The infinite sum of the cubes of reciprocal Pell numbers

Abstract

Given the sequence of Pell numbers $\{P_n\}$, we evaluate the integral part of the reciprocal of the sum ${\sum }_{k=n}^{\mathrm{\infty }}\frac{1}{P_n^3}$ explicitly in terms of the Pell numbers themselves.

MSC: 11B39.

1 Introduction

For any integer $n\ge 0$, the well-known Pell numbers $P_n$ are defined by the second-order linear recurrence sequence $P_{n+2}=2P_{n+1}+P_n$, where $P_0=0$ and $P_1=1$. The Pell-Lucas numbers $Q_n$ are defined by $Q_{n+2}=2Q_{n+1}+Q_n$, where $Q_0=2$ and $Q_1=2$. Let $\alpha =1+\sqrt{2}$ and $\beta =1-\sqrt{2}$. Then from the characteristic equations $x^2-2x-1=0$, we also have the computational formulae

$$P_n=\frac{1}{2\sqrt{2}}({\alpha }^n-{\beta }^n)\text{and}{Q}_n={\alpha }^n+{\beta }^n.$$

For example, the first few values of $P_n$ and $Q_n$ are $P_0=1,P_1=2,P_2=5,P_3=12,P_4=29,\dots$ , $Q_0=2,Q_1=2,Q_2=6,Q_3=14,Q_4=34,Q_5=82,\dots$ .

Various properties of the Pell numbers and related sequences have been studied by many authors, see [1–6]. For example, Santos and Sills [3] studied the arithmetic properties of the q-Pell sequence and obtained two identities. Kilic [4] studied the generalized order-k Fibonacci-Pell sequences and gave several congruences. Recently, the authors [7] and [8] studied the infinite sums derived from the Pell numbers and proved the following identities:

$$\begin{array}{c}\lfloor {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{P_k}\right)}^{-1}\rfloor =\{\begin{array}{ll}{P}_{n-1}+P_{n-2}& \text{if }n\text{ is even and }n\ge 2;\\ P_{n-1}+P_{n-2}-1& \text{if }n\text{ is odd and }n\ge 1,\end{array}\hfill \\ \lfloor {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{P_k^2}\right)}^{-1}\rfloor =\{\begin{array}{ll}2P_{n-1}{P}_n-1& \text{if }n\text{ is an even number;}\\ 2P_{n-1}{P}_n& \text{if }n\text{ is an odd number,}\end{array}\hfill \end{array}$$

where $\lfloor x\rfloor$ is the floor function, that is, it denotes the greatest integer less than or equal to x.

Some related works can also be found in [9] and [10]. Especially in [10], the authors studied a problem, which is little different from (1). That is, they studied the computational problem of the nearest integer function of ${({\sum }_{k=n}^{\mathrm{\infty }}\frac{1}{u_k})}^{-1}$ and proved an interesting conclusion:

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k}\right)}^{-1}\parallel =u_n-u_{n-1}\text{for all }n>n_0,$$

where $\parallel \cdot \parallel$ denotes the nearest integer, namely $\parallel x\parallel =\lfloor x+\frac{1}{2}\rfloor$, ${\{u_n\}}_{n\ge 0}$ is an integer sequence satisfying the recurrence formula

$$u_n=a{u}_{n-1}+u_{n-2}+\cdots +u_{n-s}(s\ge 2)$$

with the initial conditions $u_0\ge 0$, $u_k\in \mathbf{N}$, $1\le k\le s-1$.

Using the method in [10] seems to be very difficult to deal with ${({\sum }_{k=n}^{\mathrm{\infty }}\frac{1}{u_k^s})}^{-1}$ for all integers $s\ge 2$.

The main purpose of this paper related to the computing problem of

$$P(s,n)\equiv \lfloor {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{P_k^s}\right)}^{-1}\rfloor$$

(1)

for all integers $s\ge 3$. At the end of [7], the authors asked whether there exists a corresponding formula for $P(3,n)$.

In fact, this problem is difficult because it is quite unclear a priori what the shape of the result might be. In order to resolve the question, we carefully applied the method of undetermined coefficients and constructed a number of delicate inequalities in order to complete a proof. The result is as follows.

Theorem For any positive integer $n\ge 1$, we have the identity

$$P(3,n)=\{\begin{array}{ll}{P}_n^2{P}_{n-1}+3P_n{P}_{n-1}^2+\lfloor -\frac{61}{82}{P}_n-\frac{91}{82}{P}_{n-1}\rfloor & \mathit{\text{if}}n\mathit{\text{is even and}}n\ge 2;\\ P_n^2{P}_{n-1}+3P_n{P}_{n-1}^2+\lfloor \frac{61}{82}{P}_n+\frac{91}{82}{P}_{n-1}\rfloor & \mathit{\text{if}}n\mathit{\text{is odd and}}n\ge 1.\end{array}$$

It remains a difficult problem even to conjecture what might be an analogous expression to the formula for $P(3,n)$ in the theorem for $P(k,n)$ when $k\ge 4$.

