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The inverses of tails of the Riemann zeta function

Journal of Inequalities and Applications20182018:157

https://doi.org/10.1186/s13660-018-1743-6

  • Received: 27 March 2018
  • Accepted: 20 June 2018
  • Published:

Abstract

We present some bounds of the inverses of tails of the Riemann zeta function on \(0 < s < 1\) and compute the integer parts of the inverses of tails of the Riemann zeta function for \(s=\frac{1}{2}, \frac{1}{3}\), and \(\frac{1}{4}\).

Keywords

  • 11M06
  • 11B83

MSC

  • Riemann zeta function
  • Tails of Riemann zeta function
  • Inverses of tails of the Riemann zeta function

1 Introduction

The Riemann zeta function \(\zeta (s)\) 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 \(\zeta (s)\) in the complex variable s. In these developments, he asserted that all the non-trivial zeros of \(\zeta (s)\) are on the line \(\operatorname{Re}(s) = \frac{1}{2}\), 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 \(\zeta (s)\) on the line \(\operatorname{Re}(s) = \frac{1}{2}\), which is called the critical line, and on the strip \(0 < \operatorname{Re}(s) < 1\), 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 \(n \geq 1\) are defined, respectively, by: for \(\operatorname{Re}(s) > 1\),
$$ \zeta (s) = \sum_{k=1}^{\infty } \frac{1}{k^{s}} \quad \mbox{and} \quad \zeta_{n} (s) = \sum _{k=n}^{\infty } \frac{1}{k^{s}}, $$
and for \(0<\operatorname{Re}(s)<1\),
$$ \zeta (s) =\frac{1}{1-2^{1-s}} \sum_{k=1}^{\infty } \frac{(-1)^{k+1}}{k ^{s}} \quad \mbox{and} \quad \zeta_{n} (s) = \frac{1}{1-2^{1-s}} \sum_{k=n}^{\infty } \frac{(-1)^{k+1}}{k^{s}}. $$

To understand the values of \(\zeta (s)\), it would be helpful to understand the values of tails of \(\zeta (s)\), for example, the integer parts of their inverses \([ \zeta_{n} (s)^{-1} ] \), where \([x]\) denotes the greatest integer that is less than or equal to x.

Some values of \([ \zeta_{n} (s)^{-1} ] \) for small positive integers s have become known recently. Xin [7] showed that for \(s = 2\) and 3,
$$ \bigl[ \zeta_{n} (2)^{-1} \bigr] =n-1 \quad \mbox{and} \quad \bigl[ \zeta_{n} (3)^{-1} \bigr] =2n(n-1). $$
For \(s=4\), Xin and Xiaoxue [8] showed that
$$ \bigl[ \zeta_{n} (4)^{-1} \bigr] = 3n^{3} -5n^{2} + 4n-1+ \biggl[ \frac{(2n+1)(n-1)}{4} \biggr] $$
for any integer \(n \geq 2\), and Xu [9] showed that for \(s = 5\),
$$ \bigl[ \zeta_{n} (5)^{-1} \bigr] = 4n^{4} - 8n^{3} + 9n^{2} - 5n + \biggl[ \frac{(n+1)(n-2)}{3} \biggr] $$
for any integer \(n \geq 4\). Hwang and Song [3] provided an alternative proof of the case when \(s = 5\) and a formula when \(s = 6\) as follows. For an integer n, write \(n_{48}\) for the remainder when n is divided by 48, then
$$\begin{aligned} & \bigl[ \zeta_{n} (6)^{-1} \bigr] \\ &\quad = \textstyle\begin{cases} 5n^{5} - \frac{25}{2} n^{4} +\frac{75}{4} n^{3} -\frac{125}{8} n^{2} + \frac{185}{48}n - \frac{5 n_{48}}{48} - [ \frac{35-5 n_{48}}{48} ] ,&\mbox{if~$n$~is even}, \\ 5n^{5} - \frac{25}{2} n^{4} +\frac{75}{4} n^{3} -\frac{125}{8} n^{2} +\frac{185}{48}n - \frac{5 n_{48}+18}{48} - [ \frac{17-5 n_{48}}{48} ] ,& \mbox{if~$n$~is odd} \end{cases}\displaystyle \end{aligned}$$
for any integer \(n \geq 829\). For the integer s greater than 6, no such a formula is known.

There are other interesting results related to this theme such as bounds of \(\zeta (3)\) in greater precision in [4] and [5].

