Skip to main content

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 k = n 1 P n 3 explicitly in terms of the Pell numbers themselves.

MSC: 11B39.

1 Introduction

For any integer n0, the well-known Pell numbers P n are defined by the second-order linear recurrence sequence P n + 2 =2 P n + 1 + P n , where P 0 =0 and P 1 =1. The Pell-Lucas numbers Q n are defined by Q n + 2 =2 Q n + 1 + Q n , where Q 0 =2 and Q 1 =2. Let α=1+ 2 and β=1 2 . Then from the characteristic equations x 2 2x1=0, we also have the computational formulae

P n = 1 2 2 ( α n β n ) and Q n = α n + β 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, , Q 0 =2, Q 1 =2, Q 2 =6, Q 3 =14, Q 4 =34, Q 5 =82, .

Various properties of the Pell numbers and related sequences have been studied by many authors, see [16]. 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:

( k = n 1 P k ) 1 = { P n 1 + P n 2 if  n  is even and  n 2 ; P n 1 + P n 2 1 if  n  is odd and  n 1 , ( k = n 1 P k 2 ) 1 = { 2 P n 1 P n 1 if  n  is an even number; 2 P n 1 P n if  n  is an odd number,

where x 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 ( k = n 1 u k ) 1 and proved an interesting conclusion:

( k = n 1 u k ) 1 = u n u n 1 for all n> n 0 ,

where denotes the nearest integer, namely x=x+ 1 2 , { u n } n 0 is an integer sequence satisfying the recurrence formula

u n =a u n 1 + u n 2 ++ u n s (s2)

with the initial conditions u 0 0, u k N, 1ks1.

Using the method in [10] seems to be very difficult to deal with ( k = n 1 u k s ) 1 for all integers s2.

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

P(s,n) ( k = n 1 P k s ) 1
(1)

for all integers s3. 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 n1, we have the identity

P(3,n)= { P n 2 P n 1 + 3 P n P n 1 2 + 61 82 P n 91 82 P n 1 if n is even and n 2 ; P n 2 P n 1 + 3 P n P n 1 2 + 61 82 P n + 91 82 P n 1 if n is odd and n 1 .

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 k4.

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

P 2 m 2 P 2 m 1 + 3 P 2 m P 2 m 1 2 1 82 ( 61 P 2 m + 91 P 2 m 1 ) < ( k = 2 m 1 P k 3 ) 1 < P 2 m 2 P 2 m 1 + 3 P 2 m P 2 m 1 2 1 82 ( 61 P 2 m + 91 P 2 m 1 ) + 1 82

or

1 P 2 m 2 P 2 m 1 + 3 P 2 m P 2 m 1 2 1 82 ( 61 P 2 m + 91 P 2 m 1 ) + 1 82 < k = 2 m 1 P k 3 < 1 P 2 m 2 P 2 m 1 + 3 P 2 m P 2 m 1 2 1 82 ( 61 P 2 m + 91 P 2 m 1 ) .
(2)

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

1 P 2 k 3 + 1 P 2 k + 1 3 < 1 P 2 k 2 P 2 k 1 + 3 P 2 k P 2 k 1 2 1 82 ( 61 P 2 k + 91 P 2 k 1 ) 1 P 2 k + 2 2 P 2 k + 1 + 3 P 2 k + 2 P 2 k + 1 2 1 82 ( 61 P 2 k + 2 + 91 P 2 k + 1 ) .
(3)

It is clear that (3) holds for k=1,2,3 and 4. So, without loss of generality, we can assume that k5. Note that P 2 k 3 = 1 8 ( P 6 k 3 P 2 k ), P 2 k + 1 3 = 1 8 ( P 6 k + 3 +3 P 2 k + 1 ), P 2 k 3 + P 2 k + 1 3 = 1 8 ( P 6 k + 3 + P 6 k +3 P 2 k + 1 3 P 2 k ), P 2 k 3 P 2 k + 1 3 = 1 512 ( Q 12 k + 3 6 Q 8 k + 2 +9 Q 4 k + 1 +4) and