2 Proof of the theorem

In this section, we shall prove our theorem directly. First we consider the case that $n=2m$ is an even number. It is clear that in this case our theorem is equivalent to

$$\begin{array}{r}{P}_{2m}^2{P}_{2m-1}+3P_{2m}{P}_{2m-1}^2-\frac{1}{82}(61P_{2m}+91P_{2m-1})\\ <{\left(\sum _{k=2m}^{\mathrm{\infty }}\frac{1}{P_k^3}\right)}^{-1}<P_{2m}^2{P}_{2m-1}+3P_{2m}{P}_{2m-1}^2-\frac{1}{82}(61P_{2m}+91P_{2m-1})+\frac{1}{82}\end{array}$$

or

$$\begin{array}{r}\frac{1}{P_{2m}^2{P}_{2m-1}+3P_{2m}{P}_{2m-1}^2-\frac{1}{82}(61P_{2m}+91P_{2m-1})+\frac{1}{82}}\\ <\sum _{k=2m}^{\mathrm{\infty }}\frac{1}{P_k^3}<\frac{1}{P_{2m}^2{P}_{2m-1}+3P_{2m}{P}_{2m-1}^2-\frac{1}{82}(61P_{2m}+91P_{2m-1})}.\end{array}$$

(2)

Now we prove that for all positive integers k, we have the inequality

$$\begin{array}{rl}\frac{1}{P_{2k}^3}+\frac{1}{P_{2k+1}^3}<& \frac{1}{P_{2k}^2{P}_{2k-1}+3P_{2k}{P}_{2k-1}^2-\frac{1}{82}(61P_{2k}+91P_{2k-1})}\\ -\frac{1}{P_{2k+2}^2{P}_{2k+1}+3P_{2k+2}{P}_{2k+1}^2-\frac{1}{82}(61P_{2k+2}+91P_{2k+1})}.\end{array}$$

(3)

It is clear that (3) holds for $k=1,2,3\text{ and }4$. So, without loss of generality, we can assume that $k\ge 5$. Note that $P_{2k}^3=\frac{1}{8}(P_{6k}-3P_{2k})$, $P_{2k+1}^3=\frac{1}{8}(P_{6k+3}+3P_{2k+1})$, $P_{2k}^3+P_{2k+1}^3=\frac{1}{8}(P_{6k+3}+P_{6k}+3P_{2k+1}-3P_{2k})$, $P_{2k}^3{P}_{2k+1}^3=\frac{1}{512}(Q_{12k+3}-6Q_{8k+2}+9Q_{4k+1}+4)$ and

$$P_{2k}^2{P}_{2k-1}+3P_{2k}{P}_{2k-1}^2=\frac{1}{8}(P_{6k-1}+3P_{6k-2}+5P_{2k-1}+5P_{2k}),$$

so inequality (3) is equivalent to

$$\begin{array}{r}\frac{8(P_{6k+3}+P_{6k}+3P_{2k+1}-3P_{2k})}{Q_{12k+3}-6Q_{8k+2}+9Q_{4k+1}+4}\\ <\frac{378P_{6k-1}+154P_{6k-2}-\frac{78}{41}{P}_{2k+1}-\frac{318}{41}{P}_{2k}}{(P_{6k-1}+3P_{6k-2}-\frac{39}{41}{P}_{2k}-\frac{159}{41}{P}_{2k-1})(P_{6k+5}+3P_{6k+4}-\frac{39}{41}{P}_{2k+2}-\frac{159}{41}{P}_{2k+1})}.\end{array}$$

(4)

From the definition and properties of $P_n$ and $Q_n$, we can easily deduce the identities

$$\begin{array}{c}{P}_n{P}_k=\frac{1}{8}{Q}_{n+k}-\frac{{(-1)}^k}{8}{Q}_{n-k},n\ge k,\hfill \\ Q_n{Q}_k=Q_{n+k}+{(-1)}^k{Q}_{n-k},n\ge k,\hfill \\ P_n{Q}_k=P_{n+k}+{(-1)}^k{P}_{n-k},n\ge k.\hfill \end{array}$$

So, applying these formulae, we have

$$(P_{6k-1}+3P_{6k-2})(P_{6k+5}+3P_{6k+4})=\frac{1}{8}(8Q_{12k+3}+10Q_{12k+2}-2\text{,}772)$$

and

$$\begin{array}{r}(P_{6k-1}+3P_{6k-2}-\frac{39}{41}{P}_{2k}-\frac{159}{41}{P}_{2k-1})(P_{6k+5}+3P_{6k+4}-\frac{39}{41}{P}_{2k+2}-\frac{159}{41}{P}_{2k+1})\\ =\frac{1}{8}(8Q_{12k+3}+10Q_{12k+2}-\frac{7\text{,}344}{41}{Q}_{8k+1}-\frac{2\text{,}124}{41}{Q}_{8k}-\frac{30\text{,}226\text{,}500}{1\text{,}681}{Q}_{4k-3})\\ -\frac{1}{8}(\frac{12\text{,}507\text{,}840}{1\text{,}681}{Q}_{4k-4}+\frac{4\text{,}442\text{,}760}{1\text{,}681}).\end{array}$$