We study the inverses of tails of the Riemann zeta function \(\zeta_{n}(s)^{-1}\) 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
$$\begin{aligned} A_{n,s}= \biggl( \frac{1}{n^{s}} - \frac{1}{(n+1)^{s}} \biggr) + \biggl( \frac{1}{(n+2)^{s}} - \frac{1}{(n+3)^{s}} \biggr) + \cdots \end{aligned}$$
and
$$\begin{aligned} B_{n,s} = \biggl( -\frac{1}{n^{s}} + \frac{1}{(n+1)^{s}} \biggr) + \biggl( -\frac{1}{(n+2)^{s}} + \frac{1}{(n+3)^{s}} \biggr) + \cdots . \end{aligned}$$
Now the tail of the Riemann zeta function for \(0< s<1\) can be written as follows:
$$ \zeta_{n}(s) = \textstyle\begin{cases} - \frac{1}{1-2^{1-s}} A_{n,s}, &\mbox{if~$n$~is even}, \\ - \frac{1}{1-2^{1-s}} B_{n,s},& \mbox{if~$n$~is odd}. \end{cases} $$
(1)

In this paper, we present the bounds of \(A_{n,s}^{-1}\) and \(B_{n,s} ^{-1}\), hence the bounds of the inverses of tails of the Riemann zeta function \(\zeta_{n}(s)^{-1}\) for \(0< s<1\) in Sect. 2.1, and compute the values \([ A_{n,s}^{-1} ] \) and \([ B_{n,s}^{-1} ] \), hence the values of the inverses of tails of the Riemann zeta function \([ \frac{1}{1-2^{1-s}}\zeta_{n}(s) ^{-1} ] \) for \(s= \frac{1}{2},\, \frac{1}{3}\), and \(\frac{1}{4}\) in Sect. 2.2.

2 Main results

2.1 The bounds of the inverses of \(\zeta_{n}(s)\) for \(0< s<1\)

In this section, we present the bounds of \(A_{n,s}^{-1}\) and \(B_{n,s}^{-1}\) in Definition 1, hence the bounds of the inverses of tails of the Riemann zeta function \(\zeta_{n}(s)^{-1}\) for \(0< s<1\).

Proposition 1

Let s be a real number with \(0< s<1\). Then, for any positive even number n,
$$\begin{aligned}& 2(n-1)^{s} < A^{-1}_{n,s} < 2{n}^{s}, \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& -2{n}^{s} < B^{-1}_{n,s} < -2(n-1)^{s}. \end{aligned}$$

Proof

Let n be a positive even number. For every positive integer k, it is easy to see that
$$\begin{aligned} &\biggl( \frac{1}{(n+1+2k)^{s}} - \frac{1}{(n+2+2k)^{s}} \biggr) \\ &\quad < \biggl( \frac{1}{(n+2k)^{s}} - \frac{1}{(n+1+2k)^{s}} \biggr) \\ &\quad < \biggl( \frac{1}{(n-1+2k)^{s}} - \frac{1}{(n+2k)^{s}} \biggr) . \end{aligned}$$
The summations of each term over k give
$$\begin{aligned}& A_{n+1,s} < A_{n,s} < A_{n-1,s} \end{aligned}$$
and
$$\begin{aligned}& \frac{1}{2} (A_{n+1,s} + A_{n,s}) < A_{n,s} < \frac{1}{2} (A_{n-1,s} + A_{n,s}). \end{aligned}$$
Therefore, we have
$$\begin{aligned}& \frac{1}{2n^{s}} < A_{n,s} < \frac{1}{2(n-1)^{s}}, \end{aligned}$$
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,
$$ 2 \biggl( n-\frac{1}{2} \biggr) ^{s} < A^{-1}_{n,s}, $$
and for any positive odd number n,
$$ B^{-1}_{n,s} < -2 \biggl( n-\frac{1}{2} \biggr) ^{s} . $$