P 2 k 2 P 2 k 1 +3 P 2 k P 2 k 1 2 = 1 8 ( P 6 k 1 +3 P 6 k 2 +5 P 2 k 1 +5 P 2 k ),

so inequality (3) is equivalent to

8 ( P 6 k + 3 + P 6 k + 3 P 2 k + 1 3 P 2 k ) Q 12 k + 3 6 Q 8 k + 2 + 9 Q 4 k + 1 + 4 < 378 P 6 k 1 + 154 P 6 k 2 78 41 P 2 k + 1 318 41 P 2 k ( P 6 k 1 + 3 P 6 k 2 39 41 P 2 k 159 41 P 2 k 1 ) ( P 6 k + 5 + 3 P 6 k + 4 39 41 P 2 k + 2 159 41 P 2 k + 1 ) .
(4)

From the definition and properties of P n and Q n , we can easily deduce the identities

P n P k = 1 8 Q n + k ( 1 ) k 8 Q n k , n k , Q n Q k = Q n + k + ( 1 ) k Q n k , n k , P n Q k = P n + k + ( 1 ) k P n k , n k .

So, applying these formulae, we have

( P 6 k 1 +3 P 6 k 2 )( P 6 k + 5 +3 P 6 k + 4 )= 1 8 (8 Q 12 k + 3 +10 Q 12 k + 2 2,772)

and

( P 6 k 1 + 3 P 6 k 2 39 41 P 2 k 159 41 P 2 k 1 ) ( P 6 k + 5 + 3 P 6 k + 4 39 41 P 2 k + 2 159 41 P 2 k + 1 ) = 1 8 ( 8 Q 12 k + 3 + 10 Q 12 k + 2 7 , 344 41 Q 8 k + 1 2 , 124 41 Q 8 k 30 , 226 , 500 1 , 681 Q 4 k 3 ) 1 8 ( 12 , 507 , 840 1 , 681 Q 4 k 4 + 4 , 442 , 760 1 , 681 ) .

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

P 6 k + 3 + P 6 k + 3 P 2 k + 1 3 P 2 k Q 12 k + 3 6 Q 8 k + 2 + 9 Q 4 k + 1 + 4 < 378 P 6 k 1 + 154 P 6 k 2 78 41 P 2 k + 1 318 41 P 2 k 8 Q 12 k + 3 + 10 Q 12 k + 2 7 , 344 41 Q 8 k + 1 2 , 124 41 Q 8 k 30 , 226 , 500 1 , 681 Q 4 k 3 12 , 507 , 840 1 , 681 Q 4 k 4 4 , 442 , 760 1 , 681 .
(5)

For convenience, we let

A = ( P 6 k + 3 + P 6 k + 3 P 2 k + 1 3 P 2 k ) × ( 8 Q 12 k + 3 + 10 Q 12 k + 2 7 , 344 41 Q 8 k + 1 2 , 124 41 Q 8 k 30 , 226 , 500 1 , 681 Q 4 k 3 12 , 507 , 840 1 , 681 Q 4 k 4 4 , 442 , 760 1 , 681 )

and

B=( Q 12 k + 3 6 Q 8 k + 2 +9 Q 4 k + 1 +4) ( 378 P 6 k 1 + 154 P 6 k 2 78 41 P 2 k + 1 318 41 P 2 k ) .

Then by calculation it follows that

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

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

B A = 41 , 028 , 234 1 , 681 P 10 k + 17 , 049 , 300 1 , 681 P 10 k 1 + 348 , 025 , 790 1 , 681 P 6 k 2 + 290 , 016 , 982 1 , 681 P 6 k 3 + 39 , 772 , 560 1 , 681 P 2 k 2 + 16 , 612 , 800 1 , 681 P 2 k 3 > 0

for all integers k1. So, inequalities (3), (4) and (5) hold for all integers k1.

