Generalizations of the Reciprocal Fibonacci-Lucas Sums of Brousseau

Kunle Adegoke Department of Physics and Engineering Physics Obafemi Awolowo University 220005 Ile-Ife Nigeria

                      adegoke00@gmail.com

Abstract We derive closed form expressions for finite and infinite Fibonacci-Lucas sums having products of Fibonacci or Lucas numbers in the denominator of the summand. Our results generalize and extend those obtained by pioneer Brother Alfred Brousseau and later researchers.

1 Introduction

The Fibonacci numbers, F_{n}, and Lucas numbers, L_{n}, are defined, for n ∈ ℕ_{0}, as usual, through the recurrence relations F_{n} = F_{n−1} + F_{n−2}, with F_{0} = 0, F_{1} = 1 and L_{n} = L_{n−1} + L_{n−2}, with L_{0} = 2, L_{1} = 1.

Our main aim in this paper is to derive closed form expressions for the following sums and their corresponding alternating versions, for positive integers m, n and q:

∑_{k=1}^{∞} \frac{L_{nk+nq}L_{nk+2nq} ⋯ L_{nk+(m−1)nq}}{F_{nk}F_{nk+nq} ⋯ F_{nk+mnq}} , ∑_{k=1}^{∞} \frac{F_{nk+nq}F_{nk+2nq} ⋯ F_{nk+(m−1)nq}}{L_{nk}L_{nk+nq} ⋯ L_{nk+mnq}} , m > 1 ,

∑_{k=1}^{∞} \frac{1}{F_{nk}F_{nk+nq} ⋯ F_{nk+mnq−nq}F_{nk+mnq+nq} ⋯ F_{nk+mnq+2mnq}} ,

∑_{k=1}^{∞} \frac{1}{L_{nk}L_{nk+nq} ⋯ L_{nk+mnq−nq}L_{nk+mnq+nq} ⋯ L_{nk+mnq+2mnq}} ,

∑_{k=1}^{∞} \frac{L_{nk+mnq}}{F_{nk}F_{nk+nq} ⋯ F_{nk+2mnq}} , ∑_{k=1}^{∞} \frac{F_{nk+mnq}}{L_{nk}L_{nk+nq} ⋯ L_{nk+2mnq}} ,

∑_{k=1}^{∞} \frac{F_{2nk+mnq}}{F_{nk}^{2}F_{nk+nq}^{2}F_{nk+2nq}^{2} ⋯ F_{nk+mnq}^{2}} , ∑_{k=1}^{∞} \frac{F_{2nk+mnq}}{L_{nk}^{2}L_{nk+nq}^{2}L_{nk+2nq}^{2} ⋯ L_{nk+mnq}^{2}} ,

∑_{k=1}^{∞} \frac{L_{nk+mnq}}{F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+(m−1)nq}^{2}F_{nk+mnq}F_{nk+(m+1)q}^{2} ⋯ F_{nk+2mnq}^{2}} ,

∑_{k=1}^{∞} \frac{F_{nk+mnq}}{L_{nk}^{2}L_{nk+nq}^{2} ⋯ L_{nk+(m−1)nq}^{2}L_{nk+mnq}L_{nk+(m+1)q}^{2} ⋯ L_{nk+2mnq}^{2}} ,

∑_{k=1}^{∞} \frac{F_{2nk+mnq}L_{nk+nq}^{2}L_{nk+2nq}^{2} ⋯ L_{nk+(m−1)nq}^{2}}{F_{nk}^{2}F_{nk+nq}^{2}F_{nk+2nq}^{2} ⋯ F_{nk+mnq}^{2}} , m > 1 .

We require the following telescoping summation identities (see [1])

∑_{k=1}^{N} (f(k) − f(k + q)) = ∑_{k=1}^{q} f(k) − ∑_{k=1}^{q} f(k + N)  (1)

and

\begin{matrix}∑_{k=1}^{N} & (−1)^{k−1} (f(k) + (−1)^{q−1}f(k + q)) \\ = ∑_{k=1}^{q} (−1)^{k−1}f(k) + (−1)^{N−1} ∑_{k=1}^{q} (−1)^{k−1}f(k + N) , \end{matrix}  (2)

for N ≥ q ∈ ℕ_{0}.

In general, infinite sums are evaluated using

∑_{k=1}^{∞} (f(k) − f(k + q)) = ∑_{k=1}^{q} f(k) − ∑_{k=1}^{q} lim_{N→∞} f(k + N)  (3)

and

∑_{k=1}^{∞} (−1)^{k−1} (f(k) + (−1)^{q−1}f(k + q)) = ∑_{k=1}^{q} (−1)^{k−1}f(k) .  (4)

If f(N) approaches zero as N approaches infinity, then we have, from (1) and (2), the useful identities

∑_{k=1}^{∞} (f(k) − f(k + q)) = ∑_{k=1}^{q} f(k)  (5)

and

∑_{k=1}^{∞} (−1)^{k−1} (f(k) + (−1)^{q−1}f(k + q)) = ∑_{k=1}^{q} (−1)^{k−1}f(k) .  (6)

The golden ratio, having the numerical value of (\sqrt{5} + 1)/2, is denoted in this paper by ϕ.

We shall require the following identities (most of which can be found in the book by Vajda [10]):

L_{v}F_{u} = F_{u+v} + (−1)^{v}F_{u−v}  (7a)
F_{v}L_{u} = F_{u+v} − (−1)^{v}F_{u−v}  (7b)

2F_{u+v} = L_{v}F_{u} + L_{u}F_{v}  (8a)
(−1)^{v}2F_{u−v} = F_{u}L_{v} − L_{u}F_{v}  (8b)

L_{v}L_{u} = L_{u+v} + (−1)^{v}L_{u−v}  (9a)
5F_{v}F_{u} = L_{u+v} − (−1)^{v}L_{u−v}  (9b)
(−1)^{u−1}(F_{v+u}F_{v−u}) = F_{u}^{2}(F_{v+1}F_{v−1}) − F_{v}^{2}(F_{u+1}F_{u−1})  (10)

(−1)^{t}F_{u}F_{v} = F_{t+u}F_{t+v} − F_{t}F_{t+u+v}  (11a)
(−1)^{t+1}5F_{u}F_{v} = L_{t+u}L_{t+v} − L_{t}L_{t+u+v}  (11b)

F_{u−v}F_{u+v} = F_{u}^{2} + (−1)^{u+v−1}F_{v}^{2}  (12a)
5F_{u−v}F_{u+v} = L_{u}^{2} + (−1)^{u+v−1}L_{v}^{2}  (12b)

F_{v}F_{2u+v+p} = F_{u+v+p}F_{u+v} + (−1)^{v+1}F_{u+p}F_{u}  (13a)
F_{v}L_{2u+v+p} = L_{u+v+p}F_{u+v} + (−1)^{v+1}L_{u+p}F_{u}  (13b)

The identities (12) and (13) were proved by Howard [6].

