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.