Proof

Let n be a positive even number. We will show that
$$\begin{aligned}& A_{n,s} < \frac{1}{2 ( n-\frac{1}{2} ) ^{s}}. \end{aligned}$$
Rewriting each of the both sides as a series
$$\begin{aligned} A_{n,s} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{(2k)^{s}} - \frac{1}{(2k+1)^{s}} \biggr) \end{aligned}$$
and
$$\begin{aligned} \frac{1}{2(n-\frac{1}{2})^{s}} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{2(2k-\frac{1}{2})^{s}} - \frac{1}{2( 2k + \frac{3}{2})^{s}} \biggr) , \end{aligned}$$
we will show that for any positive integer k,
$$\begin{aligned}& \frac{1}{(2k)^{s}} - \frac{1}{(2k+1)^{s}} < \frac{1}{2 ( 2k - \frac{1}{2} ) ^{s}} - \frac{1}{2 ( 2k+\frac{3}{2} ) ^{s}}. \end{aligned}$$
For this, we let
$$ f(x) = \biggl( \frac{1}{2(2x-\frac{1}{2})^{s}} - \frac{1}{2(2x+ \frac{3}{2})^{s}} \biggr) - \biggl( \frac{1}{(2x)^{s}} - \frac{1}{(2x+1)^{s}} \biggr) $$
and will show that \(f(x)\) is positive for \(x \ge 1\) and \(0< s<1\). With
$$ g(x) = \frac{1}{2(2x-\frac{1}{2})^{s}} + \frac{1}{2(2x+\frac{1}{2})^{s}} - \frac{1}{(2x)^{s}}, $$
we have \(f(x) = g(x) -g(x+\frac{1}{2})\). Consider the derivative of \(g(x)\):
$$ g'(x) =-2s \biggl( \frac{1}{2(2x-\frac{1}{2})^{s+1}} + \frac{1}{2(2x+ \frac{1}{2})^{s+1}} - \frac{1}{(2x)^{s+1}} \biggr) . $$
Since the function \(\frac{1}{x^{s+1}}\) is convex, we obtain that
$$ \frac{1}{2(2x-\frac{1}{2})^{s+1}} + \frac{1}{2(2x+\frac{1}{2})^{s+1}} - \frac{1}{(2x)^{s+1}} \geq 0, $$
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,
$$ A^{-1}_{n,s} < 2 \biggl( n-\frac{1}{4} \biggr) ^{s}, $$
and for any positive odd number n,
$$ -2 \biggl( n-\frac{1}{4} \biggr) ^{s}< B^{-1}_{n,s}. $$

Proof

Let n be a positive even number. We will show that
$$\begin{aligned}& \frac{1}{2 ( n-\frac{1}{4} ) ^{s}} < A_{n,s}. \end{aligned}$$
Rewriting each of the both sides as a series
$$\begin{aligned} A_{n,s} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{(2k)^{s}} - \frac{1}{(2k+1)^{s}} \biggr) \end{aligned}$$
and
$$\begin{aligned} \frac{1}{2(n-\frac{1}{4})^{s}} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{2(2k-\frac{1}{4})^{s}} - \frac{1}{2(2k+\frac{7}{4})^{s}} \biggr) , \end{aligned}$$
we need to show that for any positive integer k,
$$\begin{aligned}& \frac{1}{2(2k-\frac{1}{4})^{s}} - \frac{1}{2(2k+\frac{7}{4})^{s}} < \frac{1}{(2k)^{s}} - \frac{1}{(2k+1)^{s}}. \end{aligned}$$
For this, we let
$$ f(x) = \biggl( \frac{1}{(2x)^{s}} - \frac{1}{(2x+1)^{s}} \biggr) - \biggl( \frac{1}{2(2x - \frac{1}{4})^{s}} - \frac{1}{2(2x+\frac{7}{4})^{s}} \biggr) . $$
We check that \(f(1)>0\) and now we will show that \(f(x)\) is positive for \(x \geq 2\) and \(0 < s < 1\). With
$$ g(x) = \frac{1}{(2x)^{s}} - \biggl( \frac{1}{2(2x-\frac{1}{4})^{s}} + \frac{1}{2(2x+ \frac{3}{4})^{s}} \biggr) , $$
we have \(f(x) = g(x) - g(x+\frac{1}{2})\), so we only need to show that \(g(x)\) is decreasing. Consider the derivative of \(g(x)\):
$$\begin{aligned} g'(x) & = s \biggl( - \frac{2}{(2x)^{s+1}} + \biggl( \frac{1}{ ( 2x- \frac{1}{4} ) ^{s+1}} + \frac{1}{ ( 2x+\frac{3}{4} ) ^{s+1}} \biggr) \biggr) \\ & = s \biggl( \biggl( \frac{1}{ ( 2x-\frac{1}{4} ) ^{s+1}} - \frac{1}{(2x)^{s+1}} \biggr) - \biggl( \frac{1}{(2x)^{s+1}} - \frac{1}{ ( 2x+\frac{3}{4} ) ^{s+1}} \biggr) \biggr) . \end{aligned}$$
Since the function \(\frac{1}{x^{s+1}}\) is decreasing and convex, by comparing slopes at \((2x-\frac{1}{4})\) and \((2x+\frac{3}{4})\), we obtain
$$\begin{aligned}& \frac{1}{ ( 2x-\frac{1}{4} ) ^{s+1}} - \frac{1}{(2x)^{s+1}} < \frac{1}{4} (s+1) \frac{1}{ ( 2x-\frac{1}{4} ) ^{s+2}} \end{aligned}$$
and
$$\begin{aligned}& \frac{1}{(2x)^{s+1}} - \frac{1}{ ( 2x+\frac{3}{4} ) ^{s+1}} > \frac{1}{4} (s+1) \frac{3}{ ( 2x+\frac{3}{4} ) ^{s+2}}. \end{aligned}$$
Therefore,
$$\begin{aligned}& g'(x) < \frac{1}{4} s (s+1) \biggl( \frac{1}{ ( 2x-\frac{1}{4} ) ^{s+2}} - \frac{3}{ ( 2x+\frac{3}{4} ) ^{s+2}} \biggr) . \end{aligned}$$
Consider \(h(x,s):= \frac{1}{3} ( \frac{2x+3/4}{2x-1/4} ) ^{s+2}\), which is the ratio of two terms on the right-hand side of the above expression. We check that \(h(x,s) < 1\) for \(x \geq 2\) and \(0< s<1\). Since \(h(2,1) = 6859/10\text{,}125\) and \(\lim_{x \rightarrow \infty } h(x,s) = \frac{1}{3}\) for \(0< s<1\), we obtain that \(g'(x)\) is negative and, therefore, \(g(x)\) is decreasing, which gives the statement. □

