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).
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1 Introduction and main results
Let \((a_{k} )_{k\ge1}\) be a strictly increasing positive sequence such that
Many authors study the computational formula for infinite sums of reciprocal \(a_{k}\),
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 [2] showed that
Xu and Wang [3] obtained a complex computational formula for \(a_{k}=F_{k}^{3}\).
Zhang and Wang [4] studied this problem for the Pell numbers \(P_{k}\) and showed that
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 [5–13] and references therein.
Very recently, Lin [1] investigated the related problem for the sequence \(a_{k}=k^{s}\) with integer \(s\ge2\) and showed the following two interesting identities:
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:
In [1], 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
Furthermore, for \(a_{k}=k^{5}\), we also have an analogous computational formula.
Theorem 2
For all integer \(n\ge4\), we have
2 Proof of Theorem 1
Assume that
and \(f(\infty)=g(\infty)=0\). Summing the inequalities from n to ∞, we have
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:
We will apply the multiple-correction method (see [14–16]) 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
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
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
If \(u=-\frac{3}{16}\), \(v=-\frac{1}{2}\), then we can get the approximation solution
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
Then, for \(n\ge2\),
Proof
We easily check that
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 \(n\in\mathbb{N}\),
□
Lemma 2
Let
Then, for \(n\in\mathbb{N}\),
Proof
We have
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
from n to ∞ and noting that \(f(\infty)=g(\infty)=0\), we have
Then, for \(n\ge2\),
Note that
and
For \(n\ge5\), we have \(\frac{9}{8(2n-1)}\le\frac{1}{8}\). Then
and, for \(n\in\mathbb{N}\),
It follows that, for \(n\ge5\),
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
So we have the following inequalities.
Lemma 3
Let
Then, for \(n\ge2\),
Proof
Note that
Then, for \(n\ge2\), we have
□
Lemma 4
Let
Then, for \(n\ge5\),
Proof
Similarly to the proof of Lemma 3, we have
Note that
and
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
from n to ∞ and noting that \(F(\infty)=G(\infty)=0\), we have
Next, for \(n\ge3\), we will prove the following identities:
Since
it suffices to prove that
We will consider three cases.
Case 1. If \(n=3m\), \(m\in\mathbb{N}\), then we have
Case 2. If \(n=3m+1\), \(m\in\mathbb{N}\), then we have
Case 3. If \(n=3m+2\), \(m\in\mathbb{N}\), then we have
This proves that (3.10) holds. Finally, combining (3.8) and (3.9), we prove Theorem 2. □
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Xu, H. Some computational formulas related the Riemann zeta-function tails. J Inequal Appl 2016, 132 (2016). https://doi.org/10.1186/s13660-016-1068-2
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DOI: https://doi.org/10.1186/s13660-016-1068-2