New reciprocal sums involving finite products of second order recursions

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Miskolc Mathematical Notes HU e-ISSN 1787-2413 Vol. 20 (2019), No. 2, pp. 1039–1050 DOI: 10.18514/MMN.2019.2810

NEW RECIPROCAL SUMS INVOLVING FINITE PRODUCTS OF SECOND ORDER RECURSIONS

EMRAH KILIC¸ AND DIDEM ERSANLI

Received 15 January, 2019

Abstract. In this paper, we present new kinds of reciprocal sums of finite products of general second order linear recurrences. In order to evaluate explicitly them by q-calculus, first we convert them into their q-notation and then use the methods of partial fraction decomposition and creative telescoping.

2010 Mathematics Subject Classification: 11B37; 05A30

Keywords: reciprocal sums identities, partial fraction decomposition

1. INTRODUCTION

For n > 1, define the second order linear recurrences_fUng and fVng with UnD pUn 1C Un 2and VnD pVn 1C Vn 2;

where U0D 0, U1D 1 and V0D 2, V1D p, respectively.

If pD 1, then UnD Fn(nth Fibonacci number) and VnD Ln(nth Lucas number). If pD 2, then UnD Pn(nth Pell number) and VnD Qn(nth Pell-Lucas number).

The Binet forms are UnD ˛n ˇn ˛ ˇ D ˛n 1.1 qn/ .1 q/ and VnD ˛ n C ˇnD ˛n.1_{C q}n/; where ˛; ˇD .p ˙pp2_{C 4/=2, q D ˇ=˛ and i D}p _1.

Throughout this paper we will use the following notations: the q-Pochhammer symbol .x_{I q/}nD .1 x/.1 xq/ : : : .1 xqn 1/. When xD q; we denote .qI q/nby .q/n:

Many authors [1–11] have studied both finite or infinite and alternating or non-alternating reciprocal sums including terms of certain integer sequences. More re-cently, Frontczak [3] evaluated various reciprocal sums of the Fibonacci numbers. For example, he showed that for m; n 1

N X i D1 . 1/m.i C1/ Fmi Cn1 F_{m.i 1/Cn}F_{mi Cn}F_{m.i C1/Cn} D F_m1 FnFmF2m c

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_F m.N C1/ F_{m.N C1/Cn}C FmN F_{mN Cn} Fm F_mCn . 1/m FmN F2 mFmCnFm.N C1/Cn : Kılıc¸ and Prodinger [5] consider some classes of reciprocal sums of general Fibon-acci numbers, which were computed in closed form in an earlier work and then they evaluate the same sums by a different approach: First they convert them into their q-forms and then explicitly evaluate q-versions of the sums by using partial fraction decomposition method and creative telescoping idea.

In this paper, we shall consider new kinds of reciprocal sums including finite products of the general Fibonacci and Lucas numbers. We shall summarize what we will present in this paper below.

We consider three kinds of alternating reciprocal sums including finite products of the general Fibonacci and Lucas numbers. All of them have an integer parameter to increase or decrease the indices of the general Fibonacci or Lu-cas factors in both numerator and denominator in the sums. Two of these sums are of the forms

n X kD0 . 1/k Uk d U_kCdU_{kCd C1}U_{kCd C2} and n X kD0 . 1/k VkCd C1 U_kCdU_{kCd C1}U_{kCd C2}: The other kind sums has an additional parameter m which determines the number of the general Fibonacci or Lucas factors in the numerator or de-nominator. It will be of the form

n X kD0 . 1/k UkCcUkCcC1: : : UkCcCm 1 X_kCdX_{kCd C1}: : : X_{kCd CmC1}; where Xnis either Unor Vn:

Before all, we convert all the claimed results into their q-forms. After this, we will use partial fraction decomposition (pfd) method and creative telescoping idea to prove the q-version of the claimed results.

