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# Some computational formulas related the Riemann zeta-function tails

Journal of Inequalities and Applications20162016:132

https://doi.org/10.1186/s13660-016-1068-2

• Received: 14 March 2016
• Accepted: 11 April 2016
• Published:

## Abstract

In this paper we present two computational formulae for one kind of reciprocal sums related to the Riemann zeta-function at integer points $$s=4,5$$, which answers an open problem proposed by Lin (J. Inequal. Appl. 2016:32, 2016).

## Keywords

• Riemann zeta-function
• inequality
• continued fraction
• telescoping method
• multiple-correction method

• 11B83
• 11M06
• 11J70

## 1 Introduction and main results

Let $$(a_{k} )_{k\ge1}$$ be a strictly increasing positive sequence such that
$$\sum_{k=1}^{\infty} \frac{1}{a_{k}}< +\infty.$$
(1.1)
Many authors study the computational formula for infinite sums of reciprocal $$a_{k}$$,
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{a_{k}} \Biggr)^{-1} \Biggr\rfloor ,\quad n\in\mathbb{N},$$
(1.2)
where $$\lfloor x \rfloor$$ denotes the integer part of x.
For example, let $$(F_{k})$$ be the famous Fibonacci sequence: $$F_{k+1}=F_{k}+F_{k-1}$$ with the initial values $$F_{0}=0$$ and $$F_{1}=1$$. Ohtsuka and Nakamura  showed that
\begin{aligned}& \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor = \textstyle\begin{cases} F_{n-2} &\mbox{if } n\ge2 \mbox{ is even}, \\ F_{n-2}-1 &\mbox{if } n\ge1 \mbox{ is odd}, \end{cases}\displaystyle \\& \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \textstyle\begin{cases} F_{n-1}F_{n}-1 &\mbox{if } n\ge2 \mbox{ is even}, \\ F_{n}F_{n-1} &\mbox{if } n\ge1 \mbox{ is odd}. \end{cases}\displaystyle \end{aligned}
Xu and Wang  obtained a complex computational formula for $$a_{k}=F_{k}^{3}$$.
Zhang and Wang  studied this problem for the Pell numbers $$P_{k}$$ and showed that
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{P_{k}} \Biggr)^{-1} \Biggr\rfloor = \textstyle\begin{cases} P_{n-1}+P_{n-2} &\mbox{if } n\ge2 \mbox{ is even}, \\ P_{n-1}+P_{n-2}-1 &\mbox{if } n\ge1 \mbox{ is odd}, \end{cases}$$
where the Pell numbers $$P_{k}$$ are defined by $$P_{0}=0$$, $$P_{1}=1$$, and the recurrence relation $$P_{k+1}=2P_{k}+P_{k-1}$$.

For some other results related to recursive sequences, recursive polynomials, and their promotion forms, see  and references therein.

Very recently, Lin  investigated the related problem for the sequence $$a_{k}=k^{s}$$ with integer $$s\ge2$$ and showed the following two interesting identities:
\begin{aligned} & \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{2}} \Biggr)^{-1} \Biggr\rfloor =n-1, \end{aligned}
(1.3)
\begin{aligned} & \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{3}} \Biggr)^{-1} \Biggr\rfloor =2n(n-1). \end{aligned}
(1.4)
This is an important problem, which has a close relationship with the Riemann zeta-function $$\zeta(s)$$. Lin noted that there does not exist an integer-coefficient polynomial $$q(x)$$ of degree 3 such that the following identity holds:
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{4}} \Biggr)^{-1} \Biggr\rfloor =q(n).$$
(1.5)

In , Lin declared that giving a precise calculation formula for $$(\sum_{k=n}^{\infty}\frac{1}{k^{s}} )^{-1}$$ with $$s=4$$ is a very complicated problem. In this paper, we tackle this open problem.

### Theorem 1

For all integer $$n\ge2$$, we have the identity
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{4}} \Biggr)^{-1} \Biggr\rfloor =-1+4n-5n^{2}+3n^{3} + \biggl\lfloor \frac{(2n+1)(n-1)}{4} \biggr\rfloor .$$
(1.6)

Furthermore, for $$a_{k}=k^{5}$$, we also have an analogous computational formula.

