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The inverses of tails of the generalized Riemann zeta function within the range of integers

  • In recent years, many mathematicians researched infinite reciprocal sums of various sequences and evaluated their value by the asymptotic formulas. We study the asymptotic formulas of the infinite reciprocal sums formed as (k=n1kr(k+t)s)1 for r,s,tN+, where the asymptotic formulas are polynomials.

    Citation: Zhenjiang Pan, Zhengang Wu. The inverses of tails of the generalized Riemann zeta function within the range of integers[J]. AIMS Mathematics, 2023, 8(12): 28558-28568. doi: 10.3934/math.20231461

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  • In recent years, many mathematicians researched infinite reciprocal sums of various sequences and evaluated their value by the asymptotic formulas. We study the asymptotic formulas of the infinite reciprocal sums formed as (k=n1kr(k+t)s)1 for r,s,tN+, where the asymptotic formulas are polynomials.



    Throughout the years, many mathematicians have been working on partial infinite sums of reciprocal linear recurrence sequences.

    In 2011, Takao Komatsu [8] researched the nearest integer of the sum of reciprocal Fibonacci numbers and derived

    (k=n1Fk)1=FnFn1,(k=n(1)nFk)1=(1)n(Fn+Fn+1), (1.1)

    where denoted the nearest integer; in other words, x=x12.

    In 2020, Ho-Hyeong Lee and Jong-Do Park [9] gave the concept of asymptotic formulas, which were more accurate. The conclusions were as follows:

    (k=n1FkFk+2l)1Fn+l1Fn+l(F2l+(1)l)(1)n3,(k=n1FkFk+2l1)1F2n+l1(Fl1Fl+(1)l)(1)n3, (1.2)

    where anbn meant limn(anbn)=0. For more results related to the infinite reciprocal sums of linear recurrence sequences, see [1,13,15] and references therein.

    The zeta function ζ(z) is undoubtedly the most famous function in analytic number theory. Initially studied by Euler and achieved prominence with Riemann, it abstracted the attention of many mathematicians. Another well-known sequence harmonic number Hn is the sum of the first n terms of ζ(z) when z=1, and the generating function of harmonic numbers n=1Hnxn is an important tool to study the property of Hn. Kim [4,5,6,7] derived many worthy and interesting results associated with the zeta function, harmonic number and its generating function, which inspired us deeply.

    At the same time, many researchers began to study the tails of well-known functions such as the Riemann zeta function and the Hurwitz zeta function in [2,3,10,12,14].

    For example, Kim Donggyun and Song Kyunghwan [3] studied the inverses of tails of the Riemann zeta function. Derived for s on the critical strip 0<s<1,

    (k=n1ks)1{2(121s)(n12)s,ifnis even,2(121s)(n12)s,ifnis odd. (1.3)

    Ho-Hyeong Lee and Jong-Do Park [10] dealt with the inverses of tails of Hurwitz zeta function when s2, sN and 0a<1, and derived

    (k=n1(k+a)s)1  s1j=0Aj(n+a)j, (1.4)

    where As1=s1, Asl=sl1j=1xsjAsl+j, xsj=(s2+jj)Bj and Bj are Bernoulli numbers.

    In this paper, we extend their asymptotic formulas for the methods and results by considering the tails of (k=n1k(k+t))1, (k=n1[k(k+t)]2)1, (k=n1kr(k+t)s)1 and further revealing the property of reciprocal sums of the various sequences.

    Before our conclusion, we define (i1)=0 for all iN+, which will take effect in expressing the asymptotic formulas in Theorem 3.

    Theorem 1. For all mN, we have

    (k=n1k(k+t))1n+t212.

    Theorem 2. For all mN, we have

    (k=n1[k(k+t)]2)13n3+an2+bn+c,

    where

    a=92t92,b=2720t292t+154,c=340t32740t2+158t98.