We combine the results of Proposition 2 and Proposition 3.

Theorem 1

Let s be a real number with \(0< s<1\). Then, for any positive even number n,
$$\begin{aligned}& 2 \biggl( n-\frac{1}{2} \biggr) ^{s} < A^{-1}_{n,s} < 2 \biggl( n - \frac{1}{4} \biggr) ^{s}, \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& -2 \biggl( n-\frac{1}{4} \biggr) ^{s} < B^{-1}_{n,s} < -2 \biggl( n - \frac{1}{2} \biggr) ^{s}. \end{aligned}$$

We express these bounds in terms of \(\zeta_{n}(s)\) using expression (1).

Corollary 1

Let s be a real number with \(0< s<1\). Then, for any positive even number n,
$$\begin{aligned}& 2\bigl(1-2^{1-s}\bigr) \biggl( n-\frac{1}{4} \biggr) ^{s} < \zeta_{n}(s)^{-1} < 2\bigl(1-2^{1-s} \bigr) \biggl( n - \frac{1}{2} \biggr) ^{s}, \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& -2\bigl(1-2^{1-s}\bigr) \biggl( n-\frac{1}{2} \biggr) ^{s} < \zeta_{n}(s)^{-1} < -2\bigl(1-2^{1-s} \bigr) \biggl( n - \frac{1}{4} \biggr) ^{s}. \end{aligned}$$

Furthermore, we have tighter bounds of \(A^{-1}_{n,s}\) and \(B^{-1}_{n,s}\) for a sufficiently large number n.

Theorem 2

For any positive number ϵ and any real number s with \(0< s<1\),
$$\begin{aligned}& 2 \biggl( n-\frac{1}{2} \biggr) ^{s} < A^{-1}_{n,s} < 2 \biggl( n - \frac{1}{2} + \epsilon \biggr) ^{s} \end{aligned}$$
for a sufficiently large even number n and
$$\begin{aligned}& -2 \biggl( n-\frac{1}{2} + \epsilon \biggr) ^{s} < B^{-1}_{n,s} < -2 \biggl( n - \frac{1}{2} \biggr) ^{s} \end{aligned}$$
for a sufficiently large odd number n.

