The inverses of tails of the Riemann zeta function

We present some bounds of the inverses of tails of the Riemann zeta function on 0<s<1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0 < s < 1$\end{document} and compute the integer parts of the inverses of tails of the Riemann zeta function for s=12,13\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$s=\frac{1}{2}, \frac{1}{3}$\end{document}, and 14\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\frac{1}{4}$\end{document}.


Introduction
The Riemann zeta function ζ (s) in the real variable s was introduced by Euler [2] in connection with questions about the distribution of prime numbers. Later Riemann [6] derived deeper results about a dual correspondence between the distribution of prime numbers and the complex zeros of ζ (s) in the complex variable s. In these developments, he asserted that all the non-trivial zeros of ζ (s) are on the line Re(s) = 1 2 , and this has been one of the most important unsolved problems in mathematics, called the Riemann hypothesis. A vast amount of research on calculation of ζ (s) on the line Re(s) = 1 2 , which is called the critical line, and on the strip 0 < Re(s) < 1, which is called the critical strip, has been conducted using various methods [1].
The Riemann zeta function and a tail of the Riemann zeta function from n for an integer n ≥ 1 are defined, respectively, by: for Re(s) > 1, To understand the values of ζ (s), it would be helpful to understand the values of tails of ζ (s), for example, the integer parts of their inverses [ζ n (s) -1 ], where [x] denotes the greatest integer that is less than or equal to x. Some values of [ζ n (s) -1 ] for small positive integers s have become known recently. Xin [7] showed that for s = 2 and 3, ζ n (2) -1 = n -1 and ζ n (3) -1 = 2n(n -1).
We study the inverses of tails of the Riemann zeta function ζ n (s) -1 for s on the critical strip 0 < s < 1. The following notation is needed to explain our results. Definition 1 For any positive integer n and real number s with 0 < s < 1, we define A n,s = 1 n s -1 (n + 1) s + 1 (n + 2) s -1 (n + 3) s + · · · and B n,s = -1 n s + 1 (n + 1) s + - Now the tail of the Riemann zeta function for 0 < s < 1 can be written as follows: 1-2 1-s A n,s , if n is even, - 1 1-2 1-s B n,s , if n is odd. (1) In this paper, we present the bounds of A -1 n,s and B -1 n,s , hence the bounds of the inverses of tails of the Riemann zeta function ζ n (s) -

The bounds of the inverses of ζ n (s) for 0 < s < 1
In this section, we present the bounds of A -1 n,s and B -1 n,s in Definition 1, hence the bounds of the inverses of tails of the Riemann zeta function ζ n (s) -1 for 0 < s < 1.
Proposition 1 Let s be a real number with 0 < s < 1. Then, for any positive even number n, n,s < 2n s , and for any positive odd number n, Proof Let n be a positive even number. For every positive integer k, it is easy to see that The summations of each term over k give Therefore, we have 1 2n s < A n,s < 1 2(n -1) s , which gives the first statement.
The second statement can be shown similarly.
Since every proof of the case when n is an odd number is analogous to that of the case when n is an even number, we omit all the proofs of the odd number cases in this paper. Now we find tighter bounds for A -1 n,s and B -1 n,s .

Proposition 2
Let s be a real number with 0 < s < 1. Then, for any positive even number n, and for any positive odd number n, Proof Let n be a positive even number. We will show that Rewriting each of the both sides as a series we will show that for any positive integer k, For this, we let and will show that f (x) is positive for x ≥ 1 and 0 < s < 1. With Since the function 1 x s+1 is convex, we obtain that and therefore g (x) is negative, that is, g(x) is decreasing. We conclude that f (x) is positive, which gives the statement. Proposition 3 Let s be a real number with 0 < s < 1. Then, for any positive even number n, and for any positive odd number n, Proof Let n be a positive even number. We will show that Rewriting each of the both sides as a series we need to show that for any positive integer k, For this, we let We check that f (1) > 0 and now we will show that f (x) is positive for x ≥ 2 and 0 < s < 1. With , so we only need to show that g(x) is decreasing. Consider the derivative of g(x): .
We combine the results of Proposition 2 and Proposition 3. We express these bounds in terms of ζ n (s) using expression (1). For this, let

Corollary 1 Let s be a real number with
and we will show that f (x) is positive for x ≥ x 0 , where x 0 is a sufficiently large number. With we have that f (x) = g(x)-g(x + 1 2 ), so we only need to show that g(x) is decreasing. Consider the derivative of g(x): Since 1 x s+1 is decreasing and convex, by comparing slopes at (2x -1 2 + ) and (2x + 1 2 + ), we obtain Consider h(x) := 1 2 -1 2 + ( 2x+ 1 2 + 2x-1 2 + ) s+2 , which is the ratio of two terms on the right-hand side of the above expression. We need to show that h(x) < 1 for every x > x 0 , where x 0 is a sufficiently large number. We check that For any > 0 and 0 < s < 1, we have that 1 < ( 1 2 + 1 2 -) 1/(s+2) and 2x+ 1 2 + 2x-1 2 + is larger than 1, decreasing and converges to 1 as x goes to infinity, so there is x 0 such that, for every x > x 0 , h(x) < 1. Therefore the proof is complete.
We express these bounds in terms of ζ n (s) using expression (1).

Corollary 2
For any positive number and any real number s with 0 < s < 1, we have Note that 2(n -1 4 ) 1/2 -2(n -1 2 ) 1/2 < 1 for n ≥ 2, and it implies that there is at most one integer in the open interval from 2(n - 1 2 ) 1/2 to 2(n -1 4 ) 1/2 . Suppose that there is an integer h in the open interval, i.e., 2 n -1 2 There is, however, no integer in the open interval from 4n -2 to 4n -1, therefore such an integer h does not exist. This gives the statement.

Corollary 3 For any positive integer n,
We express this result in terms of ζ n (s) using expression (1). . Note that 2(n -1 4 ) 1/4 -2(n -1 2 ) 1/4 < 1 for n ≥ 2, and it implies that there is at most one integer in the open interval from 2(n - 1 2 ) 1/4 to 2(n - This shows that the integer h 4 is one of the form 16n -7, 16n -6, or 16n -5. For any integer h, however, h 4 ≡ 0 or 1 (mod 16), hence such an integer h does not exist. Therefore this gives the statement.

Corollary 4 For any positive integer n,
We express this result in terms of ζ n (s) using expression (1).

Corollary 5
For any positive integer n, We express the results of Theorems 3, 4, and 5 in a single statement.