### Theorem 2

For all integer $$n\ge4$$, we have
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{5}} \Biggr)^{-1} \Biggr\rfloor =-5 n + 9 n^{2} - 8 n^{3} + 4 n^{4} + \biggl\lfloor \frac{(n+1)(n-2)}{3} \biggr\rfloor .$$
(1.7)

## 2 Proof of Theorem 1

Assume that
$$g(n)-g(n+1)< \frac{1}{n^{4}}< f(n)-f(n+1),$$
and $$f(\infty)=g(\infty)=0$$. Summing the inequalities from n to ∞, we have
$$g(n)< \sum_{k=n}^{\infty} \frac{1}{k^{4}}< f(n).$$
(2.1)
These inequalities allow us to study the computational formulas of Theorem 1. The problem of finding the functions $$f(n)$$, $$g(n)$$ (or $$F(n)$$, $$G(n)$$ in Section 3) is transformed into solving the finite continued fraction approximation solution of difference equation for ‘large’ n:
$$y(n)-y(n+1)=\frac{1}{n^{4}}.$$
(2.2)
We will apply the multiple-correction method (see ) and solve it as follows.

### Step 1

(The initial correction)

Choosing $$\eta_{0}(n)=\frac{b}{n^{3} + a_{2} n^{2} + a_{1} n + a_{0}}$$ and developing the expression $$\eta_{0}(n)-\eta_{0}(n+1)-\frac{1}{n^{4}}$$ into power series expansion in $$1/n$$, we easily obtain
\begin{aligned} &\eta_{0}(n)-\eta_{0}(n+1)- \frac{1}{n^{4}} \\ &\quad=(3b-1)\frac{1}{n^{4}}+(-6 b - 4 a_{2} b) \frac{1}{n^{5}}+\bigl(10 b - 5 a_{1} b + 10 a_{2} b + 5 a_{2}^{2} b\bigr)\frac{1}{n^{6}} \\ &\qquad{}+\bigl(-15 b - 6 a_{0} b + 15 a_{1} b - 20 a_{2} b + 12 a_{1} a_{2} b - 15 a_{2}^{2} b - 6 a_{2}^{3} b\bigr) \frac{1}{n^{7}}+O \biggl(\frac{1}{n^{8}} \biggr). \end{aligned}
(2.3)
If $$b=\frac{1}{3}$$, $$a_{2}=-\frac{3}{2}$$, $$a_{1}=\frac{5}{4}$$, $$a_{0}=-\frac{3}{8}$$, then we can get the approximation solution
$$g(n)=\frac{b}{n^{3} + a_{2} n^{2} + a_{1} n + a_{0}}$$
of difference equation (2.2), which is the best possible rational approximation solution of such structure as n tends to infinity.

### Step 2

(The first correction)

Choose $$\eta_{1}(n)=\frac{b}{n^{3} + a_{2} n^{2} + a_{1} n + a_{0}+\frac{u}{x+v}}$$ and developing the expression $$\eta_{1}(n)-\eta_{1}(n+1)-\frac{1}{n^{4}}$$ into power series expansion in $$1/n$$, we easily obtain
$$\eta_{1}(n)-\eta_{1}(n+1)- \frac{1}{n^{4}}= \biggl(-\frac{7}{16}-\frac{7}{3} u \biggr) \frac{1}{n^{8}}+\frac{4}{3} (u+2uv)\frac{1}{n^{9}}+O \biggl( \frac{1}{n^{10}} \biggr).$$
(2.4)
If $$u=-\frac{3}{16}$$, $$v=-\frac{1}{2}$$, then we can get the approximation solution
$$f(n)=\frac{b}{n^{3} + a_{2} n^{2} + a_{1} n + a_{0}+\frac{u}{x+v}}$$
of difference equation (2.2), which has a better approximation rate than $$g(n)$$ for ‘large’ n.

So we can get following inequalities necessary in the proofs of our theorems.

### Lemma 1

Let
$$f(n)=\frac{ \frac{8}{3} }{ -1 + 10 n - 12 n^{2} + 8 n^{3}-\frac{3}{2n-1}}.$$
(2.5)
Then, for $$n\ge2$$,
$$f(n)-f(n+1)>\frac{1}{n^{4}}.$$
(2.6)

### Proof

We easily check that
$$f(n)-f(n+1)-\frac{1}{n^{4}}=\frac{1}{n^{4} (1 + n) (1 + n + n^{2}) (-1 + 2 n - 2 n^{2} + n^{3}) }.$$
Note that $$-1 + 2 n - 2 n^{2} + n^{3}=(-2 + n)(2 + n^{2})+ 3$$, so the above polynomial is positive for $$n\ge2$$. Then, for $$n\in\mathbb{N}$$,
$$f(n)-f(n+1)-\frac{1}{n^{4}}>0.$$
□

### Lemma 2

Let
$$g(n)=\frac{8}{3 (-3 + 10 n - 12 n^{2} + 8 n^{3})}.$$
(2.7)
Then, for $$n\in\mathbb{N}$$,
$$g(n)-g(n+1)< \frac{1}{n^{4}}.$$
(2.8)