By the q-versions of three kinds of reciprocal sums, we present general cases of the original three kinds of sums with an additional integer parameter by taking a special choosing of q. By the way, we could able to derive non-alternating reciprocal sums where the indices of the general Fibonacci or Lucas numbers are in the arithmetic progressions.

Finally we shall give some applications of our results to certain alternating and non-alternating reciprocal sums with finite products of the Pell or Pell-Lucas numbers.

2. THE MAIN RESULTS

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Theorem 1. Forn; m 0 and c 2 f m C 1; : : : ; 1; 0; 1g; n X kD0 . 1/k UkCcUkCcC1: : : UkCcCm 1 X_kCdX_{kCd C1}: : : X_{kCd CmC1} D . 1/cC1 UnCcUnCcC1: : : UnCcCm U_mC1X_{d cC1}ŒX_{nCd C1}X_{nCd C2}: : : X_{nCd CmC1}; where Xn is eitherUn orVn: Note that when XnD Vn, we assume that d is any integer. WhenXnD Un, we assume thatd 1.

Proof. Let XnD Un. First we convert the LHS of the claim into its q-notation as shown n X kD0 . 1/k m 1 Q t D0 U_kCtCc mC1 Q t D0 U_kCtCd D ˛mc .mC2/d 2mC1.1 q/2 n X kD0 qk m 1 Q t D0 1 qkCcCt mC1 Q t D0 .1 qkCd Ct_/ :

Second we convert the RHS of the claim into its q-notation as shown

. 1/cC1 m Q t D0 U_nCtCc U_mC1U_{d cC1}mC1Q t D1 U_nCtCd D . 1/cC1˛.mC2/c .mC2/d 2m 1 .1 q/2 m Q t D0 1 qnCcCt .1 qmC1/.1 qd cC1/ mC1 Q t D1 .1 qnCd Ct/ :

After some simplifications, the q-version of the claimed result is

n X kD0 qk m 1 Q t D0 1 qkCcCt mC1 Q t D0 .1 qkCd Ct_/ D q1 c m Q t D0 1 qnCcCt .1 qmC1_/.1 _qd cC1_/mC1Q t D1 .1 qnCd Ct_/

or, in terms of the q-Pochhammer notation, n X kD0 qkqkCc_{I q} m qkCd_{I q} mC2 D q 1 c _qnCc_{I q} mC1 .1 qmC1_/.1 _qd cC1_{/ q}nCd C1_{I q} mC1 :

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Define SnWD n X kD0 qkqkCc_{I q} m qkCd_{I q} mC2 :

Denote the summand term of Snby Tk, that is,

TkD qkqkCc_{I q} m qkCdI q mC2 :

The partial fraction decomposition of Tk reads as qkqkCc_{I q} m qkCdI q mC2 D mC2 X t D1 q d t C1qc d t C1_{I q} m .1 qkCd Ct 1_/mC2Q i D1 i 6Dt .1 qi t_/ :

From [5], we could obtain the following identity comes from creative telescoping idea: n X kD0 ₁ 1 qkCd Cm 1 1 qkCd Ct D m t 1 X kD0 ₁ 1 qkCd CnCtC1 1 1 qkCd Ct : (2.1) In that case, by using the equality (2.1), we write

SnD mC1 X t D1 1 1 qd Ct 1 1 1 qd CnCt t X rD1 q d rC1qc d rC1_{I q} m q1 r_{I q} r 1.q/m rC2 D . 1/m 1 1 qmC1 mC1 X t D1 1 1 qd Ct 1 1 1 qd CnCt q.m t C1/c .m tC2/d Ct.t 1/Cm.m22tC1/ qc d 2_{I q} 3 t qd c_{I q} t mC1 .q/_{t 1}.q/_{m t C1} D . 1/m1 q nC1 1 qmC1 mC1 X t D1 qt . cCd mCt/ .1 qd Ct 1_/.1 _qd CnCt_/ qc 1 d C12m.mC1C2c 2d / qc d 2_{I q} 3 t qd c_{I q} t mC1 .q/_{t 1}.q/_{m t C1} D 1 q nC1 .1 qc d 1_/.1 _qmC1_/