Proof

From Theorem 1, it suffices to show that for a sufficiently large even number n,
$$\begin{aligned}& \frac{1}{2 ( n-\frac{1}{2} + \epsilon ) ^{s}} < A_{n,s}. \end{aligned}$$
Rewriting each of the both sides as a series
$$\begin{aligned} A_{n,s} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{(2k)^{s}} - \frac{1}{(2k+1)^{s}} \biggr) \end{aligned}$$
and
$$\begin{aligned} \frac{1}{2 ( n - \frac{1}{2} + \epsilon ) ^{s}} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{2 ( 2k-\frac{1}{2} + \epsilon ) ^{s}} - \frac{1}{2 ( 2k+\frac{3}{2} + \epsilon ) ^{s}} \biggr) , \end{aligned}$$
we need to show that for a sufficiently large even number n and every integer \(k \geq \frac{n}{2}\),
$$\begin{aligned}& \frac{1}{2 ( 2k - \frac{1}{2} + \epsilon ) ^{s}} - \frac{1}{2 ( 2k + \frac{3}{2} + \epsilon ) ^{s}} < \frac{1}{(2k)^{s}} - \frac{1}{(2k+1)^{s}}. \end{aligned}$$
For this, let
$$ f(x) = \biggl( \frac{1}{(2x)^{s}} - \frac{1}{(2x+1)^{s}} \biggr) - \biggl( \frac{1}{2(2x- \frac{1}{2} + \epsilon )^{s}} - \frac{1}{2(2x+\frac{3}{2} + \epsilon )^{s}} \biggr) , $$
and we will show that \(f(x)\) is positive for \(x \geq x_{0}\), where \(x_{0}\) is a sufficiently large number. With
$$ g(x)= \frac{1}{(2x)^{s}} - \biggl( \frac{1}{2(2x-\frac{1}{2} + \epsilon )^{s}} + \frac{1}{2(2x+\frac{1}{2} + \epsilon )^{s}} \biggr) , $$
we have that \(f(x) = g(x) - g(x+\frac{1}{2})\), so we only need to show that \(g(x)\) is decreasing. Consider the derivative of \(g(x)\):
$$\begin{aligned} g'(x) & = s \biggl( -\frac{2}{(2x)^{s+1}} + \frac{1}{ ( 2x - \frac{1}{2} + \epsilon ) ^{s+1}} + \frac{1}{ ( 2x + \frac{1}{2} + \epsilon ) ^{s+1}} \biggr) \\ & = s \biggl( \biggl( \frac{1}{ ( 2x - \frac{1}{2} + \epsilon ) ^{s+1}} - \frac{1}{(2x)^{s+1}} \biggr) - \biggl( \frac{1}{(2x)^{s+1}} - \frac{1}{ ( 2x + \frac{1}{2} + \epsilon ) ^{s+1}} \biggr) \biggr) . \end{aligned}$$
Since \(\frac{1}{x^{s+1}}\) is decreasing and convex, by comparing slopes at \((2x - \frac{1}{2} + \epsilon )\) and \((2x + \frac{1}{2} + \epsilon )\), we obtain
$$\begin{aligned}& \frac{1}{ ( 2x - \frac{1}{2} + \epsilon ) ^{s+1}} - \frac{1}{(2x)^{s+1}} < (s+1) \frac{\frac{1}{2} - \epsilon }{ ( 2x - \frac{1}{2} + \epsilon ) ^{s+2}} \end{aligned}$$
and
$$\begin{aligned}& \frac{1}{(2x)^{s+1}} - \frac{1}{ ( 2x+\frac{1}{2} + \epsilon ) ^{s+1}} > (s+1) \frac{\frac{1}{2} + \epsilon }{ ( 2x+\frac{1}{2} + \epsilon ) ^{s+2}}. \end{aligned}$$
Therefore
$$\begin{aligned}& g'(x) < s(s+1) \biggl( \frac{\frac{1}{2} - \epsilon }{ ( 2x - \frac{1}{2} + \epsilon ) ^{s+2}} - \frac{\frac{1}{2} + \epsilon }{ ( 2x + \frac{1}{2} + \epsilon ) ^{s+2}} \biggr) . \end{aligned}$$
Consider \(h(x):= \frac{\frac{1}{2} - \epsilon }{\frac{1}{2} + \epsilon } ( \frac{2x+\frac{1}{2} + \epsilon }{2x - \frac{1}{2} + \epsilon } ) ^{s+2}\), 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
$$\begin{aligned}& h(x) < 1 \quad \iff \quad \frac{2x+\frac{1}{2} + \epsilon }{2x - \frac{1}{2} + \epsilon } < \biggl( \frac{\frac{1}{2} + \epsilon }{\frac{1}{2} - \epsilon } \biggr) ^{\frac{1}{s+2}}. \end{aligned}$$
For any \(\epsilon > 0\) and \(0 < s < 1\), we have that \(1 < ( \frac{ \frac{1}{2} + \epsilon }{\frac{1}{2} - \epsilon } ) ^{1/(s+2)}\) and \(\frac{2x + \frac{1}{2} + \epsilon }{2x - \frac{1}{2} + \epsilon }\) 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 \(\zeta_{n}(s)\) using expression (1).

Corollary 2

For any positive number ϵ and any real number s with \(0< s<1\), we have
$$\begin{aligned}& 2\bigl(1-2^{1-s}\bigr) \biggl( n-\frac{1}{2} + \epsilon \biggr) ^{s} < \zeta_{n}(s)^{-1} < 2\bigl(1-2^{1-s} \bigr) \biggl( n-\frac{1}{2} \biggr) ^{s}, \end{aligned}$$
for a sufficiently large even number n and
$$\begin{aligned}& -2\bigl(1-2^{1-s}\bigr) \biggl( n-\frac{1}{2} \biggr) ^{s} < \zeta_{n}(s)^{-1} < -2\bigl(1-2^{1-s} \bigr) \biggl( n - \frac{1}{2} + \epsilon \biggr) ^{s} \end{aligned}$$
for a sufficiently large odd number n.