Now, applying (3) repeatedly, we have

k = 2 m 1 P k 3 = k = m ( 1 P 2 k 3 + 1 P 2 k + 1 3 ) < k = m 1 P 2 k 2 P 2 k 1 + 3 P 2 k P 2 k 1 2 1 82 ( 61 P 2 k + 91 P 2 k 1 ) k = m 1 P 2 k + 2 2 P 2 k + 1 + 3 P 2 k + 2 P 2 k + 1 2 1 82 ( 61 P 2 k + 2 + 91 P 2 k + 1 ) = 1 P 2 m 2 P 2 m 1 + 3 P 2 m P 2 m 1 2 1 82 ( 61 P 2 m + 91 P 2 m 1 ) .
(6)

On the other hand, we prove the inequality

1 P 2 k 3 + 1 P 2 k + 1 3 > 1 P 2 k 2 P 2 k 1 + 3 P 2 k P 2 k 1 2 1 82 ( 61 P 2 k + 91 P 2 k 1 ) + 1 82 1 P 2 k + 2 2 P 2 k + 1 + 3 P 2 k + 2 P 2 k + 1 2 1 82 ( 61 P 2 k + 2 + 91 P 2 k + 1 ) + 1 82 .
(7)

This inequality is equivalent to

P 6 k + 3 + P 6 k + 3 P 2 k + 1 3 P 2 k Q 12 k + 3 6 Q 8 k + 2 + 9 Q 4 k + 1 + 4 > 378 P 6 k 1 + 154 P 6 k 2 78 41 P 2 k + 1 318 41 P 2 k ( P 6 k 1 + 3 P 6 k 2 39 41 P 2 k 159 41 P 2 k 1 + 4 41 ) ( P 6 k + 5 + 3 P 6 k + 4 39 41 P 2 k + 2 159 41 P 2 k + 1 + 4 41 )

or

4 41 ( P 6 k + 3 + P 6 k + 3 P 2 k + 1 3 P 2 k ) ( 60 P 6 k + 1 + 40 P 6 k 552 41 P 2 k 396 41 P 2 k 1 + 4 41 ) > B A

or

140 Q 12 k + 3 + 140 Q 12 k + 2 + 1 , 200 41 Q 8 k + 2 + 5 , 952 41 Q 8 k + 1 + 19 , 116 41 Q 4 k + 9 , 444 41 Q 4 k 1 + 4 41 P 6 k + 3 + 4 41 P 6 k + 12 41 P 2 k + 1 12 41 P 2 k + 47 , 728 41 > 41 4 ( B A ) .
(8)

It is clear that inequality (8) holds for all integers k5, so inequality (7) is true. Now, applying (7) repeatedly, we have

k = 2 m 1 P k 3 = k = m ( 1 P 2 k 3 + 1 P 2 k + 1 3 ) > 1 P 2 m 2 P 2 m 1 + 3 P 2 m P 2 m 1 2 1 82 ( 61 P 2 m + 91 P 2 m 1 ) + 1 82 .
(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

P 2 m + 1 2 P 2 m + 3 P 2 m + 1 P 2 m 2 + 1 82 ( 61 P 2 m + 1 + 91 P 2 m ) < ( k = 2 m + 1 1 P k 3 ) 1 < P 2 m + 1 2 P 2 m + 3 P 2 m + 1 P 2 m 2 + 1 82 ( 61 P 2 m + 1 + 91 P 2 m ) + 1 82

or

1 P 2 m + 1 2 P 2 m + 3 P 2 m + 1 P 2 m 2 + 1 82 ( 61 P 2 m + 1 + 91 P 2 m ) + 1 82 < k = 2 m + 1 1 P k 3 < 1 P 2 m + 1 2 P 2 m + 3 P 2 m + 1 P 2 m 2 + 1 82 ( 61 P 2 m + 1 + 91 P 2 m ) .
(10)