    Theorem 3. If r+s1>0 and r,s,tN, then there exists the unique polynomial

    B(n,r,s,t)=br+s1nr+s1+br+s2nr+s2++b1n+b0,

    subject to

    (k=n1nr(n+t)s)1B(n,r,s,t),

    where

    br+s1=r+s1,br+s2=(r+s1)22(r+s)[(2st(r+s)],br+s3=(r+s1)212(r+s)2(r+s+1)[6s(r2sr2+2rs22rs+s3s22s)t2  6s(r+3r2sr2+3rs22rs2r+s3s22s)t  +(2r4+8r3sr3+12r2s23r2s3r2+8rs33rs26rs+2s4s33s2)],                              birs+1=k1+k2=iirs+1k1r+s2rk2r+s(sk2r)tr+sk2r+s1j=k1+1bj(jk1)3r+3si3k3+k4=iirs+2k3,k4r+s1bk3r+s1j=k4bj(jk4)3r+3si3
      +r+s1j=irs+2bj[(jirs)(r+s1)(jirs+1)]3r+3si3,                              b0=k1+k2=i0k1r+s2rk2r+s(sk2s)tr+sk2r+s1j=k1+1bj(jk1)2(r+s1)k3+k4=i1k3,k4r+s1bk3r+s1j=k4bj(jk4)+(r+s1)r+s1j=1bj2(r+s1).

    Remark. From Theorem 3, we derive that coefficients bj (0jr+s1) are determined by r,s and t. At the same time, if we calculate b0 by using the representation of birs+1, there will appear (r+s11),(r+s21),,(11). In order to make b0 the representation of bj, we give the definition (i1)=0 for iN+ in this paper. Undoubtedly, it satisfies (n+10)=(n0)+(n1) for nN+.

    Corollary 1. If s=1 and m=0, we have

    (k=n1n2)1n12.

    Corollary 2. If s=2 and t=0, we have

    (k=n1n4)13n392n2+154n98.

    Corollary 3. If s=2, we have

    (k=n1n2(n+1)2)13n3+35n,(k=n1n2(n+2)2)13n3+92n2+320n58,(k=n1n2(n+3)2)13n3+9n2+125n185.

    We need to solve several lemmas for the proof.

    Lemma 1. Let {an}n=1 and {bn}n=1 be sequences of a positive real number with limnan=limnbn=0. If an<bn+an+1 hold for any nN+, then we have

     an<k=nbn     fornN+.

    Proof. See Lemma 2.1 [11].

    Lemma 2. For all t2 and tN, we have

    1n+t212<1n(n+t)+1n+t2+12.

    Proof. It is equivalent with

    1n+t2121n(n+t)1n+t2+12<0. (3.1)
    The left side=n(n+t)(n+t212)(n+t2+12)(n+t212)(n+t2+12)n(n+t)=t24+14(n+t212)(n+t2+12)n(n+t).

    Hence, we have

    t24+14<0     for n>2,

    so

    1n+t212<1n(n+t)+1n+t2+12      for n>2.

    This completes the proof.

    Lemma 3. For all ε>0, there exists N0>2, subject to

    1n+t212ε>1n+t2+12ε+1n(n+t)     forn>N0.

    Proof. It is equivalent with

    1n+t212ε1n+t2+12ε1n(n+t)>0. (3.2)
    The left side=n2+tn(n2+tn+t24)142ε(n+t2)+ε2(n+t212ε)(nt2+12ε)n(n+t)=2εn+tε+14t24ε2(n+t212ε)(nt2+12ε)n(n+t).

    We can restrict ε<1 and fix t, then we have

    tε+14t24ε2=O(1),

    so there exists N0>2, subject to

    2εn+O(1)>0     for n>N0,

    then

    1n+t212ε>1n+t2+12ε+1n(n+t),

    which proves (3.2) and completes the proof.

    Proof of Theorem 1.

    Case 1. When  t2.