### Proof

We have
$$g(n)-g(n+1)-\frac{1}{n^{4}}=\frac{- 28 n^{2}+9}{n^{4} (-3 + 10 n - 12 n^{2} + 8 n^{3}) (3 + 10 n + 12 n^{2} + 8 n^{3})},$$
where $$h(n):=-3 + 10 n - 12 n^{2} + 8 n^{3}=(-2 + n)(18 + 4 n + 8 n^{2})+ 33>0$$ for $$n\ge2$$, and $$h(1)=3>0$$. So $$h(n)>0$$ for $$n\in\mathbb{N}$$. This completes the proof of Lemma 2. □

### Proof of Theorem 1

Summing the inequalities of the form
$$g(n)-g(n+1)< \frac{1}{n^{4}}< f(n)-f(n+1)$$
from n to ∞ and noting that $$f(\infty)=g(\infty)=0$$, we have
$$g(n)< \sum_{k=n}^{\infty}\frac{1}{k^{4}}< f(n), \quad n\ge2.$$
(2.9)
Then, for $$n\ge2$$,
$$\frac{3}{8} \biggl(\frac{-3}{2n-1}-3 + 10 n - 12 n^{2} + 8 n^{3} \biggr)< \Biggl(\sum _{k=n}^{\infty}\frac{1}{k^{4}} \Biggr)^{-1} < \frac{3}{8}\bigl(-3 + 10 n - 12 n^{2} + 8 n^{3}\bigr).$$
(2.10)
Note that
$$\frac{3}{8}\bigl(-3 + 10 n - 12 n^{2} + 8 n^{3} \bigr)=3n^{3}-5n^{2}+4n-1 + \biggl(\frac {n(2n-1)}{4}- \frac{1}{8} \biggr)$$
and
$$\frac{3}{8} \biggl(\frac{-3}{2n-1} -3+ 10 n - 12 n^{2} + 8 n^{3} \biggr)=3n^{3}-5n^{2}+4n-1 + \biggl( \frac{n(2n-1)}{4}-\frac{9}{8(2n-1)}-\frac{1}{8} \biggr).$$
For $$n\ge5$$, we have $$\frac{9}{8(2n-1)}\le\frac{1}{8}$$. Then
\begin{aligned} \biggl\lfloor \frac{n(2n-1)}{4}-\frac{1}{8} \biggr\rfloor &\ge \biggl\lfloor \frac{n(2n-1)}{4}-\frac{9}{8(2n-1)}-\frac{1}{8} \biggr\rfloor \ge \biggl\lfloor \frac{n(2n-1)}{4}-\frac{1}{4} \biggr\rfloor \\ &= \biggl\lfloor \frac{(2n+1)(n-1)}{4} \biggr\rfloor \end{aligned}
and, for $$n\in\mathbb{N}$$,
$$\biggl\lfloor \frac{n(2n-1)}{4}-\frac{1}{8} \biggr\rfloor = \biggl\lfloor \frac{(2n+1)(n-1)}{4} \biggr\rfloor .$$
It follows that, for $$n\ge5$$,
$$\biggl\lfloor \frac{n(2n-1)}{4}-\frac{1}{8} \biggr\rfloor = \biggl\lfloor \frac{n(2n-1)}{4}-\frac{9}{8(2n-1)}-\frac{1}{8} \biggr\rfloor = \biggl\lfloor \frac{(2n+1)(n-1)}{4} \biggr\rfloor .$$
(2.11)
Finally, we note that the above identities hold for $$n=2,3,4$$. Combining (2.10) and (2.11), we prove Theorem 1. □

## 3 Proof of Theorem 2

Similarly to Section 2, by the multiple-correction method we can solve the finite continued fraction approximation solution $$F(n)$$, $$G(n)$$ of the differential equation
$$y(n)-y(n+1)=\frac{1}{n^{5}}.$$
(3.1)
So we have the following inequalities.

### Lemma 3

Let
$$F(n)=\frac{9}{-2 - 48 n + 84 n^{2} - 72 n^{3} + 36 n^{4}}.$$
(3.2)
Then, for $$n\ge2$$,
$$F(n)-F(n+1)>\frac{1}{n^{5}}.$$
(3.3)

### Proof

\begin{aligned} &F(n)-F(n+1)-\frac{1}{n^{5}} \\ &\quad=\frac{-1 + 660 n^{2}}{n^{5} (-1 - 24 n + 42 n^{2} - 36 n^{3} + 18 n^{4}) (-1 +24 n + 42 n^{2} + 36 n^{3} + 18 n^{4})}. \end{aligned}
(3.4)
Note that
$$-1 - 24 n + 42 n^{2} - 36 n^{3} + 18 n^{4}=(n-2) \bigl(60 + 42 n + 18 n^{3}\bigr)+119.$$
(3.5)
Then, for $$n\ge2$$, we have
$$F(n)-F(n+1)-\frac{1}{n^{5}}>0.$$
□