2.2 The value of the inverse of \(\zeta_{n}(s)\) for \(s=\frac{1}{2}, \frac{1}{3}\), and \(\frac{1}{4}\)

We study firstly the value of the inverse of \(\zeta_{n}(\frac{1}{2})\), where \(\zeta_{n}(\frac{1}{2})\) is the tail of the Riemann zeta function from n at \(s=\frac{1}{2}\).

Theorem 3

For any positive even number n,
$$\begin{aligned}& \bigl[A^{-1}_{n, 1/2}\bigr] = \biggl[ 2 \biggl( n- \frac{1}{2} \biggr) ^{1/2} \biggr] , \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& \bigl[B^{-1}_{n, 1/2}\bigr] = \biggl[ -2 \biggl( n- \frac{1}{2} \biggr) ^{1/2} \biggr] . \end{aligned}$$

Proof

Let n be a positive even number. By Theorem 1, we have that
$$\begin{aligned}& 2 \biggl( n - \frac{1}{2} \biggr) ^{1/2} < A^{-1}_{n,1/2} < 2 \biggl( n - \frac{1}{4} \biggr) ^{1/2}. \end{aligned}$$
Note that \(2(n - \frac{1}{4})^{1/2} - 2(n - \frac{1}{2})^{1/2} < 1\) for \(n \geq 2\), and it implies that there is at most one integer in the open interval from \(2(n-\frac{1}{2})^{1/2}\) to \(2(n-\frac{1}{4})^{1/2}\). Suppose that there is an integer h in the open interval, i.e.,
$$ 2 \biggl( n - \frac{1}{2} \biggr) ^{1/2} < h < 2 \biggl( n - \frac{1}{4} \biggr) ^{1/2} \quad \mbox{or} \quad 4n-2 < h^{2} < 4n-1. $$
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. □

We express this result in terms of \(\zeta_{n}(s)\) using expression (1).

Corollary 3

For any positive integer n,
$$\begin{aligned}& \biggl[ \frac{1}{1-2^{1/2}} \zeta_{n} \biggl( \frac{1}{2} \biggr) ^{-1} \biggr] = \biggl[ (-1)^{n+1} 2 \biggl( n- \frac{1}{2} \biggr) ^{1/2} \biggr] . \end{aligned}$$

We study secondly the value of the inverse of \(\zeta_{n}(\frac{1}{3})\), where \(\zeta_{n}(\frac{1}{3})\) is the tail of the Riemann zeta function from n at \(s=\frac{1}{3}\).

Theorem 4

For any positive even number n,
$$\begin{aligned}& \bigl[A^{-1}_{n, 1/3}\bigr] = \biggl[ 2 \biggl( n- \frac{1}{2} \biggr) ^{1/3} \biggr] , \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& \bigl[B^{-1}_{n, 1/3}\bigr] = \biggl[ -2 \biggl( n- \frac{1}{2} \biggr) ^{1/3} \biggr] . \end{aligned}$$