    By Lemmas 1–3, we have for all ε>0, there exists N0>2, subject to

    n+t212ε<(k=n1n(n+t))1<n+t212     forn>N0, (3.3)

    hence

    |(k=t1k(k+t))1(n+t212)|<ε;

    in other words,

    (k=n1k(k+t))1(n+t212).

    Case 2. When t=1,

    k=n1k(k+1)=1n(n+1)+1(n+1)(n+2)+1(n+2)(n+3)+=(1n1n+1)+(1n+11n+2)+(1n+21n+3)+=1n,

    hence

    (k=n1k(k+1))1n.

    Case 3. When t=0, the proof is similar with Case 1, and we can easily deduce the result.

    Lemma 4. Let f(n,t,ε)=3n3+an2+bn+c and a,b,c are defined in Theorem 2. Then for all ε>0, there exists N1>0, subject to

    1f(n,t,ε)<1f(n+1,t,ε)+1n2(n+t)2     forn>N1.

    Proof. It is equivalent with

    1f(n,t,ε)1f(n+1,t,ε)1n2(n+t)2<0      for n>N1. (3.4)
    The left side=[f(n+1,t,ε)f(n,t,ε)]n2(n+t)2f(n,t,ε)f(n+1,t,ε)f(n,t,ε)f(n+1,t,ε)n2(n+t)2=[9n2+(2a+9)n+(a+b+3)]n2(n+t)2f(n,t,ε)f(n+1,t,ε)n2(n+t)2     (3n3+an2+bn+c)f(n,t,ε)f(n+1,t,ε)n2(n+t)2[3n3+(a+9)n2+(2a+b+9)n+(a+b+c+3+ε)]=6εn3+A(t)O(n2)f(n,t,ε)f(n+1,t,ε)n2(n+t)2,

    where A(t) is a function with variable t, then we fix t for all ε>0, and there exists N1>0, subject to

    6εn3+A(t)O(n2)=6εn3+O(n2)<0,      forn>N1,

    hence

    1f(n,t,ε)1f(n+1,t,ε)1n2(n+t)2<0,

    and this completes the proof.

    Lemma 5. Let g(n,t,ε)=3n3+an2+bn+cε, then for all ε>0 there exists N2>0, subject to

    1g(n,t,ε)>1g(n+1,t,ε)+1n2(n+t)2     forn>N2.

    Proof. The proof is similar with Lemma 4, and we can easily deduce the result.

    Proof of Theorem 2. By Lemmas 1, 4 and 5, we have for all ε>0. There exists N3=max{N1,N2}>0, subject to

    1f(n,t,ε)<k=n1n2(n+t)2<1g(n,t,ε)      forn>N3, (3.5)

    hence

    3n3+an2+bn+cε<(k=n1n2(n+t)2)1<3n3+an2+bn+c+ε,

    which is equivalent to

    (k=n1n2(n+t)2)13n3+an2+bn+c.

    Proof of Theorem 3. According to the method of proving Theorem 2, it is enough to prove that there exists polynomial

    B(n)=blnl+bl1nl1++b1n+b0,

    subject to

    1B(n)1B(n+1)1nr(n+t)s=O(nl1)B(n)B(n+1)nr(n+t)s. (3.6)
    The left side =B(n+1)B(n)B(n)B(n+1)1nr(n+t)s=[B(n+1)B(n)]nr(n+t)sB(n)B(n+1)B(n)B(n+1)nr(n+t)s.

    Let

    C(n)=B(n+1)B(n),D(n)=nr(n+t)s,E(n)=B(n),F(n)=B(n+1).