From these two identities and (4), we deduce that inequality (3) is equivalent to

$$\begin{array}{r}\frac{P_{6k+3}+P_{6k}+3P_{2k+1}-3P_{2k}}{Q_{12k+3}-6Q_{8k+2}+9Q_{4k+1}+4}\\ <\frac{378P_{6k-1}+154P_{6k-2}-\frac{78}{41}{P}_{2k+1}-\frac{318}{41}{P}_{2k}}{8Q_{12k+3}+10Q_{12k+2}-\frac{7\text{,}344}{41}{Q}_{8k+1}-\frac{2\text{,}124}{41}{Q}_{8k}-\frac{30\text{,}226\text{,}500}{1\text{,}681}{Q}_{4k-3}-\frac{12\text{,}507\text{,}840}{1\text{,}681}{Q}_{4k-4}-\frac{4\text{,}442\text{,}760}{1\text{,}681}}.\end{array}$$

(5)

For convenience, we let

$$\begin{array}{rl}A=& (P_{6k+3}+P_{6k}+3P_{2k+1}-3P_{2k})\times (8Q_{12k+3}+10Q_{12k+2}-\frac{7\text{,}344}{41}{Q}_{8k+1}\\ -\frac{2\text{,}124}{41}{Q}_{8k}-\frac{30\text{,}226\text{,}500}{1\text{,}681}{Q}_{4k-3}-\frac{12\text{,}507\text{,}840}{1\text{,}681}{Q}_{4k-4}-\frac{4\text{,}442\text{,}760}{1\text{,}681})\end{array}$$

and

$$B=(Q_{12k+3}-6Q_{8k+2}+9Q_{4k+1}+4)(378P_{6k-1}+154P_{6k-2}-\frac{78}{41}{P}_{2k+1}-\frac{318}{41}{P}_{2k}).$$

Then by calculation it follows that

$$\begin{array}{r}A=8(P_{18k+6}+P_{6k})+10(P_{18k+5}+P_{6k-1})-\frac{7\text{,}344}{41}(P_{14k+4}+P_{2k-2})\\ \phantom{A=}-\frac{2\text{,}124}{41}(P_{14k+3}+P_{2k-3})-\frac{30\text{,}226\text{,}500}{1\text{,}681}(P_{10k}-P_{2k+6})-\frac{12\text{,}507\text{,}840}{1\text{,}681}(P_{10k-1}\\ \phantom{A=}+P_{2k+7})-\frac{4\text{,}442\text{,}760}{1\text{,}681}{P}_{6k+3}+8(P_{18k+3}-P_{6k+3})+10(P_{18k+2}-P_{6k+2})\\ \phantom{A=}-\frac{7\text{,}344}{41}(P_{14k+1}-P_{2k+1})-\frac{2\text{,}124}{41}(P_{14k}-P_{2k})-\frac{30\text{,}226\text{,}500}{1\text{,}681}(P_{10k-3}-P_{2k+3})\\ \phantom{A=}-\frac{12\text{,}507\text{,}840}{1\text{,}681}(P_{10k-4}+P_{2k+4})-\frac{4\text{,}442\text{,}760}{1\text{,}681}{P}_{6k}+3\times 8(P_{14k+4}+P_{10k+2})\\ \phantom{A=}+3\times 10(P_{14k+3}+P_{10k+1})-\frac{7\text{,}344\times 3}{41}(P_{10k+2}+P_{6k})-\frac{2\text{,}124\times 3}{41}(P_{10k+1}\\ \phantom{A=}+P_{6k-1})-\frac{30\text{,}226\text{,}500\times 3}{1\text{,}681}(P_{6k-2}+P_{2k-4})-\frac{12\text{,}507\text{,}840\times 3}{1\text{,}681}(P_{6k-3}+P_{2k-5})\\ \phantom{A=}-\frac{4\text{,}442\text{,}760}{1\text{,}681}{P}_{2k+1}-3\times 8(P_{14k+3}-P_{10k+3})-3\times 10(P_{14k+2}-P_{10k+2})\\ \phantom{A=}+\frac{7\text{,}344\times 3}{41}(P_{10k+1}-P_{6k+1})+\frac{2\text{,}124\times 3}{41}(P_{10k}-P_{6k})+\frac{30\text{,}226\text{,}500\times 3}{1\text{,}681}(P_{6k-3}\\ \phantom{A=}-P_{2k-3})+\frac{12\text{,}507\text{,}840\times 3}{1\text{,}681}(P_{6k-4}-P_{2k-4})+\frac{4\text{,}442\text{,}760}{1\text{,}681}{P}_{2k}\\ \phantom{A}=154P_{18k+3}+70P_{18k+2}-\frac{95\text{,}514}{41}{P}_{14k+1}-\frac{38\text{,}910}{41}{P}_{14k}-\frac{486\text{,}612\text{,}540}{1\text{,}681}{P}_{10k-3}\\ \phantom{A=}-\frac{201\text{,}554\text{,}538}{1\text{,}681}{P}_{10k-4}-\frac{977\text{,}366\text{,}722}{1\text{,}681}{P}_{6k-3}-\frac{344\text{,}423\text{,}038}{1\text{,}681}{P}_{6k-4}\\ \phantom{A=}-\frac{285\text{,}928\text{,}452}{1\text{,}681}{P}_{2k-4}-\frac{118\text{,}454\text{,}868}{1\text{,}681}{P}_{2k-5},\\ B=378(P_{18k+2}+P_{6k+4})+154(P_{18k+1}-P_{6k+5})-\frac{78}{41}(P_{14k+4}+P_{10k+2})\\ \phantom{B=}-\frac{318}{41}(P_{14k+3}-P_{10k+3})-6\times 378(P_{14k+1}+P_{2k+3})-6\times 154(P_{14k}-P_{2k+4})\\ \phantom{B=}+\frac{78\times 6}{41}(P_{10k+3}+P_{6k+1})+\frac{318\times 6}{41}(P_{10k+2}-P_{6k+2})+9\times 378(P_{10k}-P_{2k-2})\\ \phantom{B=}+9\times 154(P_{10k-1}-P_{2k-3})-\frac{78\times 9}{41}(P_{6k+2}+P_{2k})-\frac{318\times 9}{41}(P_{6k+1}-P_{2k+1})\\ \phantom{B=}+4\times 378P_{6k-1}+4\times 154P_{6k-2}-\frac{78\times 4}{41}{P}_{2k+1}-\frac{318\times 4}{41}{P}_{2k}\\ \phantom{B}=154P_{18k+3}+70P_{18k+2}-\frac{95\text{,}514}{41}{P}_{14k+1}-\frac{38\text{,}910}{41}{P}_{14k}+\frac{158\text{,}064}{41}{P}_{10k}\\ \phantom{B=}+\frac{64\text{,}416}{41}{P}_{10k-1}+\frac{36\text{,}496}{41}{P}_{6k-1}+\frac{14\text{,}880}{41}{P}_{6k-2}-\frac{225\text{,}516}{41}{P}_{2k-2}-\frac{92\text{,}796}{41}{P}_{2k-3}.\end{array}$$

