k+\frac{m}{q} )^{1-s} } } \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left( \frac{ -q }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } l^{1-s} + O_{s,q}( l^{-s} ) \right) \left(1 + \left( \frac{- q }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } l^{1-s} + O_{s,q}( l^{-s} ) \right) l^{s-1} \mathop{\sum}\limits^{ q } \limits_{m = r} \frac{a_m} { \left( 1+ \frac{ m }{ql} \right)^{1-s} } \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left( \frac{ -q }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } l^{1-s} + O_{s,q}( l^{-s} ) \right) \left( 1+ \left( \frac{ -q }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } + O_{s,q}( l^{-1} ) \right) \left( \mathop{\sum}\limits^{q}\limits_{ m = r }a_m + O_{q, s}\left( l^{-1}\right) \right) \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left( \frac{ -q }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } l^{1-s} + O_{s,q}( l^{-s} ) \right) \left( 1-\frac{ q \mathop{\sum}\limits^{q}\limits_{ m = r }a_m }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } +O(l^{-1}) \right)^{-1} \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \left( \frac{ -q }{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m } l^{1-s} + O_{s,q}( l^{-s} ) \right) \frac{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m }{\mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m -q \mathop{\sum}\limits^{q}\limits_{ m = r }a_m } \left( 1+O(l^{-1})\right) \\ & = \frac{ (2\pi q)^{1-s}}{ 2\Gamma(1-s) } \frac{-q}{ \mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m -q \mathop{\sum}\limits^{q} \limits_{ m = r }a_m } l^{1-s} +O(l^{-s}) \\ & = \frac{ -\pi q^2 } {(2\pi q)^{s} \Gamma(1-s) \left(\mathop{\sum}\limits^{q}\limits_{ m = 1 } m a_m -q \mathop{\sum}\limits^{q} \limits_{ m = r }a_m \right) } l^{1-s} +O(l^{-s}) . \end{aligned}\end{aligned} \end{align} (5.4)

    Proof of Theorem 4. (i) First of all, since \chi is a nonprincipal character mod k, we have

    \chi(1) +\chi(2)+\cdots +\chi(k) = 0.

    Let

    n = k(l-1)+1 ,

    if

    \mathop{\sum}\limits^{k}\limits_{ j = 1 } j \chi(j)\neq 0

    holds, and the asymptotic formula of the inverse of the n -th tail of the L-function L(s, \chi) for Re(s) > 1 is as follows:

    \begin{align} \left( \,L_{n}( s, \chi)\,\right)^{-1} & = k^s \left( \sum\limits^k_{j = 1}\chi(j) \zeta_n \left( s,\frac{j}{k} \right) \right) ^{-1} \\ & = \ k^s \left( \sum\limits^{k}_{j = 1}\chi(j) \sum\limits^{\infty}_{i = l} \frac{ 1 } { ( i+\frac{j}{k} )^{s} } \right)^{-1} \\ & = k^s \left[\ \sum\limits^{\infty}_{i = l} \left( \frac{ \chi(1) } { ( i +\frac{1}{k} )^{s} } + \frac{ \chi(2) } { ( i+\frac{2}{k} )^{s}} +\cdots + \frac{ \chi(k) } { ( i+1 )^{s} } \right)\ \right]^{-1} \\ & = k^s \left[ \ \sum\limits^{\infty}_{i = l} \frac{1}{i^{s}} \left( \frac{\chi(1)}{ (1+\frac{1}{ki} )^{s}} +\frac{\chi(2)}{ (1+\frac{2}{ki} )^{s} } +\cdots + \frac{\chi(k)}{ (1+\frac{1}{i} )^{s} } \right)\ \right]^{-1} \\ & = k^s \left( \sum\limits^{\infty}_{i = l} \frac{1}{i^{s}} \left( \chi(1) +\chi(2) +\cdots+ \chi(k) -\frac{s} {ki} \sum\limits^{k}_{j = 1} j \chi(j) + O_{s, k}\left( \frac{1}{i^2} \right) \,\right)\,\right)^{-1} \\ & = k^s \left( \sum\limits^{\infty}_{i = l} \frac{1}{i^{s+1}} \right)^{-1} \left( - \frac{s}{k}\sum\limits^{k}_{j = 1} j \chi(j) + O_{s, k}\left( \frac{1}{l} \right) \right)^{-1} \\ & = k^s \times \frac{ k\left( s l^{s} + O\left( l^{s-1} \right) \right) } { - s \mathop{\sum}\limits^{k}\limits_{j = 1}j \chi(j) } \left( 1 + O_{s, k}\left( \frac{1}{l} \right) \right) \\ & = \frac{ - k^{s+1} } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) }l^{s} + O_{s,k}\left( l^{s-1} \right) . \end{align} (5.5)