First we prove the inequality

1 P 2 k + 1 3 + 1 P 2 k + 2 3 < 1 P 2 k + 1 2 P 2 k + 3 P 2 k + 1 P 2 k 2 + 1 82 ( 61 P 2 k + 1 + 91 P 2 k ) 1 P 2 k + 1 2 P 2 k + 3 P 2 k + 1 P 2 k 2 + 1 82 ( 61 P 2 k + 1 + 91 P 2 k ) .
(11)

It is easy to check that inequality (11) is correct for k=1,2 and 3. So, we can assume that k4. Note that P 2 k + 1 3 = 1 8 ( P 6 k + 3 +3 P 2 k + 1 ), P 2 k + 2 3 = 1 8 ( P 6 k + 6 3 P 2 k + 2 ), P 2 k + 1 3 + P 2 k + 2 3 = 1 8 ( P 6 k + 6 + P 6 k + 3 +3 P 2 k + 1 3 P 2 k + 2 ), P 2 k + 1 3 P 2 k + 2 3 = 1 512 ( Q 12 k + 9 +6 Q 8 k + 6 +9 Q 4 k + 3 4), P 2 k + 1 2 P 2 k +3 P 2 k + 1 P 2 k 2 = 1 8 ( P 6 k + 2 +3 P 6 k + 1 5 P 2 k + 1 5 P 2 k ), so inequality (11) is equivalent to the inequality

P 6 k + 6 + P 6 k + 3 + 3 P 2 k + 1 3 P 2 k + 2 Q 12 k + 9 + 6 Q 8 k + 6 + 9 Q 4 k + 3 4 < 378 P 6 k + 2 + 154 P 6 k + 1 + 78 41 P 2 k + 2 + 318 41 P 2 k + 1 ( P 6 k + 2 + 3 P 6 k + 1 + 39 41 P 2 k + 1 + 159 41 P 2 k ) ( P 6 k + 8 + 3 P 6 k + 7 + 39 41 P 2 k + 3 + 159 41 P 2 k + 2 ) .
(12)

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

( P 6 k + 2 +3 P 6 k + 1 )( P 6 k + 8 +3 P 6 k + 7 )=(8 Q 12 k + 9 +10 Q 12 k + 8 +2,772)/8

and

( P 6 k + 2 + 3 P 6 k + 1 + 39 41 P 2 k + 1 + 159 41 P 2 k ) ( P 6 k + 8 + 3 P 6 k + 7 + 39 41 P 2 k + 3 + 159 41 P 2 k + 2 ) = 1 8 ( 8 Q 12 k + 9 + 10 Q 12 k + 8 + 7 , 344 41 Q 8 k + 5 + 2 , 124 41 Q 8 k + 4 31 , 844 , 565 1 , 681 Q 4 k 1 13 , 186 , 923 1 , 681 Q 4 k 2 + 4 , 442 , 760 1 , 681 ) .

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

P 6 k + 6 + P 6 k + 3 + 3 P 2 k + 1 3 P 2 k + 2 Q 12 k + 9 + 6 Q 8 k + 6 + 9 Q 4 k + 3 4 < 378 P 6 k + 2 + 154 P 6 k + 1 + 78 41 P 2 k + 2 + 318 41 P 2 k + 1 8 Q 12 k + 9 + 10 Q 12 k + 8 + 7 , 344 41 Q 8 k + 5 + 2 , 124 41 Q 8 k + 4 31 , 844 , 565 1 , 681 Q 4 k 1 13 , 186 , 923 1 , 681 Q 4 k 2 + 4 , 442 , 760 1 , 681 .
(13)