Observe that the major terms of A and B (above those of order $P_{10k}$) are in total agreement. Note that $P_{n+2}=2P_{n+1}+P_n$, we have

$$\begin{array}{rl}B-A=& \frac{41\text{,}028\text{,}234}{1\text{,}681}{P}_{10k}+\frac{17\text{,}049\text{,}300}{1\text{,}681}{P}_{10k-1}\\ +\frac{348\text{,}025\text{,}790}{1\text{,}681}{P}_{6k-2}+\frac{290\text{,}016\text{,}982}{1\text{,}681}{P}_{6k-3}\\ +\frac{39\text{,}772\text{,}560}{1\text{,}681}{P}_{2k-2}+\frac{16\text{,}612\text{,}800}{1\text{,}681}{P}_{2k-3}>0\end{array}$$

for all integers $k\ge 1$. So, inequalities (3), (4) and (5) hold for all integers $k\ge 1$.

Now, applying (3) repeatedly, we have

$$\begin{array}{rl}\sum _{k=2m}^{\mathrm{\infty }}\frac{1}{P_k^3}=& \sum _{k=m}^{\mathrm{\infty }}(\frac{1}{P_{2k}^3}+\frac{1}{P_{2k+1}^3})\\ <& \sum _{k=m}^{\mathrm{\infty }}\frac{1}{P_{2k}^2{P}_{2k-1}+3P_{2k}{P}_{2k-1}^2-\frac{1}{82}(61P_{2k}+91P_{2k-1})}\\ -\sum _{k=m}^{\mathrm{\infty }}\frac{1}{P_{2k+2}^2{P}_{2k+1}+3P_{2k+2}{P}_{2k+1}^2-\frac{1}{82}(61P_{2k+2}+91P_{2k+1})}\\ =& \frac{1}{P_{2m}^2{P}_{2m-1}+3P_{2m}{P}_{2m-1}^2-\frac{1}{82}(61P_{2m}+91P_{2m-1})}.\end{array}$$

(6)

On the other hand, we prove the inequality

$$\begin{array}{rl}\frac{1}{P_{2k}^3}+\frac{1}{P_{2k+1}^3}>& \frac{1}{P_{2k}^2{P}_{2k-1}+3P_{2k}{P}_{2k-1}^2-\frac{1}{82}(61P_{2k}+91P_{2k-1})+\frac{1}{82}}\\ -\frac{1}{P_{2k+2}^2{P}_{2k+1}+3P_{2k+2}{P}_{2k+1}^2-\frac{1}{82}(61P_{2k+2}+91P_{2k+1})+\frac{1}{82}}.\end{array}$$

(7)