The following limiting values are readily established using Binet’s formula:

lim_{N→∞} \frac{F_{N+m}}{F_{N+n}} = lim_{N→∞} \frac{L_{N+m}}{L_{N+n}} = ϕ^{m−n},  (14a)
lim_{N→∞} \frac{F_{N+m}}{L_{N+n}} = \frac{1}{5} lim_{N→∞} \frac{L_{N+m}}{F_{N+n}} = \frac{ϕ^{m−n}}{\sqrt{5}} .  (14b)

We shall adopt the following conventions for empty sums and empty products:

∑_{k=1}^{0} f(k) = 0, ∏_{k=1}^{0} f(k) = 1 .

2 Results: Generalizations

2.1 Telescoping summation identities

Lemma 1. If m, n, q and N are positive integers and f(k) is a real sequence, then

\begin{aligned}∑_{k=1}^{N} & \{(f(nk) − f(nk + mnq)) ∏_{j=1}^{m−1} f(nk + jnq)\} \\ & = ∑_{k=1}^{q} \{∏_{j=0}^{m−1} f(nk + jnq)\} − ∑_{k=1}^{q} \{∏_{j=0}^{m−1} f(nk + nN + jnq)\} . \end{aligned}

If the sequence f(k) is convergent and we denote by f_{∞} the limiting value of f(nN) as N approaches infinity, we have

\begin{aligned}∑_{k=1}^{∞} & \{(f(nk) − f(nk + mnq)) ∏_{j=1}^{m−1} f(nk + jnq)\} \\ & = ∑_{k=1}^{q} \{∏_{j=0}^{m−1} f(nk + jnq)\} − f_{∞}^{m}q . \end{aligned}  (15)

Proof. We have

(f(nk) − f(nk + mnq)) ∏_{j=1}^{m−1} f(nk + jnq)

= f(nk) ∏_{j=1}^{m−1} f(nk + jnq) − f(nk + mnq) ∏_{j=1}^{m−1} f(nk + jnq)

= ∏_{j=0}^{m−1} f(nk + jnq) − ∏_{j=1}^{m} f(nk + jnq)

\begin{matrix}= ∏_{j=0}^{m−1} f(nk + jnq) − ∏_{j=0}^{m−1} f(nk + jnq + nq) \\ ≡ ∏_{j=0}^{m−1} f(nk + jnq) − ∏_{j=0}^{m−1} f(nk + jnq)|_{k→k+q} . \end{matrix}  (16)

The result follows by summing both sides of identity (16), using the identity (1) to perform the telescopic summation on the right hand side.

Lemma 2. If f(k) is a real sequence and m, n, q and N are positive integers such that q is even, then

∑_{k=1}^{N} (−1)^{k−1} {(f(nk) − f(nk + mnq)) ∏_{j=1}^{m−1} f(nk + jnq)}

= ∑_{k=1}^{q} (−1)^{k−1} {∏_{j=0}^{m−1} f(nk + jnq)}

+ (−1)^{N−1} ∑_{k=1}^{q} (−1)^{k−1} {∏_{j=0}^{m−1} f(nk + nN + jnq)} .

Proof. Multiply through the identity (16) by (−1)^{k−1} and use identity (2).

Lemma 3. If f(k) is a real sequence and m, n, q and N are positive integers such that q is odd, then

∑_{k=1}^{N} (−1)^{k−1} {(f(nk) + f(nk + mnq)) ∏_{j=1}^{m−1} f(nk + jnq)}

= ∑_{k=1}^{q} (−1)^{k−1} {∏_{j=0}^{m−1} f(nk + jnq)}

+ (−1)^{N−1} ∑_{k=1}^{q} (−1)^{k−1} {∏_{j=0}^{m−1} f(nk + nN + jnq)} .

Proof. We have the identity

\begin{matrix}(f(nk) + f(nk + mnq)) & ∏_{j=1}^{m−1} & f(nk + jnq) \\ = ∏_{j=0}^{m−1} f(nk + jnq) + ∏_{j=0}^{m−1} f(nk + jnq + nq) \\ ≡ ∏_{j=0}^{m−1} f(nk + jnq) + ∏_{j=0}^{m−1} f(nk + jnq)|_{k→k+q} , \end{matrix}  (17)

from which the result follows after multiplying through by (−1)^{k−1} and summing over k, making use of the identity (2).

If the sequence f(k) is convergent and f(2Nn) and f((2N − 1)n) both have the same limiting value as N approaches infinity, we have

\begin{matrix}∑_{k=1}^{∞} & (−1)^{k−1} \{(f(nk) + (−1)^{q−1}f(nk + mnq)) ∏_{j=1}^{m−1} f(nk + jnq)\} \\ = ∑_{k=1}^{q} (−1)^{k−1} \{∏_{j=0}^{m−1} f(nk + jnq)\} . \end{matrix}  (18)

2.2 Sums with F_{nk}F_{nk+nq} ⋯ F_{nk+mnq} or F_{nk}F_{nk+nq} ⋯ F_{nk+(m−1)nq}F_{nk+(m+1)nq} ⋯ F_{nk+2mnq} in the denominator

Theorem 4. If m, n and q are positive integers, then

∑_{k=1}^{∞} ((−1)^{nk−1} \begin{matrix}∏ & {m−1 \atop j=1} & L_{nk+jnq} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{matrix} ) = \frac{q\sqrt{5^{m}}}{2F_{mnq}} − \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} ,

so that

∑_{k=1}^{∞} (\begin{matrix}∏ & {m−1 \atop j=1} & L_{nk+jnq} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{matrix} ) = \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} − \frac{q\sqrt{5^{m}}}{2F_{mnq}} , n even  (19)

and

∑_{k=1}^{∞} ((−1)^{k−1} \begin{matrix}∏ & {m−1 \atop j=1} & L_{nk+jnq} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{matrix} ) = \frac{q\sqrt{5^{m}}}{2F_{mnq}} − \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} , n odd . (20)

In particular,

∑_{k=1}^{∞} \frac{(−1)^{nk−1}}{F_{nk}F_{nk+nq}} = \frac{q\sqrt{5}}{2F_{nq}} − \frac{1}{2F_{nq}} ∑_{k=1}^{q} \frac{L_{nk}}{F_{nk}} .  (21)

Brousseau’s result ([3], equation (3), also rederived in various equivalent forms by other authors, see for example reference [9]) corresponds to setting n = 1 in (21), but with a different, but equivalent, form for the right hand side. Bruckman and Good’s result ([4], equation (19)) is also a special case of (21), corresponding to setting q = 1.

Theorem 5. If m, n and q are integers such that n is odd and q is even, then

∑_{k=1}^{∞} (\begin{matrix}∏ & {m−1 \atop j=1} & L_{nk+jnq} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{matrix} ) = \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ((−1)^{k} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} ) .

