Open Access

The infinite sum of the cubes of reciprocal Pell numbers

Advances in Difference Equations20132013:184

https://doi.org/10.1186/1687-1847-2013-184

Received: 17 May 2013

Accepted: 4 June 2013

Published: 26 June 2013

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.

Keywords

Pell numbers floor function identity

1 Introduction

For any integer n 0 , 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 2 x 1 = 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 ( s 2 )

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

Using the method in [10] seems to be very difficult to deal with ( k = n 1 u k s ) 1 for all integers s 2 .

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 s 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 1 , 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 k 4 .

2 Proof of the theorem

In this section, we shall prove our theorem directly. First we consider the case that n = 2 m 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 k 5 . 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 k 1 . So, inequalities (3), (4) and (5) hold for all integers k 1 .

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 k 5 , 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 = 2 m + 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 k 4 . 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 k 4 . So, inequalities (11), (12) and (13) hold for all integers k 4 .

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 k 4 . 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 k 4 , 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.

Declarations

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.

Authors’ Affiliations

(1)
Department of Mathematics, Northwest University

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© Xu and Wang; licensee Springer 2013

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