This inequality is equivalent to

$$\begin{array}{r}\frac{P_{6k+3}+P_{6k}+3P_{2k+1}-3P_{2k}}{Q_{12k+3}-6Q_{8k+2}+9Q_{4k+1}+4}\\ >\frac{378P_{6k-1}+154P_{6k-2}-\frac{78}{41}{P}_{2k+1}-\frac{318}{41}{P}_{2k}}{(P_{6k-1}+3P_{6k-2}-\frac{39}{41}{P}_{2k}-\frac{159}{41}{P}_{2k-1}+\frac{4}{41})(P_{6k+5}+3P_{6k+4}-\frac{39}{41}{P}_{2k+2}-\frac{159}{41}{P}_{2k+1}+\frac{4}{41})}\end{array}$$

or

$$\begin{array}{r}\frac{4}{41}(P_{6k+3}+P_{6k}+3P_{2k+1}-3P_{2k})(60P_{6k+1}+40P_{6k}-\frac{552}{41}{P}_{2k}-\frac{396}{41}{P}_{2k-1}+\frac{4}{41})\\ >B-A\end{array}$$

or

$$\begin{array}{r}140Q_{12k+3}+140Q_{12k+2}+\frac{1\text{,}200}{41}{Q}_{8k+2}+\frac{5\text{,}952}{41}{Q}_{8k+1}+\frac{19\text{,}116}{41}{Q}_{4k}+\frac{9\text{,}444}{41}{Q}_{4k-1}\\ +\frac{4}{41}{P}_{6k+3}+\frac{4}{41}{P}_{6k}+\frac{12}{41}{P}_{2k+1}-\frac{12}{41}{P}_{2k}+\frac{47\text{,}728}{41}>\frac{41}{4}(B-A).\end{array}$$

(8)

It is clear that inequality (8) holds for all integers $k\ge 5$, so inequality (7) is true. Now, applying (7) repeatedly, we have

$$\begin{array}{rl}\sum _{k=2m}^{\mathrm{\infty }}\frac{1}{P_k^3}=& \sum _{k=m}^{\mathrm{\infty }}(\frac{1}{P_{2k}^3}+\frac{1}{P_{2k+1}^3})\\ >& \frac{1}{P_{2m}^2{P}_{2m-1}+3P_{2m}{P}_{2m-1}^2-\frac{1}{82}(61P_{2m}+91P_{2m-1})+\frac{1}{82}}.\end{array}$$

(9)

Combining (6) and (9), we may immediately deduce inequality (2).

Now we consider that $n=2m+1$ is an odd number. It is clear that in this case our theorem is equivalent to

$$\begin{array}{r}{P}_{2m+1}^2{P}_{2m}+3P_{2m+1}{P}_{2m}^2+\frac{1}{82}(61P_{2m+1}+91P_{2m})\\ <{\left(\sum _{k=2m+1}^{\mathrm{\infty }}\frac{1}{P_k^3}\right)}^{-1}<P_{2m+1}^2{P}_{2m}+3P_{2m+1}{P}_{2m}^2+\frac{1}{82}(61P_{2m+1}+91P_{2m})+\frac{1}{82}\end{array}$$

or

$$\begin{array}{r}\frac{1}{P_{2m+1}^2{P}_{2m}+3P_{2m+1}{P}_{2m}^2+\frac{1}{82}(61P_{2m+1}+91P_{2m})+\frac{1}{82}}\\ <\sum _{k=2m+1}^{\mathrm{\infty }}\frac{1}{P_k^3}<\frac{1}{P_{2m+1}^2{P}_{2m}+3P_{2m+1}{P}_{2m}^2+\frac{1}{82}(61P_{2m+1}+91P_{2m})}.\end{array}$$

(10)

First we prove the inequality

$$\begin{array}{rl}\frac{1}{P_{2k+1}^3}+\frac{1}{P_{2k+2}^3}<& \frac{1}{P_{2k+1}^2{P}_{2k}+3P_{2k+1}{P}_{2k}^2+\frac{1}{82}(61P_{2k+1}+91P_{2k})}\\ -\frac{1}{P_{2k+1}^2{P}_{2k}+3P_{2k+1}{P}_{2k}^2+\frac{1}{82}(61P_{2k+1}+91P_{2k})}.\end{array}$$

(11)

It is easy to check that inequality (11) is correct for $k=1,2\text{ and }3$. So, we can assume that $k\ge 4$. Note that $P_{2k+1}^3=\frac{1}{8}(P_{6k+3}+3P_{2k+1})$, $P_{2k+2}^3=\frac{1}{8}(P_{6k+6}-3P_{2k+2})$, $P_{2k+1}^3+P_{2k+2}^3=\frac{1}{8}(P_{6k+6}+P_{6k+3}+3P_{2k+1}-3P_{2k+2})$, $P_{2k+1}^3{P}_{2k+2}^3=\frac{1}{512}(Q_{12k+9}+6Q_{8k+6}+9Q_{4k+3}-4)$, $P_{2k+1}^2{P}_{2k}+3P_{2k+1}{P}_{2k}^2=\frac{1}{8}(P_{6k+2}+3P_{6k+1}-5P_{2k+1}-5P_{2k})$, so inequality (11) is equivalent to the inequality

$$\begin{array}{r}\frac{P_{6k+6}+P_{6k+3}+3P_{2k+1}-3P_{2k+2}}{Q_{12k+9}+6Q_{8k+6}+9Q_{4k+3}-4}\\ <\frac{378P_{6k+2}+154P_{6k+1}+\frac{78}{41}{P}_{2k+2}+\frac{318}{41}{P}_{2k+1}}{(P_{6k+2}+3P_{6k+1}+\frac{39}{41}{P}_{2k+1}+\frac{159}{41}{P}_{2k})(P_{6k+8}+3P_{6k+7}+\frac{39}{41}{P}_{2k+3}+\frac{159}{41}{P}_{2k+2})}.\end{array}$$