Proof

Let n be a positive even number. By Theorem 1, we have that
$$\begin{aligned}& 2 \biggl( n - \frac{1}{2} \biggr) ^{1/3} < A^{-1}_{n,1/3} < 2 \biggl( n - \frac{1}{4} \biggr) ^{1/3}. \end{aligned}$$
Note that \(2(n - \frac{1}{4})^{1/3} - 2(n - \frac{1}{2})^{1/3} < 1\) for \(n \geq 2\), and it implies that there is at most one integer in the open interval from \(2(n-\frac{1}{2})^{1/3}\) to \(2(n-\frac{1}{4})^{1/3}\). Suppose that there is an integer h in the open interval, i.e.,
$$ 2\biggl(n - \frac{1}{2}\biggr)^{1/3} < h < 2\biggl(n - \frac{1}{4}\biggr)^{1/3} \quad \mbox{or} \quad 8n - 4 < h^{3} < 8n - 2. $$
This shows that the integer h is of the form \(h = 2(n-\frac{3}{8})^{1/3}\). If we show \(A^{-1}_{n,1/3} < 2(n-\frac{3}{8})^{1/3}\) or, equivalently, \(\frac{1}{2(n-\frac{3}{8})^{1/3}} < A_{n,1/3}\), then our proof will be done. Let us rewrite
$$\begin{aligned} A_{n,1/3} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{(2k)^{1/3}} - \frac{1}{(2k+1)^{1/3}} \biggr) \end{aligned}$$
and
$$\begin{aligned} \frac{1}{2(n-\frac{3}{8})^{1/3}} & = \sum_{k=\frac{n}{2}}^{\infty } \biggl( \frac{1}{2(2k-\frac{3}{8})^{1/3}} - \frac{1}{2(2k+\frac{13}{8})^{1/3}} \biggr) . \end{aligned}$$
Now it suffices to show that for any positive integer k,
$$\begin{aligned}& \frac{1}{2 ( 2k - \frac{3}{8} ) ^{1/3}} - \frac{1}{2 ( 2k + \frac{13}{8} ) ^{1/3}} < \frac{1}{(2k)^{1/3}} - \frac{1}{2 ( 2k +1 ) ^{1/3}}. \end{aligned}$$
For this, we let
$$ f(x) = \biggl( \frac{1}{(2x)^{1/3}} - \frac{1}{(2x+1)^{1/3}} \biggr) - \biggl( \frac{1}{2(2x - \frac{3}{8})^{1/3}} - \frac{1}{2(2x+ \frac{13}{8})^{1/3}} \biggr) , $$
and we will show that \(f(x)\) is positive for any positive integer x.
We check that \(f(1) = 0.00053\cdots\) and \(f(2) = 0.00081\cdots\) , so it suffices to show \(f(x) > 0\) for \(x \geq 3\). With
$$ g(x) = \frac{1}{(2x)^{1/3}} - \biggl( \frac{1}{2(2x - \frac{3}{8})^{1/3}} + \frac{1}{2(2x + \frac{5}{8})^{1/3}}\biggr) , $$
we have that \(f(x) = g(x) - g(x+\frac{1}{2})\), so we only need to show that \(g(x)\) is decreasing for \(x \ge 3\). Consider the derivative of \(g(x)\):
$$\begin{aligned} g'(x) & = \frac{1}{3} \biggl( -\frac{2}{(2x)^{4/3}} + \frac{1}{ ( 2x - \frac{3}{8} ) ^{4/3}} + \frac{1}{ ( 2x + \frac{5}{8} ) ^{4/3}} \biggr) \\ & = \frac{1}{3} \biggl( \biggl( \frac{1}{ ( 2x -\frac{3}{8} ) ^{4/3}} - \frac{1}{(2x)^{4/3}} \biggr) - \biggl( \frac{1}{(2x)^{4/3}} - \frac{1}{ ( 2x + \frac{5}{8} ) ^{4/3}} \biggr) \biggr) . \end{aligned}$$
Since \(\frac{1}{x^{4/3}}\) is decreasing and convex, by comparing slopes at \((2x - \frac{3}{8})\) and \((2x + \frac{5}{8})\), we obtain
$$\begin{aligned}& \frac{1}{ ( 2x - \frac{3}{8} ) ^{4/3}} - \frac{1}{(2x)^{4/3}} < 2\cdot \frac{3}{16} \cdot \frac{4}{3} \cdot \frac{1}{ ( 2x - \frac{3}{8} ) ^{7/3}} \end{aligned}$$
and
$$\begin{aligned}& \frac{1}{(2x)^{4/3}} - \frac{1}{ ( 2x + \frac{5}{8} ) ^{4/3}} > 2\cdot \frac{5}{16} \cdot \frac{4}{3} \cdot \frac{1}{ ( 2x + \frac{5}{8} ) ^{7/3}}. \end{aligned}$$
Therefore
$$\begin{aligned}& g'(x) < \frac{1}{18} \biggl( \frac{3}{ ( 2x - \frac{3}{8} ) ^{7/3}} - \frac{5}{ ( 2x + \frac{5}{8} ) ^{7/3}} \biggr) . \end{aligned}$$
Consider \(h(x):= \frac{3}{5} ( \frac{2x + 5/8}{2x - 3/8} ) ^{7/3}\), which is the ratio of two terms of the right-hand side of the above expression. We check that \(h(x) < 1\) for \(x \geq 3\) because \(h(3) = 0.87\cdots \) and \(\lim_{x \rightarrow \infty } h(x) = \frac{3}{5}\) and \(h'(x) < 0\) for \(x \geq 3\). Hence we obtain that \(g'(x)\) is negative and so \(g(x)\) is decreasing for \(x \geq 3\), which proves the statement. □

We express this result in terms of \(\zeta_{n}(s)\) using expression (1).

Corollary 4

For any positive integer n,
$$\begin{aligned}& \biggl[ \frac{1}{1-2^{2/3}} \zeta_{n} \biggl( \frac{1}{3} \biggr) ^{-1} \biggr] = \biggl[ (-1)^{n+1} 2 \biggl( n- \frac{1}{2} \biggr) ^{1/3} \biggr] . \end{aligned}$$

We study lastly the value of the inverse of \(\zeta_{n}(\frac{1}{4})\), which is the tail of the Riemann zeta function from n at \(s= \frac{1}{4}\).