    Therefore, it is enough to prove

    C(n)D(n)E(n)F(n)=O(nl1), (3.7)

    hence we have the necessary condition (1):

    α(C(n)D(n))=α(E(n)F(n)),

    the notation α(f(x)) means the order of f(x), then we have

    l=r+s1. (3.8)

    We note the number of coefficients is r+s, then we have

    C(n)D(n)=(r+s1i=1(bii1j=0(ij)nj))(nr(sp=0(sp)tspnp))=(r+s2i=0(r+s1j=i+1bj(ji))ni)(r+sp=r((spr)tr+sp)np)=2r+2s2i=r(k1+k2=i,0k1r+s2,rk2r+s(sk2r)tr+sk2r+s1j=k1+1bj(jk1))ni=2r+2s2i=rk1+k2=i,0k1r+s2,rk2r+s(sk2r)tr+sk2r+s1j=k1+1bj(jk1)ni

    and

    E(n)F(n)=(r+s1i=0bini)(r+s1q=0bq(qj=0(qj)nj))=(r+s1i=0bini)(r+s1q=0(r+s1j=qbj(jq))nq)=2r+2s2i=0(k3+k4=i,0k3,k4r+s1bk3r+s1j=k4bj(jk4))ni=2r+2s2i=0k3+k4=i,0k3,k4r+s1bk3r+s1j=k4bj(jk4)ni.

    We get the necessary condition (2): If r+s1i2r+2s2, the coefficients of C(n)D(n) and E(n)F(n) are equal, which is equivalent to the system of equations as follows:

    k1+k2=i,0k1r+s2,rk2r+s(sk2r)tr+sk2r+s1j=k1+1bj(jk1)=k3+k4=i,0k3,k4r+s1bk3r+s1j=k4bj(jk4), (3.9)

    where i=2r+2s2, 2r+2s3, , r+s and r+s1.

    We rewrite the system of (3.9) as

    {                    (r+s1)br+s1=b2r+s1,                                           [(r+s2)2br+s1]br+s2=f1(br+s1),                                       [(r+s3)2br+s1]br+s3=f2(br+s1,br+s2),                                                                                                                                            2br+s1b0=fs+r(br+s1,br+s2,,b1). (3.10)

    Clearly the first equation of (3.10) has two solutions, br+s1=0 and br+s1=r+s1, and combined with (3.8) l=r+s1, we have

    br+s1=r+s1. (3.11)

    Substitute (3.11) into the second equation of (3.10). We have the coefficient in the left side [(r+s2)2br+s1] as not equal to zero, and the right side f1(br+s1) as a constant, then the second equation of (3.10) has unique solution.

    Repeat the above process for every equation of (3.10). The coefficient in the left side is never equal to zero, and the right side is always a constant, which implies the system of equations has a unique solution, denoted it by (b2s1, b2s2, b2s3,  b1, b0) with

    br+s1=r+s1,br+s2=(r+s1)22(r+s)[(2st(r+s)],br+s3=(r+s1)212(r+s)2(r+s+1)[6s(r2sr2+2rs22rs+s3s22s)t2    6s(r+3r2sr2+3rs22rs2r+s3s22s)t    +(2r4+8r3sr3+12r2s23r2s3r2+8rs33rs26rs+2s4s33s2)],                              
    birs+1=k1+k2=iirs+1k1r+s2rk2r+s(sk2r)tr+sk2r+s1j=k1+1bj(jk1)3r+3si3k3+k4=iirs+2k3,k4r+s1bk3r+s1j=k4bj(jk4)3r+3si3    +r+s1j=irs+2bj[(jirs)(r+s1)(jirs+1)]3r+3si3,                    b0=k1+k2=i0k1r+s2rk2r+s(sk2s)tr+sk2r+s1j=k1+1bj(jk1)2(r+s1)k3+k4=i1k3,k4r+s1bk3r+s1j=k4bj(jk4)+(r+s1)r+s1j=1bj2(r+s1),

    where we define (j1)=0 for all jN+.

    It is clear that the coefficient bi is determined by r,s and t, and the solution is corresponded to a polynomial with (r+s-1)-order denoted by B(n,r,s,t). We can easily prove B(n,r,s,t) satisfies (3.6), and then we have

    (k=n1nr(n+t)s)1B(n,r,s,t).

    In this paper we discussed the reciprocal sums of the generalized Riemann zeta function within the range of integers, and we also considered other functions within the range of integers.

    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 conflict of interest.



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