(12)

From the definition and properties of the Pell-Lucas numbers, we have

$$(P_{6k+2}+3P_{6k+1})(P_{6k+8}+3P_{6k+7})=(8Q_{12k+9}+10Q_{12k+8}+2\text{,}772)/8$$

and

$$\begin{array}{r}(P_{6k+2}+3P_{6k+1}+\frac{39}{41}{P}_{2k+1}+\frac{159}{41}{P}_{2k})(P_{6k+8}+3P_{6k+7}+\frac{39}{41}{P}_{2k+3}+\frac{159}{41}{P}_{2k+2})\\ =\frac{1}{8}(8Q_{12k+9}+10Q_{12k+8}+\frac{7\text{,}344}{41}{Q}_{8k+5}+\frac{2\text{,}124}{41}{Q}_{8k+4}-\frac{31\text{,}844\text{,}565}{1\text{,}681}{Q}_{4k-1}\\ -\frac{13\text{,}186\text{,}923}{1\text{,}681}{Q}_{4k-2}+\frac{4\text{,}442\text{,}760}{1\text{,}681}).\end{array}$$

By these two identities and (12), we deduce that inequality (11) is equivalent to

$$\begin{array}{r}\frac{P_{6k+6}+P_{6k+3}+3P_{2k+1}-3P_{2k+2}}{Q_{12k+9}+6Q_{8k+6}+9Q_{4k+3}-4}\\ <\frac{378P_{6k+2}+154P_{6k+1}+\frac{78}{41}{P}_{2k+2}+\frac{318}{41}{P}_{2k+1}}{8Q_{12k+9}+10Q_{12k+8}+\frac{7\text{,}344}{41}{Q}_{8k+5}+\frac{2\text{,}124}{41}{Q}_{8k+4}-\frac{31\text{,}844\text{,}565}{1\text{,}681}{Q}_{4k-1}-\frac{13\text{,}186\text{,}923}{1\text{,}681}{Q}_{4k-2}+\frac{4\text{,}442\text{,}760}{1\text{,}681}}.\end{array}$$

(13)

For convenience, we let

$$\begin{array}{rl}{A}^{\prime }=& (8Q_{12k+9}+10Q_{12k+8}+\frac{7\text{,}344}{41}{Q}_{8k+5}+\frac{2\text{,}124}{41}{Q}_{8k+4}-\frac{31\text{,}844\text{,}565}{1\text{,}681}{Q}_{4k-1}\\ -\frac{13\text{,}186\text{,}923}{1\text{,}681}{Q}_{4k-2}+\frac{4\text{,}442\text{,}760}{1\text{,}681})\times (P_{6k+6}+P_{6k+3}+3P_{2k+1}-3P_{2k+2})\end{array}$$

and $B^{\prime }=(Q_{12k+9}+6Q_{8k+6}+9Q_{4k+3}-4)(378P_{6k+2}+154P_{6k+1}+\frac{78}{41}{P}_{2k+2}+\frac{318}{41}{P}_{2k+1})$. Then we have