In particular,

∑_{k=1}^{∞} \frac{1}{F_{nk}F_{nk+nq}} = \frac{1}{2F_{nq}} ∑_{k=1}^{q} ((−1)^{k} \frac{L_{nk}}{F_{nk}} ), n odd, q even ,  (22)

which generalizes the result obtained by Rabinowitz ([9], the second of equation (26)), the latter corresponding to the special case n = 1 in the identity (22), but with a different, but equivalent, form for the right hand side.

Proof of Theorem 4 and Theorem 5

Dividing through the identity (8b) by F_{u}F_{v} and setting u = nk + mnq and v = nk, the following identity is established for k, m, n and q positive integers:

\frac{(−1)^{nk−1}2F_{mnq}}{F_{nk}F_{nk+mnq}} = \frac{L_{nk+mnq}}{F_{nk+mnq}} − \frac{L_{nk}}{F_{nk}} .

Using f(k) = L_{k}/F_{k} in Lemma 1 we get the finite summation identity

\begin{matrix}2F_{mnq} ∑_{k=1}^{N} ((−1)^{nk−1} \begin{matrix}∏ & {m−1 \atop j=1} & L_{nk+jnq} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{matrix} ) \\ = ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{L_{nk+nN+jnq}}{F_{nk+nN+jnq}} ) − ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} ) , \end{matrix}  (23)

which yields Theorem 4 in the limit N approaches infinity. Theorem 5 is proved by using f(k) = L_{k}/F_{k} in identity (18).

Theorem 6. If m, n and q are positive \underline{odd} integers, then

∑_{k=1}^{∞} (\frac{1}{∏_{j=0}^{m−1} F_{nk+njq} ∏_{j=m+1}^{2m} F_{nk+njq}} ) = \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{2m−1} F_{nk+njq}} ) .

Below are a few explicit examples from Theorem 6:

\begin{matrix}At m = 1 : \\ ∑_{k=1}^{∞} \frac{1}{F_{nk}F_{nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{q} \frac{1}{F_{nk}F_{nk+nq}} , nq odd . \end{matrix}  (24)

\begin{matrix}At (m, n, q) = (1, 1, 1) and (m, n, q) = (1, 1, 3): \\ ∑_{k=1}^{∞} \frac{1}{F_{k}F_{k+2}} = 1, ∑_{k=1}^{∞} \frac{1}{F_{k}F_{k+6}} = \frac{143}{960} , \end{matrix}

corresponding to Formulas (4) and (6) of Brousseau [2].

\begin{matrix}At (m, n, q) = (3, 1, 3): \\ ∑_{k=1}^{∞} \frac{1}{F_{k}F_{k+3}F_{k+6}F_{k+12}F_{k+15}F_{k+18}} = \frac{938359017897442612}{5579104720519492358676480} . \end{matrix}

At (m, n, q) = (5, 3, 1):

∑_{k=1}^{∞} \frac{1}{F_{3k}F_{3k+3}F_{3k+6}F_{3k+9}F_{3k+12}F_{3k+18}F_{3k+21}F_{3k+24}F_{3k+27}F_{3k+30}}

= \frac{1}{13970032097862115517068710877593600} .

Theorem 7. If q, m and n are positive integers such that q is odd and nm is even, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+njq} ∏_{j=m+1}^{2m} F_{nk+njq}} ) = \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{2m−1} F_{nk+njq}} ) .

Examples from Theorem 7 include

\begin{matrix}At m = 2 : \\ ∑_{k=1}^{∞} \frac{(−1)^{k−1}}{F_{nk}F_{nk+nq}F_{nk+3nq}F_{nk+4nq}} = \frac{1}{L_{2nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{F_{nk}F_{nk+nq}F_{nk+2nq}F_{nk+3nq}} , q odd . \end{matrix}  (25)

\begin{matrix}At (m, n, q) = (1, 2, 1) and (m, n, q) = (2, 1, 1): \\ ∑_{k=1}^{∞} \frac{(−1)^{k−1}}{F_{2k}F_{2k+4}} = \frac{1}{9}, ∑_{k=1}^{∞} \frac{(−1)^{k−1}}{F_{k}F_{k+1}F_{k+3}F_{k+4}} = \frac{1}{18} . \end{matrix}

\begin{matrix}At (m, n, q) = (2, 6, 1): \\ ∑_{k=1}^{∞} \frac{(−1)^{k−1}}{F_{6k}F_{6k+6}F_{6k+18}F_{6k+24}} = \frac{1}{44444622716928} . \end{matrix}

Proof of Theorem 6 and Theorem 7 With v = mnq and u = nk + mnq in identity (7a), the following identity is established:

\frac{L_{mnq}F_{nk+mnq}}{F_{nk}F_{nk+2mnq}} = \frac{1}{F_{nk}} + \frac{(−1)^{mnq}}{F_{nk+2mnq}} ,  (26)

so that

\frac{L_{mnq}F_{nk+mnq}}{F_{nk}F_{nk+2mnq}} = \frac{1}{F_{nk}} − \frac{1}{F_{nk+2mnq}} , mnq odd  (27)

and

\frac{L_{mnq}F_{nk+mnq}}{F_{nk}F_{nk+2mnq}} = \frac{1}{F_{nk}} + \frac{1}{F_{nk+2mnq}} , mnq even .  (28)

If m, n and q are positive \underline{odd} integers, then from (27) and using f(k) = 1/F_{k} in Lemma 1 (with m → 2m), we have the following definite summation identity

\begin{aligned}∑_{k=1}^{N} & (\frac{1}{∏_{j=0}^{m−1} F_{nk+jnq} ∏_{j=m+1}^{2m} F_{nk+jnq}} ) \\ & = \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{2m−1} F_{nk+jnq}} ) − \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{2m−1} F_{nk+nN+jnq}} ) , \end{aligned}  (29)

from which Theorem 6 follows as N approaches infinity.

If q is an odd positive integer and either m or n is even, then from (28) and Lemma 3 (with m → 2m) we have the summation identity

\begin{aligned}∑_{k=1}^{N} & (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+njq} ∏_{j=m+1}^{2m} F_{nk+njq}} ) \\ & = \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{2m−1} F_{nk+njq}} ) + \frac{(−1)^{N−1}}{L_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{2m−1} F_{nk+nN+njq}} ) , \end{aligned} (30)

from which Theorem 7 follows in the limit that N approaches infinity.

Theorem 8. If m, n and q are positive integers such that q is odd, then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq} ∏_{j=1}^{m−1} L_{nk+jnq}}{∏_{j=0}^{m} F_{nk+jnq}} = \frac{1}{2} ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} .

In particular,

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+nq}}{F_{nk}F_{nk+nq}} = \frac{1}{2} ∑_{k=1}^{q} (−1)^{k−1} \frac{L_{nk}}{F_{nk}} , q odd .  (31)

Corollary 9. If m, n and q are positive integers such that q is odd, then

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{nk+mnq}^{2} ∏_{j=1}^{m−1} L_{nk+jnq} ∏_{j=m+1}^{2m−1} L_{nk+jnq}}{∏_{j=0}^{m−1} F_{nk+jnq} ∏_{j=m+1}^{2m} F_{nk+jnq}} = \frac{1}{2} ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{2m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} .

