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 [2] 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 [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

$$\Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{P_{k}} \Biggr)^{-1} \Biggr\rfloor = \textstyle\begin{cases} P_{n-1}+P_{n-2} &\mbox{if } n\ge2 \mbox{ is even}, \\ P_{n-1}+P_{n-2}-1 &\mbox{if } n\ge1 \mbox{ is odd}, \end{cases} $$

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 [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:

$$\begin{aligned} & \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{2}} \Biggr)^{-1} \Biggr\rfloor =n-1, \end{aligned}$$
(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 [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

$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{k^{4}} \Biggr)^{-1} \Biggr\rfloor =-1+4n-5n^{2}+3n^{3} + \biggl\lfloor \frac{(2n+1)(n-1)}{4} \biggr\rfloor . $$
(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 [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

$$\begin{aligned} &\eta_{0}(n)-\eta_{0}(n+1)- \frac{1}{n^{4}} \\ &\quad=(3b-1)\frac{1}{n^{4}}+(-6 b - 4 a_{2} b) \frac{1}{n^{5}}+\bigl(10 b - 5 a_{1} b + 10 a_{2} b + 5 a_{2}^{2} b\bigr)\frac{1}{n^{6}} \\ &\qquad{}+\bigl(-15 b - 6 a_{0} b + 15 a_{1} b - 20 a_{2} b + 12 a_{1} a_{2} b - 15 a_{2}^{2} b - 6 a_{2}^{3} b\bigr) \frac{1}{n^{7}}+O \biggl(\frac{1}{n^{8}} \biggr). \end{aligned}$$
(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. □