$$\begin{array}{r}{A}^{\prime }=8(P_{18k+15}-P_{6k+3})+10(P_{18k+14}-P_{6k+2})+\frac{7\text{,}344}{41}(P_{14k+11}-P_{2k-1})\\ \phantom{A^{\prime }=}+\frac{2\text{,}124}{41}(P_{14k+10}-P_{2k-2})-\frac{31\text{,}844\text{,}565}{1\text{,}681}(P_{10k+5}-P_{2k+7})-\frac{13\text{,}186\text{,}923}{1\text{,}681}(P_{10k+4}\\ \phantom{A^{\prime }=}+P_{2k+8})+\frac{4\text{,}442\text{,}760}{1\text{,}681}{P}_{6k+6}+8(P_{18k+12}+P_{6k+6})+10(P_{18k+11}+P_{6k+5})\\ \phantom{A^{\prime }=}+\frac{7\text{,}344}{41}(P_{14k+8}+P_{2k+2})+\frac{2\text{,}124}{41}(P_{14k+7}+P_{2k+1})-\frac{31\text{,}844\text{,}565}{1\text{,}681}(P_{10k+2}-P_{2k+4})\\ \phantom{A^{\prime }=}-\frac{13\text{,}186\text{,}923}{1\text{,}681}(P_{10k+1}+P_{2k+5})+\frac{4\text{,}442\text{,}760}{1\text{,}681}{P}_{6k+3}+3\times 8(P_{14k+10}+P_{10k+8})\\ \phantom{A^{\prime }=}+3\times 10(P_{14k+9}+P_{10k+7})+\frac{7\text{,}344\times 3}{41}(P_{10k+6}+P_{6k+4})+\frac{2\text{,}124\times 3}{41}(P_{10k+5}\\ \phantom{A^{\prime }=}+P_{6k+3})-\frac{31\text{,}844\text{,}565\times 3}{1\text{,}681}(P_{6k}+P_{2k-2})-\frac{13\text{,}186\text{,}923\times 3}{1\text{,}681}(P_{6k-1}+P_{2k-3})\\ \phantom{A^{\prime }=}+\frac{4\text{,}442\text{,}760\times 3}{1\text{,}681}{P}_{2k+1}-3\times 8(P_{14k+11}-P_{10k+7})-3\times 10(P_{14k+10}-P_{10k+6})\\ \phantom{A^{\prime }=}-\frac{7\text{,}344\times 3}{41}(P_{10k+7}-P_{6k+3})-\frac{2\text{,}124\times 3}{41}(P_{10k+6}-P_{6k+2})+\frac{31\text{,}844\text{,}565\times 3}{1\text{,}681}(P_{6k+1}\\ \phantom{A^{\prime }=}-P_{2k-3})+\frac{13\text{,}186\text{,}923\times 3}{1\text{,}681}(P_{6k}-P_{2k-4})-\frac{4\text{,}442\text{,}760\times 3}{1\text{,}681}{P}_{2k+2}\\ \phantom{A^{\prime }}=154P_{18k+12}+70P_{18k+11}+\frac{95\text{,}514}{41}{P}_{14k+8}+\frac{38\text{,}910}{41}{P}_{14k+7}-\frac{506\text{,}756\text{,}250}{1\text{,}681}{P}_{10k+2}\\ \phantom{A^{\prime }=}-\frac{209\text{,}906\text{,}976}{1\text{,}681}{P}_{10k+1}+\frac{983\text{,}915\text{,}086}{1\text{,}681}{P}_{6k}\\ \phantom{A^{\prime }=}+\frac{407\text{,}692\text{,}984}{1\text{,}681}{P}_{6k-1}-\frac{771\text{,}966\text{,}210}{1\text{,}681}{P}_{2k-3}\\ \phantom{A^{\prime }=}-\frac{304\text{,}496\text{,}118}{1\text{,}681}{P}_{2k-4},\\ B^{\prime }=378(P_{18k+11}-P_{6k+7})+154(P_{18k+10}+P_{6k+8})+\frac{78}{41}(P_{14k+11}-P_{10k+7})\\ \phantom{B^{\prime }=}+\frac{318}{41}(P_{14k+10}+P_{10k+8})+6\times 378(P_{14k+8}-P_{2k+4})+924(P_{14k+7}+P_{2k+5})\\ \phantom{B^{\prime }=}+\frac{468}{41}(P_{10k+8}-P_{6k+4})+\frac{1\text{,}908}{41}(P_{10k+7}+P_{6k+5})+9\times 378(P_{10k+5}-P_{2k-1})\\ \phantom{B^{\prime }=}+1\text{,}386(P_{10k+4}-P_{2k-2})+\frac{702}{41}(P_{6k+5}-P_{2k+1})+\frac{318\times 9}{41}(P_{6k+4}+P_{2k+2})\\ \phantom{B^{\prime }=}-4\times 378P_{6k+2}-4\times 154P_{6k+1}-\frac{78\times 4}{41}{P}_{2k+2}-\frac{318\times 4}{41}{P}_{2k+1}\\ \phantom{B^{\prime }}=154P_{18k+12}+70P_{18k+11}+\frac{95\text{,}514}{41}{P}_{14k+8}+\frac{38\text{,}910}{41}{P}_{14k+7}+\frac{158\text{,}064}{41}{P}_{10k+5}\\ \phantom{B^{\prime }=}+\frac{64\text{,}416}{41}{P}_{10k+4}-\frac{36\text{,}496}{41}{P}_{6k+2}-\frac{14\text{,}880}{41}{P}_{6k+1}-\frac{225\text{,}516}{41}{P}_{2k-1}-\frac{92\text{,}796}{41}{P}_{2k-2}.\end{array}$$

Note that $P_{n+2}=2P_{n+1}+P_n$, we have

$$\begin{array}{rcl}{B}^{\prime }-A^{\prime }& =& \frac{597\text{,}729\text{,}018}{1\text{,}681}{P}_{10k+2}+\frac{247\text{,}592\text{,}208}{1\text{,}681}{P}_{10k+1}-\frac{992\text{,}616\text{,}926}{1\text{,}681}{P}_{6k}\\ -\frac{411\text{,}295\text{,}736}{1\text{,}681}{P}_{6k-1}+\frac{718\text{,}126\text{,}158}{1\text{,}681}{P}_{2k-3}+\frac{282\text{,}199\text{,}170}{1\text{,}681}{P}_{2k-4}\\ =& (\frac{3\text{,}483\text{,}829\text{,}506}{1\text{,}681}{P}_{10k}-\frac{992\text{,}616\text{,}926}{1\text{,}681}{P}_{6k})+(\frac{1\text{,}443\text{,}050\text{,}244}{1\text{,}681}{P}_{10k-1}\\ -\frac{411\text{,}295\text{,}736}{1\text{,}681}{P}_{6k-1})+\frac{718\text{,}126\text{,}158}{1\text{,}681}{P}_{2k-3}+\frac{282\text{,}199\text{,}170}{1\text{,}681}{P}_{2k-4}>0\end{array}$$

for all integers $k\ge 4$. So, inequalities (11), (12) and (13) hold for all integers $k\ge 4$.