For convenience, we let

A = ( 8 Q 12 k + 9 + 10 Q 12 k + 8 + 7 , 344 41 Q 8 k + 5 + 2 , 124 41 Q 8 k + 4 31 , 844 , 565 1 , 681 Q 4 k 1 13 , 186 , 923 1 , 681 Q 4 k 2 + 4 , 442 , 760 1 , 681 ) × ( P 6 k + 6 + P 6 k + 3 + 3 P 2 k + 1 3 P 2 k + 2 )

and B =( Q 12 k + 9 +6 Q 8 k + 6 +9 Q 4 k + 3 4)(378 P 6 k + 2 +154 P 6 k + 1 + 78 41 P 2 k + 2 + 318 41 P 2 k + 1 ). Then we have

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

Note that P n + 2 =2 P n + 1 + P n , we have

B A = 597 , 729 , 018 1 , 681 P 10 k + 2 + 247 , 592 , 208 1 , 681 P 10 k + 1 992 , 616 , 926 1 , 681 P 6 k 411 , 295 , 736 1 , 681 P 6 k 1 + 718 , 126 , 158 1 , 681 P 2 k 3 + 282 , 199 , 170 1 , 681 P 2 k 4 = ( 3 , 483 , 829 , 506 1 , 681 P 10 k 992 , 616 , 926 1 , 681 P 6 k ) + ( 1 , 443 , 050 , 244 1 , 681 P 10 k 1 411 , 295 , 736 1 , 681 P 6 k 1 ) + 718 , 126 , 158 1 , 681 P 2 k 3 + 282 , 199 , 170 1 , 681 P 2 k 4 > 0

for all integers k4. So, inequalities (11), (12) and (13) hold for all integers k4.

Now, applying (11) repeatedly, we have

k = 2 m + 1 1 P k 3 = k = m ( 1 P 2 k + 1 3 + 1 P 2 k + 2 3 ) < k = m 1 P 2 k + 1 2 P 2 k + 3 P 2 k + 1 P 2 k 2 + 1 82 ( 61 P 2 k + 1 + 91 P 2 k ) k = m 1 P 2 k + 3 2 P 2 k + 2 + 3 P 2 k + 3 P 2 k + 2 2 + 1 82 ( 61 P 2 k + 3 + 91 P 2 k + 2 ) .
(14)

On the other hand, we prove the inequality

1 P 2 k + 1 3 + 1 P 2 k + 2 3 > 1 P 2 k + 1 2 P 2 k + 3 P 2 k + 1 P 2 k 2 + 1 82 ( 61 P 2 k + 1 + 91 P 2 k ) + 1 82 1 P 2 k + 3 2 P 2 k + 2 + 3 P 2 k + 3 P 2 k + 2 2 + 1 82 ( 61 P 2 k + 3 + 91 P 2 k + 2 ) + 1 82 .
(15)

It is easy to check that inequality (15) is correct for k=1,2 and 3. So, we can assume that k4. This time, inequality (15) is equivalent to

P 6 k + 6 + P 6 k + 3 + 3 P 2 k + 1 3 P 2 k + 2 Q 12 k + 9 + 6 Q 8 k + 6 + 9 Q 4 k + 3 4 > 378 P 6 k + 2 + 154 P 6 k + 1 + 78 41 P 2 k + 2 + 318 41 P 2 k + 1 ( P 6 k + 2 + 3 P 6 k + 1 + 39 41 P 2 k + 1 + 159 41 P 2 k + 4 41 ) ( P 6 k + 8 + 3 P 6 k + 7 + 39 41 P 2 k + 3 + 159 41 P 2 k + 2 + 4 41 )

or

4 41 ( P 6 k + 6 + P 6 k + 3 + 3 P 2 k + 1 3 P 2 k + 2 ) ( 60 P 6 k + 4 + 40 P 6 k + 3 + 552 41 P 2 k + 1 + 396 41 P 2 k + 4 41 ) > B A

or

140 Q 12 k + 9 + 140 Q 12 k + 8 17 , 904 41 Q 8 k + 4 8 , 352 41 Q 8 k + 3 + 19 , 116 41 Q 4 k + 2 + 9 , 444 41 Q 4 k + 1 + 4 41 P 6 k + 6 + 4 41 P 6 k + 3 + 12 41 P 2 k + 1 12 41 P 2 k + 2 + 21 , 152 41 > 41 4 ( B A ) .
(16)