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mC1 X t D1 . 1/t C1 q.tC12 / 1 qc d 1I q 2 t qc dI q m t C1 .1 qd Ct 1_/.1 _qd CnCt_{/ .q/} t 1.q/m t C1 D 1 q nC1 .1 qc d 1_/.1 _qmC1_/ m X t D0 . 1/t q.tC22 / 1 qc d 1_{I q} 1 t qc d_{I q} m t .1 qd Ct/.1 qd CnCtC1/ .q/t.q/m t D 1 q nC1 .1 qmC1_/.1 _qc d 1_/ m X t D0 . 1/t q. t 2/ .q t _qd_/.q t _qd CnC1_/ .1 q c d t 1_/.1 _qc d t_{/ : : : .1} _qc d Cm t 2_/.1 _qc d Cm t 1_/ .q/_t.q/_{m t} : Now we define h.´/WD .1 ´q c d 1_{/ : : : .1} _´qc d Cm 1_/ .´ qd_/.´ _qd CnC1_/.1 _´/.1 _{´q/ : : : .1} _´qm_/: The partial fraction decomposition of h.´/ is

h.´/ D .1 q c 1_{/ : : : .1} _qcCm 1_/ .´ qd_/.qd _qd CnC1_/.1 _qd_/.1 _qd C1_{/ : : : .1} _qd Cm_/ C .1 q cCn_{/ : : : .1} _qcCnCm_/ .qd CnC1 _qd_/.´ _qd CnC1_/.1 _qd CnC1_/.1 _qd CnC2_{/ : : : .1} _qd CnCmC1_/ C m X t D0 . 1/tq.tC12 / .1 q t Cc d 1_{/ : : : .1} _q t Cc d Cm 1_/ .q t _qd_/.q t _qd CnC1_{/ .q/} t.q/m t.1 ´qt/ : If we multiply h.´/ by ´ and let ´! 1, then we get

0D q d_.1 _qc 1_{/ : : : .1} _qcCm 1_/ .1 qnC1_/.1 _qd_/.1 _qd C1_{/ : : : .1} _qd Cm_/ q d.1 qcCn/ : : : .1 qcCnCm/ .1 qnC1_/.1 _qd CnC1_/.1 _qd CnC2_{/ : : : .1} _qd CnCmC1_/ C m X t D0 . 1/t C1q.2t/ .1 q t Cc d 1_{/ : : : .1} _q t Cc d Cm 1_/ .q t _qd_/.q t _qd CnC1_{/ .q/} t.q/m t : Since c2 f m C 1; : : : ; 1; 0; 1g; we write m X t D0 . 1/tq.2t/ .1 q t Cc d 1_{/ : : : .1} _q t Cc d Cm 1_/ .q t _qd_/.q t _qd CnC1_{/ .q/} t.q/m t

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D q

d_.1 _qnCc_{/ : : : .1} _qcCmCn_/

.1 qnC1_/.1 _qd CnC1_/.1 _qd CnC2_{/ : : : .1} _qd CmCnC1_/: Taking into the constant factor, we write

1 qnC1 .1 qmC1_/.1 _qc d 1_/ m X t D0 . 1/tq.2t/C1.1 ´q c d 1_{/ : : : .1} _´qc d Cm 1_/ .´ qd/.´ qd CnC1/ .q/t.q/m t D q d_.1 _qnCc_{/ : : : .1} _qcCmCn_/ .1 qmC1_/.1 _qc d 1_/.1 _qd CnC1_{/ : : : .1} _qd CmCnC1_/ D q d _qnCc_{I q} mC1 .1 qmC1/.1 qc d 1/ qnCd C1I q mC1 D q 1 c _qnCc_{I q} mC1 .1 qmC1/.1 qd cC1/ qnCd C1I q mC1 ;

which completes the proof.