In particular,

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{nk+nq}^{2}}{F_{nk}F_{nk+2nq}} = \frac{1}{2} ∑_{k=1}^{q} (−1)^{k−1} \frac{L_{nk}L_{nk+nq}}{F_{nk}F_{nk+nq}} , q odd .  (32)

Proof of Theorem 8 and Corollary 9

Dividing through the identity (8a) by F_{u}F_{v} and setting u = nk + mnq and v = nk we have the identity

2\frac{F_{2nk+mnq}}{F_{nk}F_{nk+mnq}} = \frac{L_{nk+mnq}}{F_{nk+mnq}} + \frac{L_{nk}}{F_{nk}} .  (33)

If q is odd, then from (33) and with f(k) = L_{k}/F_{k} in Lemma 3 we have

\begin{matrix}2 ∑_{k=1}^{N} \frac{(−1)^{k−1}F_{2nk+mnq} ∏_{j=1}^{m−1} L_{nk+jnq}}{∏_{j=0}^{m} F_{nk+jnq}} \\ = ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}}{F_{nk+jnq}} + (−1)^{N−1} ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{m−1} \frac{L_{nk+nN+jnq}}{F_{nk+nN+jnq}} , \end{matrix}  (34)

from which Theorem (8) follows in the limit as N approaches infinity. Corollary (9) is obtained by specifically requiring m to be even in Theorem (8).

Theorem 10. If m, n, q and p are positive integers, then

\begin{aligned}∑_{k=1}^{∞} & \{\frac{(−1)^{nk−1} ∏_{j=1}^{m−1} F_{nk+jnq+np}}{∏_{j=0}^{m} F_{nk+jnq}} \} \\ & = \frac{ϕ^{mnp}q}{F_{mnq}F_{np}} − \frac{1}{F_{mnq}F_{np}} ∑_{k=1}^{q} \{∏_{j=0}^{m−1} \frac{F_{nk+jnq+np}}{F_{nk+jnq}} \} . \end{aligned}

In particular we have

∑_{k=1}^{∞} \{\frac{(−1)^{nk−1} ∏_{j=m+1}^{p+m−1} F_{nk+jn}}{∏_{j=0}^{p} F_{nk+jn}} \} = \frac{ϕ^{mnp}}{F_{mn}F_{pn}} − \frac{1}{F_{mn}F_{pn}} ∏_{j=0}^{m−1} \frac{F_{jn+np+n}}{F_{jn+n}}  (35)

and

∑_{k=1}^{∞} \frac{(−1)^{nk−1}}{F_{nk}F_{nk+nq}} = \frac{ϕ^{n}q}{F_{nq}F_{n}} − \frac{1}{F_{nq}F_{n}} ∑_{k=1}^{q} \frac{F_{nk+n}}{F_{nk}} .  (36)

Observe that identity (36) is equivalent to identity (21) but with a different form for the right hand side. Since 2ϕ = \sqrt{5} + 1, F_{n−1} + F_{n+1} = L_{n} and ϕ^{n} = ϕF_{n} + F_{n−1}, both identities can be combined to yield the following interesting summation identity which is valid for all non-zero integers n and non-negative integers q:

\frac{1}{F_{n}} ∑_{k=1}^{q} (\frac{F_{nk+n}}{F_{nk}} ) − \frac{1}{2} ∑_{k=1}^{q} (\frac{L_{nk}}{F_{nk}} ) = \frac{q}{2} \frac{L_{n}}{F_{n}} .  (37)

Theorem 11. If m, n, q and p are positive integers such that n is odd and q is even, then

\begin{aligned}∑_{k=1}^{∞} & \begin{cases}∏ & {m−1 \atop j=1} & F_{nk+jnq+np} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{cases} \} \\ & = \frac{1}{F_{mnq}F_{np}} ∑_{k=1}^{q} \{(−1)^{k} ∏_{j=0}^{m−1} \frac{F_{nk+jnq+np}}{F_{nk+jnq}} \} . \end{aligned}

In particular,

∑_{k=1}^{∞} \frac{1}{F_{nk}F_{nk+nq}} = \frac{1}{F_{n}F_{nq}} ∑_{k=1}^{q} ((−1)^{k} \frac{F_{nk+n}}{F_{nk}} ), n odd, q even .  (38)

From identity (22) and identity (38) we have the interesting result

\frac{1}{2} ∑_{k=1}^{q} ((−1)^{k} \frac{L_{nk}}{F_{nk}} ) = \frac{1}{F_{n}} ∑_{k=1}^{q} ((−1)^{k} \frac{F_{nk+n}}{F_{nk}} ) , q even .  (39)

Dividing through identity (11a) by F_{t+u}F_{t} and choosing t = nk, u = mnq and v = np we obtain the identity:

\frac{(−1)^{nk−1}F_{mnq}F_{np}}{F_{nk}F_{nk+mnq}} = \frac{F_{nk+mnq+np}}{F_{nk+mnq}} − \frac{F_{nk+np}}{F_{nk}} .  (40)

With f(k) = F_{k+np}/F_{k} in Lemma 1 and using the identity (40), we have the finite summation identity

\begin{matrix}F_{mnq}F_{np} ∑_{k=1}^{N} \{\frac{(−1)^{nk−1} ∏_{j=1}^{m−1} F_{nk+jnq+np}}{∏_{j=0}^{m} F_{nk+jnq}} \} \\ = ∑_{k=1}^{q} \{∏_{j=0}^{m−1} \frac{F_{nk+nN+jnq+np}}{F_{nk+nN+jnq}} \} − ∑_{k=1}^{q} \{∏_{j=0}^{m−1} \frac{F_{nk+jnq+np}}{F_{nk+jnq}} \} , \end{matrix}  (41)

from which Theorem 10 follows in the limit as N approaches infinity.

Using f(k) = F_{k+np}/F_{k} in Lemma 2 gives

\begin{matrix}F_{mnq}F_{np} ∑_{k=1}^{N} \begin{cases}∏ & {m−1 \atop j=1} & F_{nk+jnq+np} \\ ∏ & {m \atop j=0} & F_{nk+jnq} \end{cases} \} \\ = (−1)^{N} ∑_{k=1}^{q} \{(−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+nN+jnq+np}}{F_{nk+nN+jnq}} \} \\ − ∑_{k=1}^{q} \{(−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+jnq+np}}{F_{nk+jnq}} \} , \end{matrix}  (42)

from which Theorem 11 follows.