It is clear that inequality (16) holds for all integers k4, so inequality (15) is true. Now, applying (15) repeatedly, we have

k = 2 m + 1 1 P k 3 = k = m ( 1 P 2 k + 1 3 + 1 P 2 k + 2 3 ) > 1 P 2 m + 1 2 P 2 m + 3 P 2 m + 1 P 2 m 2 + 1 82 ( 61 P 2 m + 1 + 91 P 2 m ) + 1 82 .
(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.

References

  1. 1.

    Kilic E, Altunkaynak B, Tasci D: On the computing of the generalized order- k Pell numbers in log time. Appl. Math. Comput. 2006, 181: 511-515. 10.1016/j.amc.2005.12.053

    MathSciNet  Article  Google Scholar 

  2. 2.

    Hao P: Arithmetic properties of q -Fibonacci numbers and q -Pell numbers. Discrete Math. 2006, 306: 2118-2127. 10.1016/j.disc.2006.03.067

    MathSciNet  Article  Google Scholar 

  3. 3.

    Santos JPO, Sills AV: q -Pell sequences and two identities of V.A. Lebesgue. Discrete Math. 2002, 257: 125-142. 10.1016/S0012-365X(01)00475-7

    MathSciNet  Article  Google Scholar 

  4. 4.

    Kilic E: The generalized order- k Fibonacci-Pell sequences by matrix methods. J. Comput. Appl. Math. 2007, 209: 133-145. 10.1016/j.cam.2006.10.071

    MathSciNet  Article  Google Scholar 

  5. 5.

    Egge ES, Mansour T: 132-avoiding two-stack sortable permutations, Fibonacci numbers, and Pell numbers. Discrete Appl. Math. 2004, 143: 72-83. 10.1016/j.dam.2003.12.007

    MathSciNet  Article  Google Scholar 

  6. 6.

    Mansour T, Shattuck M: Restricted partitions and q -Pell numbers. Cent. Eur. J. Math. 2011, 9: 346-355. 10.2478/s11533-011-0002-6

    MathSciNet  Article  Google Scholar 

  7. 7.

    Zhang W, Wang T: The infinite sum of reciprocal Pell numbers. Appl. Math. Comput. 2012, 218: 6164-6167. 10.1016/j.amc.2011.11.090

    MathSciNet  Article  Google Scholar 

  8. 8.

    Zhang W, Wang T: The infinite sum of reciprocal of the square of the Pell numbers. J. Weinan Teach. Univ. 2011, 26: 39-42.

    Google Scholar 

  9. 9.

    Ohtsuka H, Nakamura S: On the sum of reciprocal Fibonacci numbers. Fibonacci Q. 2008/2009, 46/47: 153-159.

    Google Scholar 

  10. 10.

    Komatsu T, Laohakosol V: On the sum of reciprocals of numbers satisfying a recurrence relation of order s . J. Integer Seq. 2010., 13: Article ID 10.5.8

    Google Scholar 

Download references

Acknowledgements

The authors express their gratitude to the referee for his very helpful and detailed comments. This work is supported by the N.S.F. (11001218, 11071194) of P.R. China and the Research Fund for the Doctoral Program of Higher Education (20106101120001) of P.R. China and the G.I.C.F. (YZZ12065) of NWU.

Author information

Affiliations

Authors

Corresponding author

Correspondence to Tingting Wang.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

ZX drafted the manuscript. TW participated in its design and coordination and helped to draft the manuscript. All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Cite this article

Xu, Z., Wang, T. The infinite sum of the cubes of reciprocal Pell numbers. Adv Differ Equ 2013, 184 (2013). https://doi.org/10.1186/1687-1847-2013-184

Download citation

Keywords

  • Pell numbers
  • floor function
  • identity