Also, when XnD Vn; the proof is similarly obtained. Now we will present some interesting corollaries of Theorem 1. When mD 2; XnD Ln, UnD Fn; dD 3 and c D 0 in Theorem1, it gives us

n X kD0 . 1/k FkFkC1 L_kC3L_kC4L_kC5L_kC6 D FnFnC1FnC2 F3L4LnC4LnC5LnC6 :

When mD 3; XnD UnD Pn; dD 5 and c D 1 in Theorem1, we get n X kD0 . 1/k PkC1PkC2PkC3 P_kC5P_kC6P_kC7P_kC8P_kC9 D P_nC1P_nC2P_nC3P_nC4 P4P5PnC6PnC7PnC8PnC9 :

When mD 4; XnD Qn, UnD Pn; d D 4 and c D 2 in Theorem1, we get n X kD0 . 1/k Pk 2Pk 1PkPkC1 Q_kC4Q_kC5Q_kC6Q_kC7Q_kC8Q_kC9 D Pn 2Pn 1PnPnC1PnC2 P5Q7QnC5QnC6QnC7QnC8QnC9 : When mD 5; XnD UnD Fn; d D 5 and c D 3 in Theorem1, we get

n X kD0

. 1/k Fk 3Fk 2Fk 1FkFkC1

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D Fn 3Fn 2Fn 1FnFnC1FnC2 F6F9FnC6FnC7FnC8FnC9FnC10FnC11

: Now we shall give our second result. The first main result stands for the finite product of the general Fibonacci or Lucas numbers. But the next two results are valid for special cases.

Theorem 2. Ford > 0, n X kD0 . 1/k Uk d U_kCdU_{kCd C1}U_{kCd C2} D . 1/ d C1 U2nC2 U2U_{nCd C1}U_{nCd C2} : (2.2)

Proof. First we convert the LHS of (2.2) into its q-notation as shown n X kD0 . 1/k Uk d U_kCdU_{kCd C1}U_{kCd C2} D ˛ 4d 1.1 q/2 n X kD0 qk.1 qk d/ .1 qkCd_/.1 _qkCd C1_/.1 _qkCd C2_/: Second we convert the RHS of (2.2) into its q-notation as shown

. 1/d C1 U2nC2

U2UnCd C1UnCd C2 D ˛

2d 1_{.1 q/}2 . 1/d C1.1 q2nC2/

1 q2_{.1 q}d CnC1_/.1 _qd CnC2_/: After some simplifications, q-version of the claimed result is

n X kD0 qk.1 qk d/ .1 qkCd_/.1 _qkCd C1_/.1 _qkCd C2_/D q d.1 q2nC2/ 1 q2_{.1 q}d CnC1_/.1 _qd CnC2_/: Define SnWD n X kD0 ´.1 ´q d/ .1 ´qd_/.1 _´qd C1_/.1 _´qd C2_/: Denote the summand term of Snby T .´/, that is,

T .´/D ´.1 ´q

d_/

.1 ´qd_/.1 _´qd C1_/.1 _´qd C2_/: The partial fraction decomposition of T .´/ reads as

´.1 ´q d/ .1 ´qd_/.1 _´qd C1_/.1 _´qd C2_/ D 1 q3d C1_.1 _q/2_.1_{C q/} q.1 q2d/ 1 ´qd C .1_{C q/.1 q}2d C1/ 1 ´qd C1 1 q2d C2 1 ´qd C2 ! D 1 q3d C1_.1 _q/2_.1_{C q/} q.1 q2d/ ₁ 1 ´qd C2 1 1 ´qd

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.1_{C q/.1 q}2d C1/ 1 1 ´qd C2 1 1 ´qd C1 : And so SnD 1 q3d C1.1 q/2.1C q/ " q.1 q2d/ n X kD0 1 1 ´qd C2 1 1 ´qd (2.3) .1_{C q/.1 q}2d C1/ n X kD0 1 1 ´qd C2 1 1 ´qd C1 # :