2.3 Sums with L_{nk}L_{nk+nq} ⋯ L_{nk+mnq} or L_{nk}L_{nk+nq} ⋯ L_{nk+mnq−nq}L_{nk+mnq+nq} ⋯ L_{nk+2mnq} in the denominator

The derivations here proceed in the same fashion as in the previous section. The theorems will therefore be stated without proof. The analogous identity to (23) is

\begin{matrix}2F_{mnq} ∑_{k=1}^{N} ((−1)^{nk−1} \begin{matrix}∏ & {m−1 \atop j=1} & F_{nk+jnq} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{matrix} ) \\ = ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} ) − ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{F_{nk+nN+jnq}}{L_{nk+nN+jnq}} ) . \end{matrix}  (43)

Theorem 12. If m, n and q are positive integers, then

∑_{k=1}^{∞} ((−1)^{nk−1} \begin{matrix}∏ & {m−1 \atop j=1} & F_{nk+jnq} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{matrix} ) = \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} − \frac{q}{2F_{mnq}\sqrt{5^{m}}} ,

so that

∑_{k=1}^{∞} (\begin{matrix}∏ & {m−1 \atop j=1} & F_{nk+jnq} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{matrix} ) = \frac{q}{2F_{mnq}\sqrt{5^{m}}} − \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} , n even  (44)

and

∑_{k=1}^{∞} ((−1)^{k−1} \begin{matrix}∏ & {m−1 \atop j=1} & F_{nk+jnq} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{matrix} ) = \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} − \frac{q}{2F_{mnq}\sqrt{5^{m}}} , n odd . (45)

In particular,

∑_{k=1}^{∞} \frac{(−1)^{nk−1}}{L_{nk}L_{nk+nq}} = \frac{1}{2F_{nq}} ∑_{k=1}^{q} \frac{F_{nk}}{L_{nk}} − \frac{q}{2F_{nq}\sqrt{5}}, n, q ∈ ℤ^{+} .  (46)

The case n = 3, q = 1 in (46) was mentioned by Brousseau ([2], equation (14)). Theorem 13. If m, n and q are integers such that n is odd and q is even, then

∑_{k=1}^{∞} (\begin{matrix}∏ & {m−1 \atop j=1} & F_{nk+jnq} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{matrix} ) = \frac{1}{2F_{mnq}} ∑_{k=1}^{q} ((−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} ) .

In particular,

∑_{k=1}^{∞} \frac{1}{L_{nk}L_{nk+nq}} = \frac{1}{2F_{nq}} ∑_{k=1}^{q} ((−1)^{k−1} \frac{F_{nk}}{L_{nk}} ), n odd, q even .  (47)

Theorem 14. If m, n and q are positive \underline{odd} integers, then

∑_{k=1}^{∞} (\frac{1}{∏_{j=0}^{m−1} L_{nk+njq} ∏_{j=m+1}^{2m} L_{nk+njq}} ) = \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{2m−1} L_{nk+njq}} ) .

Theorem 15. If q, m and n are positive integers such that q is odd and nm is even, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+njq} ∏_{j=m+1}^{2m} L_{nk+njq}} ) = \frac{1}{L_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{2m−1} L_{nk+njq}} ) .

Analogous identity to identity (34) is

\begin{matrix}2 ∑_{k=1}^{N} \frac{(−1)^{k−1}F_{2nk+mnq} ∏_{j=1}^{m−1} F_{nk+jnq}}{∏_{j=0}^{m} L_{nk+jnq}} \\ = ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} + (−1)^{N−1} ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+nN+jnq}}{L_{nk+nN+jnq}} , \end{matrix}  (48)

from which we get the following theorem in the limit as N approaches infinity. Theorem 16. If m, n and q are positive integers such that q is odd, then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq} ∏_{j=1}^{m−1} F_{nk+jnq}}{∏_{j=0}^{m} L_{nk+jnq}} = \frac{1}{2} ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} .

Corollary 17. If m, n and q are positive integers such that q is odd, then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{nk+mnq}^{2} ∏_{j=1}^{m−1} F_{nk+jnq} ∏_{j=m+1}^{2m−1} F_{nk+jnq}}{∏_{j=0}^{m−1} L_{nk+jnq} ∏_{j=m+1}^{2m} L_{nk+jnq}} = \frac{1}{2} ∑_{k=1}^{q} (−1)^{k−1} ∏_{j=0}^{2m−1} \frac{F_{nk+jnq}}{L_{nk+jnq}} .

Corresponding to identity (40) of section 2.2 we have (from identity (11b))

\frac{(−1)^{nk−1}5F_{mnq}F_{np}}{L_{nk}L_{nk+mnq}} = −\frac{L_{nk+mnq+np}}{L_{nk+mnq}} + \frac{L_{nk+np}}{L_{nk}} ,  (49)

leading to the summation identities

\begin{matrix}5F_{mnq}F_{np} ∑_{k=1}^{N} \{\frac{(−1)^{nk−1} ∏_{j=1}^{m−1} L_{nk+jnq+np}}{∏_{j=0}^{m} L_{nk+jnq}} \} \\ = − ∑_{k=1}^{q} \{∏_{j=0}^{m−1} \frac{L_{nk+nN+jnq+np}}{L_{nk+nN+jnq}} \} + ∑_{k=1}^{q} \{∏_{j=0}^{m−1} \frac{L_{nk+jnq+np}}{L_{nk+jnq}} \} \end{matrix}  (50)

and

\begin{matrix}5F_{mnq}F_{np} ∑_{k=1}^{N} \begin{cases}∏ & {m−1 \atop j=1} & L_{nk+jnq+np} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{cases} \} \\ = (−1)^{N−1} ∑_{k=1}^{q} \{(−1)^{k−1} ∏_{j=0}^{m−1} \frac{L_{nk+nN+jnq+np}}{L_{nk+nN+jnq}} \} \\ + ∑_{k=1}^{q} \{(−1)^{k−1} ∏_{j=0}^{m−1} \frac{L_{nk+jnq+np}}{L_{nk+jnq}} \} , \end{matrix}  (51)

from which Theorem 18 and Theorem 19 follow.

Theorem 18. If m, n, q and p are positive integers, then

\begin{aligned}∑_{k=1}^{∞} & \{\frac{(−1)^{nk−1} ∏_{j=1}^{m−1} L_{nk+jnq+np}}{∏_{j=0}^{m} L_{nk+jnq}} \} \\ & = −\frac{ϕ^{mnp}q}{5F_{mnq}F_{np}} + \frac{1}{5F_{mnq}F_{np}} ∑_{k=1}^{q} \{∏_{j=0}^{m−1} \frac{L_{nk+jnq+np}}{L_{nk+jnq}} \} . \end{aligned}

In particular we have

∑_{k=1}^{∞} \{\frac{(−1)^{nk−1} ∏_{j=m+1}^{p+m−1} L_{nk+jn}}{∏_{j=0}^{p} L_{nk+jn}} \} = −\frac{ϕ^{mnp}}{5F_{mn}F_{pn}} + \frac{1}{5F_{mn}F_{pn}} ∏_{j=0}^{m−1} \frac{L_{jn+np+n}}{L_{jn+n}}  (52)

and

∑_{k=1}^{∞} \frac{(−1)^{nk−1}}{L_{nk}L_{nk+nq}} = −\frac{ϕ^{n}q}{5F_{nq}F_{n}} + \frac{1}{5F_{nq}F_{n}} ∑_{k=1}^{q} \frac{L_{nk+n}}{L_{nk}} .  (53)