    (ii) Let

    n = k(l-1)+r

    for some k \geq r\geq 1 and l, r \in \mathbb{N_+}, if

    \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j)- k \mathop{\sum}\limits^k\limits_{j = r} \chi(j) \neq 0

    and

    \mathop{\sum}\limits^{k}\limits_{ j = 1 } j \chi(j)\neq 0

    hold, we have

    \begin{align} \begin{aligned} \left( L_{n}( s, \chi) \right)^{-1} & = \left[ k^{-s} \left(\, \sum\limits^{r-1}_{j = 1} \chi(j) \zeta_{l+1} \left( s, \frac{j}{k}\right) + \sum\limits^{k}_{j = r} \chi(j) \zeta_{l} \left( s, \frac{j}{k}\right)\, \right)\ \right]^{-1} \\ & = \ k^s \left( \sum\limits^{r-1}_{j = 1}\chi(j) \sum\limits^{\infty}_{i = l+1} \frac{ 1 } { ( i+\frac{j}{k} )^{s} } + \sum\limits^{k}_{j = r}\chi(j) \sum\limits^{\infty}_{i = l} \frac{ 1 } { ( i+\frac{j}{k} )^{s} } \right)^{-1} \\ & = k^s \left(\ \sum\limits^{\infty}_{i = l+1} \sum\limits^{ k }_{j = 1} \frac{ \chi{(j)} } { ( i +\frac{j}{k} )^{s} } + \left( \frac{\chi(r)}{ (l+\frac{r}{k} )^{s}} + \frac{\chi(r+1)}{ (l+\frac{r+1}{k} )^{s} } +\cdots + \frac{\chi(k)}{ (l+ 1 )^{s} } \right)\ \right)^{-1} \\ & = k^s \left[ \ \sum\limits^{\infty}_{i = l+1} \sum\limits^{ k }_{j = 1} \frac{ \chi(j) } { ( i +\frac{j}{k} )^{s} } \left( 1+ \frac{ \mathop{\sum}\limits^{ k }\limits_{j = r } \frac{\chi(j) }{ \left(l+\frac{j}{k} \right)^{s} } }{ \mathop{\sum}\limits^{\infty}\limits_{i = l+1} \mathop{\sum}\limits^{ k }\limits_{j = 1} \frac{ \chi(j) } { ( i +\frac{j}{k} )^{s} } }\right)\ \right]^{-1} \\ & = k^s \left( \frac{ - k } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) }l^{s} + O_{s,k}\left( l^{s-1} \right) \right) \left( 1+ \left( \frac{ - k } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) }l^{s} + O_{s,k}\left( l^{s-1} \right) \right)\ \left( \mathop{\sum}\limits^{ k }\limits_{j = r } \frac{\chi(j) }{ \left(1+\frac{j}{kl} \right)^{s} } \right) \right)^{-1} \\ & = k^s \left( \frac{ - k } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) }l^{s} + O_{s,k}\left( l^{s-1} \right) \right) \left( 1- \frac{ k \mathop{\sum}\limits^k\limits_{j = r} \chi(j) } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) } + O_{s,k}\left( l^{-1} \right) \right)^{-1} \\ & = k^s \times \frac{ -k } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) - k \mathop{\sum}\limits^k\limits_{j = r} \chi(j) } l^{s} +O_{s,k}( l^{s-1} ) \\ & = \frac{ -k^{1+s} } { \mathop{\sum}\limits^{ k }\limits_{j = 1} j\chi(j) - k \mathop{\sum}\limits^k\limits_{j = r} \chi(j) } l^{s} +O_{s,k}( l^{s-1} ) .\end{aligned} \end{align} (5.6)