If we take mD 2, t D 0; and, m D 2, t D 1 in (2.1), respectively, then we rewrite Sn given in (2.3) as 1 q3d C1_.1 _q/2_.1_{C q/} " q.1 q2d/ 1 X kD0 ₁ 1 qkCd C1Cn 1 1 qkCd .1_{C q/.1 q}2d C1/ ₁ 1 qd CnC2 1 1 qd C1 D 1C qnC1 1 qd C1 .1 qnC1/ qd ₁ _qd C1 1 qd CnC1 1 qd CnC2_{.1 q/ .1 C q/} D q d_.1 _q2nC2_/ 1 q2_{.1 q}d CnC1_/.1 _qd CnC2_/:

Thus, we have the conclusion.

When UnD Fnand dD 3 in (2.2), we obtain n X kD0 . 1/k Fk 3 F_kC3F_kC4F_kC5 D F_2nC2 F_nC4F_nC5: Now we going to give our third result:

Theorem 3. Ford > 0, n X kD0 . 1/k VkCd C1 U_kCdU_{kCd C1}U_{kCd C2} D U_nC1U_{nC2.d C1/} U1UdUd C1UnCd C1UnCd C2 : (2.4)

Proof. After required converting and simplifications, we find the q-version of the claimed result as follows

n X kD0 qk.1_{C q}kCd C1/ .1 qkCd_/.1 _qkCd C1_/.1 _qkCd C2_/ D .1 q nC1_/.1 _qnC2.d C1/_/ .1 q/ 1 qd 1 qd C1_{.1 q}nCd C1_/.1 _qnCd C2_/:

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Define SnWD n X kD0 ´.1C ´qd C1/ .1 ´qd_/.1 _´qd C1_/.1 _´qd C2_/: Denote the summand term of Snby T .´/, that is,

T .´/D ´.1C ´q

d C1_/

.1 ´qd_/.1 _´qd C1_/.1 _´qd C2_/: The partial fraction decomposition of T .´/ reads as

´.1_{C ´q}d C1/ .1 ´qd_/.1 _´qd C1_/.1 _´qd C2_/ D 1 qd_.1_{C q/.1 q/}2 ₁ C q 1 ´qd 2.1_{C q/} 1 ´qd C1C 1_{C q} 1 ´qd C2 D 1 qd_.1_{C q/.1 q/}2 .1_{C q/} ₁ 1 ´qd C2 1 1 ´qd 2.1_{C q/} 1 1 ´qd C2 1 1 ´qd C1 : And so SnD 1 qd_.1_{C q/.1 q/}2 " .1C q/ n X kD0 ₁ 1 ´qd C2 1 1 ´qd 2.1_{C q/} n X kD0 ₁ 1 ´qd C2 1 1 ´qd C1 # :

If we take mD 2, t D 0; and, m D 2, t D 1 in (2.1), respectively, then we rewrite the last equality as SnD 1 qd_.1_{C q/.1 q/}2 " .1_{C q/} 1 X kD0 1 1 qkCd C1Cn 1 1 qkCd 2.1_{C q/} 1 1 qd CnC2 1 1 qd C1 D .1 q nC1_/.1 _qnC2d C2_/ .1 q/.1 qd_/.1 _qd C1_/.1 _qnCd C1_/.1 _qnCd C2_/:

Thus, we have the conclusion.

When UnD Fn; VnD Lnand dD 5 in (2.4), we obtain n X kD0 . 1/k LkC6 F_kC5F_kC6F_kC7 D F_nC1F_nC12 F5F6FnC6FnC7 :

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3. GENERAL CASES

In the previous section, we found the q-versions of the claimed results in Theorems

1-3 while proving them. Now we shall present more general cases of Theorems

1-3 by taking q D ˇs=˛s for any integer s in the q-forms of them without proof, respectively.