Theorem 19. If m, n, q and p are positive integers such that n is odd and q is even, then

\begin{aligned}∑_{k=1}^{∞} & \begin{cases}∏ & {m−1 \atop j=1} & L_{nk+jnq+np} \\ ∏ & {m \atop j=0} & L_{nk+jnq} \end{cases} \} \\ & = \frac{1}{5F_{mnq}F_{np}} ∑_{k=1}^{q} \{(−1)^{k−1} ∏_{j=0}^{m−1} \frac{L_{nk+jnq+np}}{L_{nk+jnq}} \} . \end{aligned}

In particular,

∑_{k=1}^{∞} \frac{1}{L_{nk}L_{nk+nq}} = \frac{1}{5F_{n}F_{nq}} ∑_{k=1}^{q} ((−1)^{k−1} \frac{L_{nk+n}}{L_{nk}} ), n odd, q even .  (54)

From identity (47) and identity (54) we have

\frac{1}{2} ∑_{k=1}^{q} ((−1)^{k−1} \frac{F_{nk}}{L_{nk}} ) = \frac{1}{5F_{n}} ∑_{k=1}^{q} ((−1)^{k−1} \frac{L_{nk+n}}{L_{nk}} ) , q even .  (55)

2.4 Sums with F_{nk}F_{nk+nq} ⋯ F_{nk+2mnq} in the denominator

The results in this section are obtained from identity (7b). We have

\frac{F_{v}L_{u}}{F_{u−v}F_{u+v}} = \frac{1}{F_{u−v}} − \frac{(−1)^{v}}{F_{u+v}} ,

from which, by setting v = mnq and u = nk + mnq, we get

\frac{F_{mnq}L_{nk+mnq}}{F_{nk}F_{nk+2mnq}} = \frac{1}{F_{nk}} − \frac{(−1)^{mnq}}{F_{nk+2mnq}} ,  (56)

so that

\frac{F_{mnq}L_{nk+mnq}}{F_{nk}F_{nk+2mnq}} = \frac{1}{F_{nk}} − \frac{1}{F_{nk+2mnq}} , mnq even (57)

and

\frac{F_{mnq}L_{nk+mnq}}{F_{nk}F_{nk+2mnq}} = \frac{1}{F_{nk}} + \frac{1}{F_{nk+2mnq}} , mnq odd .  (58)

The derivations then proceed as in the previous sections. Theorem 20. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} (\frac{L_{nk+mnq}}{∏_{j=0}^{2m} F_{nk+njq}} ) = \frac{1}{F_{mnq}} ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{2m−1} F_{nk+njq}} )

Examples from Theorem 20 include:

\begin{matrix}At (m, n, q) = (1, 2, 1) and (m, n, q) = (1, 1, 2): \\ ∑_{k=1}^{∞} \frac{L_{2k+2}}{F_{2k}F_{2k+2}F_{2k+4}} = \frac{1}{3}, ∑_{k=1}^{∞} \frac{L_{k+2}}{F_{k}F_{k+2}F_{k+4}} = \frac{5}{6} . \end{matrix}  (59)

The first of the identities in (59) was also derived by Melham ( [7], equation 3.7).

\begin{matrix}At (m, n, q) = (2, 7, 1): \\ ∑_{k=1}^{∞} \frac{L_{7k+14}}{F_{7k}F_{7k+7}F_{7k+14}F_{7k+21}F_{7k+28}} = \frac{1}{6427623373464462} . \end{matrix}

Theorem 21. If m, n and q are positive integers such that q is even or mnq is odd, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}L_{nk+mnq}}{∏_{j=0}^{2m} F_{nk+njq}} ) = \frac{1}{F_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{2m−1} F_{nk+njq}} )

Examples from Theorem 21 include:

\begin{matrix}At (m, n, q) = (1, 1, 2) and (m, n, q) = (1, 3, 2): \\ ∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{k+2}}{F_{k}F_{k+2}F_{k+4}} = \frac{1}{6} , ∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{3k+6}}{F_{3k}F_{3k+6}F_{3k+12}} = \frac{271}{156672} . \end{matrix}

\begin{matrix}At (m, n, q) = (2, 4, 2): \\ ∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{4k+16}}{F_{4k}F_{4k+8}F_{4k+16}F_{4k+24}F_{4k+32}} = \frac{177072540680427}{166704475185956548320480} . \end{matrix}

2.5 Sums with L_{nk}L_{nk+nq} ⋯ L_{nk+2mnq} in the denominator

The results here follow from the identity (9b). Theorem 22. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} (\frac{F_{nk+mnq}}{∏_{j=0}^{2m} L_{nk+jnq}} ) = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} ( ∏_{j=0}^{2m−1} \frac{1}{L_{nk+jnq}} ) .

Theorem 23. If m, n and q are positive integers such that q is even or mnq is odd, then

∑_{k=1}^{∞} (−1)^{k−1} (\frac{F_{nk+mnq}}{∏_{j=0}^{2m} L_{nk+jnq}} ) = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} ((−1)^{k−1} ∏_{j=0}^{2m−1} \frac{1}{L_{nk+jnq}} ) .

2.6 Sums with F_{nk}F_{nk+2nq}F_{nk+4nq}F_{nk+6nq} ⋯ F_{nk+2mnq} in the denomina-

tor

Theorem 24. If m, n and q are positive \underline{odd} integers, then

∑_{k=1}^{∞} \frac{(±1)^{k−1}F_{nk+mnq}}{∏_{j=0}^{m} F_{nk+2jnq}} = \frac{1}{L_{mnq}} ∑_{k=1}^{2q} \frac{(±1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{F_{nk+nq}}{F_{nk}F_{nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{2q} \frac{1}{F_{nk}} , nq odd .  (60)

Proof. From identity (27) and with f(k) = 1/F_{k} (and q → 2q) in Lemma 1 we have the finite summation identity

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{F_{nk+mnq}}{∏_{j=0}^{m} F_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} F_{nk+2jnq}}  (mnq odd) \\ & − ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} F_{nk+nN+2jnq}} . \end{aligned}  (61)

From identity (27) with m, n and q positive odd intergers and with f(k) = 1/F_{k} (and q → 2q) in Lemma 2 we have the alternating finite summation identity

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}F_{nk+mnq}}{∏_{j=0}^{m} F_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+2jnq}}  (mnq odd) \\ & + (−1)^{N−1} ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+nN+2jnq}} \end{aligned}  (62)

Theorem 24 follows from identities (61) and (62) in the limit as N approaches infinity.

Theorem 25. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} \frac{(±1)^{k−1}L_{nk+mnq}}{∏_{j=0}^{m} F_{nk+2jnq}} = \frac{1}{F_{mnq}} ∑_{k=1}^{2q} \frac{(±1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{L_{nk+nq}}{F_{nk}F_{nk+2nq}} = \frac{1}{F_{nq}} ∑_{k=1}^{2q} \frac{1}{F_{nk}} , nq even .  (63)

Proof. As in Theorem 24, with the identity (57), with mnq even.