    (iii) Let

    n = k^2(l-1)+1

    for some l \in \mathbb{N_+} , if

    \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k}\limits_{m = 1}m \chi(j) a_{j,m}\neq 0

    holds, in combination with

    a_{j,m} = \cos \left( \frac{\pi (1-s)}{2}- \frac{2\pi mj}{k} \right),

    we obtain

    \begin{aligned} \sum\limits^{k}_{j = 1} \sum\limits^{k}_{m = 1} \chi(j) a_{j,m} = \sum\limits^{k-1}_{j = 1} \sum\limits^{k}_{m = 1} \chi(j) a_{j,m} = 0. \end{aligned}

    Then, we have

    \begin{align} \begin{aligned} \left( L_{n}( s, \chi) \right)^{-1} & = \left( \frac{ \Gamma(1-s) }{ 2^{-s} \pi^{1-s} k } \sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1} \chi(j) a_{j,m} \zeta_{l} \left( 1-s, \frac{m}{k}\right) \right)^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s) }\left[ \sum\limits^{ \infty } _{i = l} \left(\sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1} \chi(j) a_{j,m} \right) \frac{ 1 }{ \left( i+ \frac{m}{k} \right)^{1-s} } \right]^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s) }\left[ \sum\limits^{ \infty } _{i = l} \frac{ 1 }{ i^{1-s} } \left(\sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1} \frac{ \chi(j) a_{j,m}} { \left( 1+ \frac{m}{ki} \right)^{1-s} } \right) \ \right]^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s) }\left[ \sum\limits^{ \infty } _{i = l} \frac{ 1 }{ i^{1-s} } \left( \sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1} \chi(j) a_{j,m} + \frac{(s-1)} { ki } \sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1}m \chi(j) a_{j,m} +O_{s, k} \left( \frac{1}{i^2} \right) \right)\ \right]^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s) }\left[ \sum\limits^{ \infty } _{i = l} \frac{ 1 }{ i^{2-s} } \left( \frac{(s-1)} { k } \sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1}m \chi(j) a_{j,m} +O_{s, k} \left( \frac{1}{i} \right) \right)\ \right]^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s) }\left( \sum\limits^{ \infty } _{i = l} \frac{ 1 }{ i^{2-s} } \right)^{-1} \left( \frac{(s-1)} { k } \sum\limits^k_{j = 1} \sum\limits^{k}_{m = 1}m \chi(j) a_{j,m} \right) ^{-1} \left( 1+O_{s,k}( l^{-1} ) \right) \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s) }\times \left( (1-s) l^{1-s}+ O( l^{-s} ) \right) \frac{k}{(s-1) \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k}\limits_{m = 1}m \chi(j) a_{j,m} } \left( 1+O_{s,k}( l^{-1} ) \right) \\ & = \frac{ -\pi k^2 }{ (2\pi)^s \Gamma(1-s)\mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k}\limits_{m = 1}m \chi(j) a_{j,m} } l^{1-s} + O_{ s,k}( l^{-s} ) . \end{aligned} \end{align} (5.7)

    (iv) Let

    n = k^2(l-1)+ k (u-1) + r

    for some k \geq u \geq1 , k \geq r\geq 1 , and l, u, r \in \mathbb{N_+} , if

    \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k}\limits_{m = 1}m \chi(j) a_{j,m}\neq 0

    and

    \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k} \limits_{m = 1}m \chi(j) a_{j,m} -k \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\mathop{\sum}\limits^k\limits_{j = u+1} \mathop{\sum}\limits^{k}\limits_{m = 1} \right) \chi(j) a_{j,m} \neq 0