Theorem 4. Forn; m_{0 and c 2 f m C 1; : : : ; 1; 0; 1g;}

n X kD0 . 1/sk m 1 Q t D0 U_s.kCcCt/ mC1 Q t D0 X_{s.kCd Ct/} D . 1/s.cC1/ m Q t D0 U_s.nCtCc/ U_s.mC1/X_{s.d cC1/}mC1Q t D1 X_{s.nCtCd /} ; (3.1) where Xn is eitherUn orVn: Note that when XnD Vn, we assume that d is any integer. WhenXnD Un, we assume thatd 1.

In (3.1), if we take mD 3; s D 2; UnD Pn; XnD Qn; d D 4 and c D 1; then we obtain n X kD0 P_2kC2P_2kC4P_2kC6 Q_2kC8Q_2kC10Q_2kC12Q_2kC14Q_2kC16D P_2nC2P_2nC4P_2nC6P_2nC8 P8Q8Q2nC10Q2nC12Q2nC14Q2nC16 :

When m_{D 4; s D 1; U}nD XnD Fn; d D 6 and c D 0 in Theorem4, then, by F nD . 1/nC1Fn; we obtain n X kD0 . 1/k F kF k 1F k 2F k 3 F k 6F k 7F k 8F k 9F k 10F k 11 D n X kD0 . 1/k FkFkC1FkC2FkC3 F_kC6F_kC7F_kC8F_kC9F_kC10F_kC11 D FnFnC1FnC2FnC3FnC4 F5F7FnC7FnC8FnC9FnC10FnC11 :

When mD 5; s D 3; UnD Fn; XnD Ln; d D 2 and c D 3 in Theorem 4, then we obtain n X kD0 . 1/k F3k 9F3k 6F3k 3F3kF3kC3 L_3kC6L_3kC9L_3kC12L_3kC15L_3kC18L_3kC21L_3kC24 D F3n 9F3n 6F3n 3F3nF3nC3F3nC6 F18L18L3nC9L3nC12L3nC15L3nC18L3nC21L3nC24 :

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Theorem 5. Ford > 0, n X kD0 . 1/sk Us.k d / U_{s.kCd /}U_{s.kCd C1/}U_{s.kCd C2/} D . 1/ sd C1 Us.2nC2/ U2sUs.nCd C1/Us.nCd C2/ : (3.2) If we take UnD Fn; dD 2 and s D 5 in (3.2), then we obtain

n X kD0 . 1/k F5k 10 F_5kC10F_5kC15F_5kC20 D F_10nC10 F10F5nC15F5nC20 : Theorem 6. Ford > 0, n X kD0 . 1/sk Vs.kCd C1/ U_{s.kCd /}U_{s.kCd C1/}U_{s.kCd C2/} D U_s.nC1/U_{s.nC2d C2/} UsUsdUs.d C1/Us.nCd C1/Us.nCd C2/ : (3.3) If we take UnD Pn; Vn D Qn; d D 4 and s D 3 in (3.3), then, by P nD . 1/nC1Pnand Q nD . 1/nQn; we obtain n X kD0 . 1/k Q 3k 15 P 3k 12P 3k 15P 3k 18 D n X kD0 . 1/k Q3kC15 P_3kC12P_3kC15P_3kC18 D P3nC3P3nC30 P3P12P15P3nC15P3nC18 : 4. CONCLUSIONS

In the last section, we derived more general results, where the sign functions and the indices of Fibonacci or Lucas factors are depend on the integer parameter s. By the way, one can see that these generalized reciprocal sums are alternating for odd integer s; and, non-alternating for even integer s while the original sums in Theorems

1-3are always alternating sums.

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Authors’ addresses

Emrah Kılıc¸

TOBB University of Economics and Technology, Mathematics Department, 06560, Ankara, Turkey E-mail address: ekilic@etu.edu.tr

Didem Ersanlı

TOBB University of Economics and Technology, Mathematics Department, 06560, Ankara, Turkey E-mail address: didemersanli@gmail.com