The corresponding finite summation identities are

\begin{aligned}F_{mnq} ∑_{k=1}^{N} \frac{L_{nk+mnq}}{∏_{j=0}^{m} F_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} F_{nk+2jnq}}  (mnq even) \\ & − ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} F_{nk+nN+2jnq}} \end{aligned}  (64)

and

\begin{aligned}F_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}L_{nk+mnq}}{∏_{j=0}^{m} F_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+2jnq}}  (mnq even) \\ & + (−1)^{N−1} ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+nN+2jnq}} . \end{aligned}  (65)

2.7 Sums with L_{nk}L_{nk+2nq}L_{nk+4nq}L_{nk+6nq} ⋯ L_{nk+2mnq} in the denominator

In this section we state the Lucas versions of the results given in section 2.6.

Here the basic identities (from identities (9)) are:

\frac{L_{mnq}L_{nk+mnq}}{L_{nk}L_{nk+2mnq}} = \frac{1}{L_{nk}} − \frac{1}{L_{nk+2mnq}} , mnq odd  (66)

\frac{5F_{mnq}F_{nk+mnq}}{L_{nk}L_{nk+2mnq}} = \frac{1}{L_{nk}} − \frac{1}{L_{nk+2mnq}} , mnq even .  (67)

Theorem 26. If m, n and q are positive odd integers, then

∑_{k=1}^{∞} \frac{(±1)^{k−1}L_{nk+mnq}}{∏_{j=0}^{m} L_{nk+2jnq}} = \frac{1}{L_{mnq}} ∑_{k=1}^{2q} \frac{(±1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{L_{nk+nq}}{L_{nk}L_{nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{2q} \frac{1}{L_{nk}} , nq odd .  (68)

Theorem 27. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} \frac{(±1)^{k−1}F_{nk+mnq}}{∏_{j=0}^{m} L_{nk+2jnq}} = \frac{1}{5F_{mnq}} ∑_{k=1}^{2q} \frac{(±1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{F_{nk+nq}}{L_{nk}L_{nk+2nq}} = \frac{1}{5F_{nq}} ∑_{k=1}^{2q} \frac{1}{F_{nk}} , nq even .  (69)

We have the following finite summation identities:

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{L_{nk+mnq}}{∏_{j=0}^{m} L_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} L_{nk+2jnq}}  (mnq odd) \\ & − ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} L_{nk+nN+2jnq}} . \end{aligned}  (70)

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}L_{nk+mnq}}{∏_{j=0}^{m} L_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+2jnq}}  (mnq odd) \\ & + (−1)^{N−1} ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+nN+2jnq}} . \end{aligned}  (71)

\begin{aligned}5F_{mnq} ∑_{k=1}^{N} \frac{F_{nk+mnq}}{∏_{j=0}^{m} L_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} L_{nk+2jnq}}  (mnq even) \\ & − ∑_{k=1}^{2q} \frac{1}{∏_{j=0}^{m−1} L_{nk+nN+2jnq}} . \end{aligned}  (72)

\begin{aligned}5F_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}F_{nk+mnq}}{∏_{j=0}^{m} L_{nk+2jnq}} & = ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+2jnq}}  (mnq even) \\ & + (−1)^{N−1} ∑_{k=1}^{2q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+nN+2jnq}} . \end{aligned}  (73)

2.8 Sums with F_{2nk}F_{2nk+2nq}F_{2nk+4nq}F_{2nk+6nq} ⋯ F_{2nk+2mnq} in the denom-

inator

Theorem 28. If m, n and q are positive \underline{odd} integers, then

∑_{k=1}^{∞} \frac{F_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} = \frac{1}{L_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{2nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{F_{2nk+nq}}{F_{2nk}F_{2nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{q} \frac{1}{F_{2nk}} ,  (74)
∑_{k=1}^{∞} \frac{F_{2nk+3nq}}{F_{2nk}F_{2nk+2nq}F_{2nk+4nq}F_{2nk+6nq}} = \frac{1}{L_{3nq}} ∑_{k=1}^{q} \frac{1}{F_{2nk}F_{2nk+2nq}F_{2nk+4nq}} .  (75)

The identity derived by Frontczak ([5], identity (18)), corresponds to setting q = 1 in identity (74). Proof. From identity (7a) with v = mnq and u = 2nk + mnq comes the identity

\frac{L_{mnq}F_{2nk+mnq}}{F_{2nk}F_{2nk+2mnq}} = \frac{1}{F_{2nk}} − \frac{1}{F_{2nk+2mnq}} , mnq odd .  (76)

From identity (76) and Lemma 1 with f(k) = 1/F_{2k} we have the finite summation identity:

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{F_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{2nk+2jnq}} \\ & − ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{2nk+2nN+2jnq}} , \end{aligned}  (77)

from which Theorem 28 follows in the limit as N approaches infinity. Theorem 29. If m, n and q are positive integers such that q is odd and mn is even , then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} = \frac{1}{L_{mnq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{2nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+nq}}{F_{2nk}F_{2nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{F_{2nk}} .  (78)

Proof. From identity (7a) with v = mnq and u = 2nk + mnq comes the identity

\frac{L_{mnq}F_{2nk+mnq}}{F_{2nk}F_{2nk+2mnq}} = \frac{1}{F_{2nk}} + \frac{1}{F_{2nk+2mnq}} , mnq even .  (79)

From identity (79) and Lemma 3 with f(k) = 1/F_{2k} we have the finite summation identity:

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}F_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{2nk+2jnq}} \\ & + (−1)^{N−1} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{2nk+2nN+2jnq}} , \end{aligned}  (80)

from which Theorem 29 follows in the limit as N approaches infinity. Theorem 30. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} \frac{L_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{2nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{L_{2nk+nq}}{F_{2nk}F_{2nk+2nq}} = \frac{1}{F_{nq}} ∑_{k=1}^{q} \frac{1}{F_{2nk}} ,  (81)
∑_{k=1}^{∞} \frac{L_{2nk+3nq}}{F_{2nk}F_{2nk+2nq}F_{2nk+4nq}F_{2nk+6nq}} = \frac{1}{F_{3nq}} ∑_{k=1}^{q} \frac{1}{F_{2nk}F_{2nk+2nq}F_{2nk+4nq}} .  (82)

Proof. From identity (7b) with v = mnq and u = 2nk + mnq comes the identity

\frac{F_{mnq}L_{2nk+mnq}}{F_{2nk}F_{2nk+2mnq}} = \frac{1}{F_{2nk}} − \frac{1}{F_{2nk+2mnq}} , mnq even .  (83)

From identity (83) and Lemma 1 with f(k) = 1/F_{2k} we have the finite summation identity:

\begin{aligned}F_{mnq} ∑_{k=1}^{N} \frac{L_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{2nk+2jnq}} \\ & − ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{2nk+2nN+2jnq}} , \end{aligned}  (84)

from which Theorem 30 follows in the limit as N approaches infinity. Theorem 31. If m, n and q are positive integers such that q is even \underline{or} mnq is odd , then

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{2nk+2jnq}} .