    hold, we have

    \begin{align} \begin{aligned} \left( L_{n}( s, \chi)\right)^{-1} & = \left[ \frac{ \Gamma(1-s) }{ 2^{-s} \pi^{1-s} k } \left( \left( \sum\limits^{u-1}_{j = 1} \sum\limits^{{ k }}_{m = 1} + \mathop{\mathop{\sum}\limits^{ r-1 }\limits_{m = 1} }\limits_{j = u} \right) \chi(j) a_{j,m} \zeta_{l+1} \left( 1-s, \frac{m}{k}\right) + \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\sum\limits^k_{j = u+1} \sum\limits^{k}_{m = 1} \right) \chi(j) a_{j,m} \zeta_l \left( 1-s, \frac{m}{k}\right) \right)\ \right]^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s)} \left( \sum\limits^{k}_{j = 1} \sum\limits^{{k}}_{m = 1} \chi(j) a_{j,m} \zeta_{l+1} \left( 1-s, \frac{m}{k}\right)+ \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\sum\limits^k_{j = u+1} \sum\limits^{k}_{m = 1} \right) \frac{ \chi(j) a_{j,m} }{ \left(l+ \frac{m}{k} \right)^{1-s} } \right)^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s)} \left( \sum\limits^{k}_{j = 1} \sum\limits^{{k}}_{m = 1} \chi(j) a_{j,m} \zeta_{l+1} \left( 1-s, \frac{m}{k}\right) \right) ^{-1} \left( 1+ \frac { \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\mathop{\sum}\limits^k\limits_{j = u+1} \mathop{\sum}\limits^{k}\limits_{m = 1} \right) \frac{ \chi(j) a_{j,m} } { \left(l+ \frac{m}{k} \right)^{1-s} } }{ \mathop{\sum}\limits^{k}\limits_{j = 1} \mathop{\sum}\limits^{{k}}\limits_{m = 1} \chi(j) a_{j,m} \zeta_{l+1} \left( 1-s, \frac{m}{k}\right) } \right)^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s)} \left( \frac{-k}{ \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k} \limits_{m = 1}m \chi(j) a_{j,m} } l^{1-s} +O_{s,k}( l^{-s} ) \right) \left( 1- \frac{ k \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\mathop{\sum}\limits^k\limits_{j = u+1} \mathop{\sum}\limits^{k}\limits_{m = 1} \right) \chi(j) a_{j,m} } { \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k} \limits_{m = 1}m \chi(j) a_{j,m} } +O_{s,k}\left( \frac{1}{l}\right) \right)^{-1} \\ & = \frac{ 2^{-s} \pi^{1-s} k } { \Gamma(1-s)} \times \frac{ -k }{ \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k} \limits_{m = 1}m \chi(j) a_{j,m} -k \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\mathop{\sum}\limits^k\limits_{j = u+1} \mathop{\sum}\limits^{k}\limits_{m = 1} \right) \chi(j) a_{j,m} }l^{1-s} +O_{s,k}(l^{-s} ) \\ & = \frac{ -\pi k^2 }{ (2 \pi)^s \Gamma(1-s) \left( \mathop{\sum}\limits^k\limits_{j = 1} \mathop{\sum}\limits^{k} \limits_{m = 1}m \chi(j) a_{j,m} -k \left( \mathop{\mathop{\sum}\limits^{ k }\limits_{m = r} }\limits_{j = u} +\mathop{\sum}\limits^k\limits_{j = u+1} \mathop{\sum}\limits^{k}\limits_{m = 1} \right) \chi(j) a_{j,m} \right) } l^{1-s} +O_{s,k} ( l^{-s}) . \end{aligned} \end{align} (5.8)

    In this paper, we derive the asymptotic formulas for the reciprocal sums of the Riemann zeta function, the Hurwitz zeta function, and the Dirchlet L-function for Re(s) > 1 and Re(s) < 0 .

    Zhenjiang Pan: writing-review and editing, writing-original draft, validation, resources, methodology, formal analysis, conceptualization.

    Zhengang Wu: writing-review and editing, resources, methodology, supervision, validation, formal analysis, funding acquisition.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors express their gratitude to the referee for very helpful and detailed comments. Supported by the National Natural Science Foundation of China (Grant No. 11701448).

    The authors declare no conflicts of interest.



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