Now, applying (11) repeatedly, we have

$$\begin{array}{rl}\sum _{k=2m+1}^{\mathrm{\infty }}\frac{1}{P_k^3}=& \sum _{k=m}^{\mathrm{\infty }}(\frac{1}{P_{2k+1}^3}+\frac{1}{P_{2k+2}^3})\\ <& \sum _{k=m}^{\mathrm{\infty }}\frac{1}{P_{2k+1}^2{P}_{2k}+3P_{2k+1}{P}_{2k}^2+\frac{1}{82}(61P_{2k+1}+91P_{2k})}\\ -\sum _{k=m}^{\mathrm{\infty }}\frac{1}{P_{2k+3}^2{P}_{2k+2}+3P_{2k+3}{P}_{2k+2}^2+\frac{1}{82}(61P_{2k+3}+91P_{2k+2})}.\end{array}$$

(14)

On the other hand, we prove the inequality

$$\begin{array}{rl}\frac{1}{P_{2k+1}^3}+\frac{1}{P_{2k+2}^3}>& \frac{1}{P_{2k+1}^2{P}_{2k}+3P_{2k+1}{P}_{2k}^2+\frac{1}{82}(61P_{2k+1}+91P_{2k})+\frac{1}{82}}\\ -\frac{1}{P_{2k+3}^2{P}_{2k+2}+3P_{2k+3}{P}_{2k+2}^2+\frac{1}{82}(61P_{2k+3}+91P_{2k+2})+\frac{1}{82}}.\end{array}$$

(15)

It is easy to check that inequality (15) is correct for $k=1,2\text{ and }3$. So, we can assume that $k\ge 4$. This time, inequality (15) is equivalent to

$$\begin{array}{r}\frac{P_{6k+6}+P_{6k+3}+3P_{2k+1}-3P_{2k+2}}{Q_{12k+9}+6Q_{8k+6}+9Q_{4k+3}-4}\\ >\frac{378P_{6k+2}+154P_{6k+1}+\frac{78}{41}{P}_{2k+2}+\frac{318}{41}{P}_{2k+1}}{(P_{6k+2}+3P_{6k+1}+\frac{39}{41}{P}_{2k+1}+\frac{159}{41}{P}_{2k}+\frac{4}{41})(P_{6k+8}+3P_{6k+7}+\frac{39}{41}{P}_{2k+3}+\frac{159}{41}{P}_{2k+2}+\frac{4}{41})}\end{array}$$

or

$$\begin{array}{r}\frac{4}{41}(P_{6k+6}+P_{6k+3}+3P_{2k+1}-3P_{2k+2})(60P_{6k+4}+40P_{6k+3}+\frac{552}{41}{P}_{2k+1}+\frac{396}{41}{P}_{2k}+\frac{4}{41})\\ >B^{\prime }-A^{\prime }\end{array}$$

or

$$\begin{array}{r}140Q_{12k+9}+140Q_{12k+8}-\frac{17\text{,}904}{41}{Q}_{8k+4}-\frac{8\text{,}352}{41}{Q}_{8k+3}+\frac{19\text{,}116}{41}{Q}_{4k+2}+\frac{9\text{,}444}{41}{Q}_{4k+1}\\ +\frac{4}{41}{P}_{6k+6}+\frac{4}{41}{P}_{6k+3}+\frac{12}{41}{P}_{2k+1}-\frac{12}{41}{P}_{2k+2}+\frac{21\text{,}152}{41}>\frac{41}{4}(B^{\prime }-A^{\prime }).\end{array}$$

(16)

It is clear that inequality (16) holds for all integers $k\ge 4$, so inequality (15) is true. Now, applying (15) repeatedly, we have

$$\begin{array}{rl}\sum _{k=2m+1}^{\mathrm{\infty }}\frac{1}{P_k^3}=& \sum _{k=m}^{\mathrm{\infty }}(\frac{1}{P_{2k+1}^3}+\frac{1}{P_{2k+2}^3})\\ >& \frac{1}{P_{2m+1}^2{P}_{2m}+3P_{2m+1}{P}_{2m}^2+\frac{1}{82}(61P_{2m+1}+91P_{2m})+\frac{1}{82}}.\end{array}$$

(17)

Combining (14) and (17), we may immediately deduce inequality (10).

Now our theorem follows from inequalities (2) and (10). This completes the proof of our theorem.