### Lemma 4

Let
$$G(n)=\frac{9}{-1/4 - 48 n + 84 n^{2} - 72 n^{3} + 36 n^{4}}.$$
(3.6)
Then, for $$n\ge5$$,
$$G(n)-G(n+1)< \frac{1}{n^{5}}.$$
(3.7)

### Proof

Similarly to the proof of Lemma 3, we have
\begin{aligned} &G(n)-G(n+1)-\frac{1}{n^{5}} \\ &\quad=-\frac{1-37\text{,}536 n^{2} +2\text{,}016 n^{4}}{n^{5} (-1 - 192 n + 336 n^{2} - 288 n^{3} + 144 n^{4}) (-1 + 192 n +336 n^{2} + 288 n^{3} + 144 n^{4})}. \end{aligned}
Note that
$$1-37\text{,}536 n^{2} +2\text{,}016 n^{4}=(-5 + n) \bigl(64\text{,}320 + 12\text{,}864 n + 10\text{,}080 n^{2} + 2\text{,}016 n^{3}\bigr)+ 321\text{,}601$$
and
$$-1 - 192 n + 336 n^{2} - 288 n^{3} + 144 n^{4}=(-4 + n) \bigl(5\text{,}760 + 1\text{,}488 n + 288 n^{2} + 144 n^{3}\bigr)+ 23\text{,}039.$$
Then, for $$n\ge5$$, the inequality $$G(n)-G(n+1)-\frac{1}{n^{5}}<0$$ holds. □

### Proof of Theorem 2

We assume that $$n\ge5$$ in the following proof. Summing the inequalities of the form
$$G(n)-G(n+1)< \frac{1}{n^{4}}< F(n)-F(n+1)$$
from n to ∞ and noting that $$F(\infty)=G(\infty)=0$$, we have
\begin{aligned} &{-} 5 n + 9 n^{2} - 8 n^{3} + 4n^{4}+ \biggl(-\frac{2}{9} -\frac{n}{3} + \frac {n^{2}}{3} \biggr) \\ &\quad< \Biggl(\sum_{k=n}^{\infty}\frac{1}{k^{5}} \Biggr)^{-1} \\ &\quad< - 5 n + 9 n^{2} - 8 n^{3} + 4n^{4}+ \biggl(- \frac{1}{36} -\frac{n}{3} + \frac {n^{2}}{3} \biggr). \end{aligned}
(3.8)
Next, for $$n\ge3$$, we will prove the following identities:
$$\biggl\lfloor -\frac{2}{9} -\frac{n}{3} + \frac{n^{2}}{3} \biggr\rfloor = \biggl\lfloor -\frac{1}{36} - \frac{n}{3} + \frac{n^{2}}{3} \biggr\rfloor = \biggl\lfloor \frac{(n+1)(n-2)}{3} \biggr\rfloor .$$
(3.9)
Since
$$\frac{(n+1)(n-2)}{3}< -\frac{2}{9} -\frac{n}{3} + \frac{n^{2}}{3}< -\frac {1}{36} -\frac{n}{3} + \frac{n^{2}}{3},$$
it suffices to prove that
$$\biggl\lfloor -\frac{1}{36} -\frac{n}{3} + \frac{n^{2}}{3} \biggr\rfloor = \biggl\lfloor \frac{(n+1)(n-2)}{3} \biggr\rfloor .$$
(3.10)
We will consider three cases.
Case 1. If $$n=3m$$, $$m\in\mathbb{N}$$, then we have
$$\biggl\lfloor -\frac{1}{36} -\frac{n}{3} + \frac{n^{2}}{3} \biggr\rfloor = \biggl\lfloor \frac{(n+1)(n-2)}{3} \biggr\rfloor =-1-m+3m^{2}.$$
Case 2. If $$n=3m+1$$, $$m\in\mathbb{N}$$, then we have
$$\biggl\lfloor -\frac{1}{36} -\frac{n}{3} + \frac{n^{2}}{3} \biggr\rfloor = \biggl\lfloor \frac{(n+1)(n-2)}{3} \biggr\rfloor =-1+m + 3 m^{2}.$$
Case 3. If $$n=3m+2$$, $$m\in\mathbb{N}$$, then we have
$$\biggl\lfloor -\frac{1}{36} -\frac{n}{3} + \frac{n^{2}}{3} \biggr\rfloor = \biggl\lfloor \frac{(n+1)(n-2)}{3} \biggr\rfloor =3m + 3 m^{2}.$$
This proves that (3.10) holds. Finally, combining (3.8) and (3.9), we prove Theorem 2. □

## Declarations 