Theorem 5

For any positive even number n,
$$\begin{aligned}& \bigl[A^{-1}_{n, 1/4}\bigr] = \biggl[ 2 \biggl( n- \frac{1}{2} \biggr) ^{1/4} \biggr] , \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& \bigl[B^{-1}_{n, 1/4}\bigr] = \biggl[ -2 \biggl( n- \frac{1}{2} \biggr) ^{1/4} \biggr] . \end{aligned}$$

Proof

Let n be a positive even number. By Theorem 1, we have that
$$\begin{aligned}& 2 \biggl( n - \frac{1}{2} \biggr) ^{1/4} < A^{-1}_{n,1/4} < 2 \biggl( n - \frac{1}{4} \biggr) ^{1/4}. \end{aligned}$$
Note that \(2(n - \frac{1}{4})^{1/4} - 2(n - \frac{1}{2})^{1/4} < 1\) for \(n \geq 2\), and it implies that there is at most one integer in the open interval from \(2(n-\frac{1}{2})^{1/4}\) to \(2(n-\frac{1}{4})^{1/4}\). Suppose that there is an integer h in the open interval, i.e.,
$$ 2\biggl(n-\frac{1}{2}\biggr)^{1/4} < h < 2\biggl(n - \frac{1}{4}\biggr)^{1/4}\quad \mbox{or}\quad 16n - 8 < h^{4} < 16n - 4. $$
This shows that the integer \(h^{4}\) is one of the form \(16n - 7\), \(16n - 6\), or \(16n - 5\). For any integer h, however, \(h^{4} \equiv 0\) or \(1 \pmod{16}\), hence such an integer h does not exist. Therefore this gives the statement. □

We express this result in terms of \(\zeta_{n}(s)\) using expression (1).

Corollary 5

For any positive integer n,
$$\begin{aligned}& \biggl[ \frac{1}{1-2^{3/4}} \zeta_{n} \biggl( \frac{1}{4} \biggr) ^{-1} \biggr] = \biggl[ (-1)^{n+1} 2 \biggl( n- \frac{1}{2} \biggr) ^{1/4} \biggr] . \end{aligned}$$

We express the results of Theorems 3, 4, and 5 in a single statement.

Theorem 6

For \(s = \frac{1}{2},\, \frac{1}{3}\), or \(\frac{1}{4}\), and for any positive even number n,
$$\begin{aligned}& \bigl[A^{-1}_{n, s}\bigr] = \biggl[ 2 \biggl( n- \frac{1}{2} \biggr) ^{s} \biggr] , \end{aligned}$$
and for any positive odd number n,
$$\begin{aligned}& \bigl[B^{-1}_{n, s}\bigr] = \biggl[ -2 \biggl( n- \frac{1}{2} \biggr) ^{s} \biggr] . \end{aligned}$$

We express the results of Corollaries 3, 4, and 5 in a single statement.

Corollary 6

For any positive integer n and \(s = \frac{1}{2},\, \frac{1}{3}\), or \(\frac{1}{4}\),
$$\begin{aligned}& \biggl[ \frac{1}{1-2^{1-s}} \zeta_{n} ( s ) ^{-1} \biggr] = \biggl[ (-1)^{n+1} 2 \biggl( n-\frac{1}{2} \biggr) ^{s} \biggr] . \end{aligned}$$

3 Conclusion

In this paper, we have presented the bounds of \(A_{n,s}^{-1}\) and \(B_{n,s}^{-1}\), hence the bounds of the inverses of tails of the Riemann zeta function \(\zeta_{n}(s)^{-1}\) for \(0< s<1\), and computed the values \([ A_{n,s}^{-1} ] \) and \([ B_{n,s}^{-1} ] \), hence the values of the inverses of tails of the Riemann zeta function \([ \frac{1}{1-2^{1-s}}\zeta_{n}(s) ^{-1} ] \) for \(s= \frac{1}{2},\, \frac{1}{3}\), and \(\frac{1}{4}\). For other values of s, for example \(s=\frac{1}{5}\) or \(\frac{2}{3} \), the values of \(A_{n,s}\) and \(B_{n,s}\) do not seem to have simple expressions.

Declarations

Acknowledgements

The authors would like to express their thanks to the referees and the editors for their helpful comments and advice.

Availability of data and materials

Not applicable.

Funding

We have not received any funding.

Authors’ contributions

All authors equally contributed to the manuscript. All authors read and approved the final manuscript.

Competing interests

We declare that none of the authors has any competing interests in the manuscript.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Korea University, Seoul, Republic of Korea

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