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

1.   Introduction

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

where $\left\Vert \cdot \right\Vert$ denoted the nearest integer; in other words, $\left\Vert x \right\Vert = \lfloor x-\frac{1}{2} \rfloor.$

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:

where $a_{n}\sim b_{n}$ meant $\lim\limits_{n\rightarrow \infty}(a_{n}-b_{n}) = 0.$ For more results related to the infinite reciprocal sums of linear recurrence sequences, see ^[1,13,15] and references therein.

The zeta function $\zeta(z)$ 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 $H_n$ is the sum of the first n terms of $\zeta(z)$ when $z = 1$, and the generating function of harmonic numbers $\sum^{\infty}_{n = 1} H_nx^n$ is an important tool to study the property of $H_n$. 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 $0 < s < 1$,

Ho-Hyeong Lee and Jong-Do Park ^[10] dealt with the inverses of tails of Hurwitz zeta function when $s\geq$2, $s\in \mathbb{N}$ and $0\leq a < 1$, and derived

where $A^*_{s-1} = s-1$, $A^s_{l} = -\sum^{s-l-1}_{j = 1}x^s_jA^s_{l+j}$, $x^s_j = \binom{s-2+j}{j}B_j$ and $B_j$ are Bernoulli numbers.

In this paper, we extend their asymptotic formulas for the methods and results by considering the tails of $\left(\sum^{\infty}_{k = n} \frac{1}{k(k+t)}\right)^{-1}$, $\left(\sum^{\infty}_{k = n} \frac{1}{[k(k+t)]^{2}} \right) ^{-1}$, $\left(\sum^{\infty}_{k = n} \frac{1}{k^r(k+t)^s}\right)^{-1}$ and further revealing the property of reciprocal sums of the various sequences.

2.   Main results

Before our conclusion, we define $\binom{i}{-1} = 0$ for all $i \in \mathbb{N^+}$, which will take effect in expressing the asymptotic formulas in Theorem 3.

Theorem 1. For all $m \in \mathbb{N}$, we have

Theorem 2. For all $m\in \mathbb{N}$, we have

where

Theorem 3. If $r+s-1 > 0$ and $r, s, t \in \mathbb{N}$, then there exists the unique polynomial

subject to

where

Remark. From Theorem 3, we derive that coefficients $b_{j} \ (0 \leq j \leq r+s-1)$ are determined by $r, s$ and $t$. At the same time, if we calculate $b_0$ by using the representation of $b_{i-r-s+1}$, there will appear $\binom{r+s-1}{-1}, \binom{r+s-2}{-1}, \cdots, \binom{1}{-1}$. In order to make $b_0$ the representation of $b_j$, we give the definition $\binom{i}{-1} = 0$ for $i \in \mathbb{N^+}$ in this paper. Undoubtedly, it satisfies $\binom{n+1}{0} = \binom{n}{0}+\binom{n}{-1}$ for $n \in \mathbb{N^+}$.

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

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

Corollary 3. If $s = 2$, we have

3.   Proof of theorem

3.1.   Proof of Theorem 1

We need to solve several lemmas for the proof.

Lemma 1. Let $\{a_n\}^{\infty}_{n = 1}$ and $\{b_n\}^{\infty}_{n = 1}$ be sequences of a positive real number with $\underset{n\mapsto\infty}\lim a_n = \underset{n\mapsto\infty}\lim b_n = 0$. If $a_{n} < b_{n}+a_{n+1}$ hold for any $n\in N^{+}$, then we have

Proof. See Lemma 2.1 ^[11]. □

Lemma 2. For all $t\geq2$ and $t\in \mathbb{N}$, we have

Proof. It is equivalent with

Hence, we have

so

This completes the proof. □

Lemma 3. For all $\varepsilon > 0$, there exists $N_{0} > 2$, subject to

Proof. It is equivalent with

We can restrict $\varepsilon < 1$ and fix $t$, then we have

so there exists $N_{0} > 2$, subject to

then

which proves (3.2) and completes the proof. □

Proof of Theorem 1.

Case 1. When $\ t\geq 2$.

By Lemmas 1–3, we have for all $\varepsilon > 0$, there exists $N_{0} > 2$, subject to

hence

in other words,

Case 2. When $t = 1$,

hence

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

□

3.2.   Proof of Theorem 2

Lemma 4. Let $f(n, t, \varepsilon) = 3n^3+an^2+bn+c$ and $a, b, c$ are defined in Theorem 2. Then for all $\varepsilon > 0$, there exists $N_{1} > 0$, subject to

Proof. It is equivalent with

where $A(t)$ is a function with variable $t$, then we fix $t$ for all $\varepsilon > 0$, and there exists $N_{1} > 0$, subject to

hence

and this completes the proof.□

Lemma 5. Let $g(n, t, \varepsilon) = 3n^3+an^2+bn+c-\varepsilon$, then for all $\varepsilon > 0$ there exists $N_{2} > 0$, subject to

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 $\varepsilon > 0$. There exists $N_{3} = max \{ N_{1}, N_{2} \} > 0$, subject to

hence

which is equivalent to

□

3.3.   Proof of Theorem 3

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

subject to

Let

Therefore, it is enough to prove

hence we have the necessary condition (1):

the notation $\alpha(f(x))$ means the order of $f(x)$, then we have

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

and

We get the necessary condition (2): If $r+s-1 \leq i \leq 2r+2s-2$, the coefficients of $C(n)D(n)$ and $E(n)F(n)$ are equal, which is equivalent to the system of equations as follows:

where $i = 2r+2s-2, \ 2r+2s-3, \ \cdots, \ r+s\ and\ r+s-1$.

We rewrite the system of (3.9) as

Clearly the first equation of (3.10) has two solutions, $b_{r+s-1} = 0$ and $b_{r+s-1} = r+s-1,$ and combined with (3.8) $l = r+s-1$, we have

Substitute (3.11) into the second equation of (3.10). We have the coefficient in the left side $\mathop{[} (r+s$$-2) -2 b_{r+s-1} \mathop{]}$ as not equal to zero, and the right side $f_1(b_{r+s-1})$ 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 $( b_{2s-1}, \ b_{2s-2}, \ b_{2s-3}, \ \cdots\, \ b_1, \ b_0 )$ with

where we define $\binom{j}{-1} = 0$ for all $j \in \mathbb{N^+}.$

It is clear that the coefficient $b_{i}$ 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

□

4.   Conclusions

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.

Use of AI tools declaration

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

Acknowledgments

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).

Conflict of interest

The authors declare no conflict of interest.