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{2nk+nq}}{F_{2nk}F_{2nk+2nq}} = \frac{1}{F_{nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{F_{2nk}} , q even \underline{or} nq odd .  (85)

The alternating summation identity here, valid for q even or mnq odd, is

\begin{aligned}F_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}L_{2nk+mnq}}{∏_{j=0}^{m} F_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{2nk+2jnq}} \\ & + (−1)^{N−1} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{2nk+2nN+2jnq}} . \end{aligned}  (86)

2.9 Sums with L_{2nk}L_{2nk+2nq}L_{2nk+4nq}L_{2nk+6nq} ⋯ L_{2nk+2mnq} in the denom-

inator

The results in this section are derived from identities (9). The proofs are identical to those in section 2.8 and are therefore omitted.

Theorem 32. If m, n and q are positive \underline{odd} integers, then

∑_{k=1}^{∞} \frac{L_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} = \frac{1}{L_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{2nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{L_{2nk+nq}}{L_{2nk}L_{2nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{q} \frac{1}{L_{2nk}} , nq odd ,  (87)
∑_{k=1}^{∞} \frac{L_{2nk+3nq}}{L_{2nk}L_{2nk+2nq}L_{2nk+4nq}L_{2nk+6nq}} = \frac{1}{L_{3nq}} ∑_{k=1}^{q} \frac{1}{L_{2nk}L_{2nk+2nq}L_{2nk+4nq}} .  (88)

The finite summation identity is

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{L_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{2nk+2jnq}}  mnq odd \\ & − ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{2nk+2nN+2jnq}} . \end{aligned}  (89)

Theorem 33. If m, n and q are positive integers such that q is odd and mn is even , then

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} = \frac{1}{L_{mnq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{2nk+2jnq}} .

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{2nk+nq}}{L_{2nk}L_{2nk+2nq}} = \frac{1}{L_{nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{L_{2nk}} ,  q odd, n even .  (90)

The alternating finite summation identity is

\begin{aligned}L_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}L_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{2nk+2jnq}}  q odd, mn even \\ & + (−1)^{N−1} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{2nk+2nN+2jnq}} . \end{aligned}  (91)

Theorem 34. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} \frac{F_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{2nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{F_{2nk+nq}}{L_{2nk}L_{2nk+2nq}} = \frac{1}{5F_{nq}} ∑_{k=1}^{q} \frac{1}{L_{2nk}} ,  nq even ,  (92)

∑_{k=1}^{∞} \frac{F_{2nk+3nq}}{L_{2nk}L_{2nk+2nq}L_{2nk+4nq}L_{2nk+6nq}} = \frac{1}{5F_{3nq}} ∑_{k=1}^{q} \frac{1}{L_{2nk}L_{2nk+2nq}L_{2nk+4nq}} .  (93)

The finite summation identity is

\begin{aligned}5F_{mnq} ∑_{k=1}^{N} \frac{F_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{2nk+2jnq}}  mnq even \\ & − ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{2nk+2nN+2jnq}} , \end{aligned}  (94)

Theorem 35. If m, n and q are positive integers such that q is even \underline{or} mnq is odd , then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{2nk+2jnq}} .

In particular,

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+nq}}{L_{2nk}F_{2nk+2nq}} = \frac{1}{5F_{nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{L_{2nk}} , q even \underline{or} nq odd .  (95)

The alternating summation identity here, valid for q even or mnq odd, is

\begin{aligned}5F_{mnq} ∑_{k=1}^{N} \frac{(−1)^{k−1}F_{2nk+mnq}}{∏_{j=0}^{m} L_{2nk+2jnq}} & = ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{2nk+2jnq}} \\ & + (−1)^{N−1} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{2nk+2nN+2jnq}} . \end{aligned}  (96)

2.10 Sums with F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+mnq−nq}^{2}F_{nk+mnq}F_{nk+mnq+nq}^{2} ⋯ F_{nk+2mnq}^{2} or F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+mnq}^{2} in the denominator

Theorem 36. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} (\frac{F_{2nk+mnq}}{∏_{j=0}^{m} F_{nk+jnq}^{2}} ) = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{nk+jnq}^{2}} .

Explicitly,

∑_{k=1}^{∞} \frac{F_{2nk+mnq}}{F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+mnq}^{2}} = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{1}{F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+(m−1)nq}^{2}} .

Examples include:

\begin{matrix}At m = 1 \\ ∑_{k=1}^{∞} \frac{F_{2nk+nq}}{F_{nk}^{2}F_{nk+nq}^{2}} = \frac{1}{F_{nq}} ∑_{k=1}^{q} \frac{1}{F_{nk}^{2}} , nq even . \end{matrix}  (97)

\begin{matrix}At (m, n, q) = (1, 1, 2) and (m, n, q) = (1, 2, 1) : \\ ∑_{k=1}^{∞} \frac{F_{2k+2}}{F_{k}^{2}F_{k+2}^{2}} = 2, ∑_{k=1}^{∞} \frac{F_{4k+2}}{F_{2k}^{2}F_{2k+2}^{2}} = 1 . \end{matrix}  (98)

\begin{matrix}At (m, n, q) = (3, 2, 2) : \\ ∑_{k=1}^{∞} \frac{F_{4k+12}}{F_{2k}^{2}F_{2k+4}^{2}F_{2k+8}^{2}F_{2k+12}^{2}} = \frac{1288981}{35850395750400} . \end{matrix}  (99)

Corollary 37. If m, n and q are positive integers, then

∑_{k=1}^{∞} (\frac{L_{nk+mnq}}{F_{nk+mnq} ∏_{j=0}^{m−1} F_{nk+jnq}^{2} ∏_{j=m+1}^{2m} F_{nk+jnq}^{2}} ) = \frac{1}{F_{2mnq}} ∑_{k=1}^{q} ( ∏_{j=0}^{2m−1} \frac{1}{F_{nk+jnq}^{2}} ) .

Explicitly, Corollary 37 is

\begin{aligned}∑_{k=1}^{∞} & \frac{L_{nk+mnq}}{F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+(m−1)nq}^{2}F_{nk+mnq}F_{nk+(m+1)nq}^{2} ⋯ F_{nk+2mnq}^{2}} \\ & = \frac{1}{F_{2mnq}} ∑_{k=1}^{q} \frac{1}{F_{nk}^{2}F_{nk+nq}^{2} ⋯ F_{nk+(2m−1)nq}^{2}} . \end{aligned}

Below are a couple of examples:

\begin{matrix}At (m, n, q) = (1, 1, 1) and (m, n, q) = (2, 1, 1) : \\ ∑_{k=1}^{∞} \frac{L_{k+1}}{F_{k}^{2}F_{k+1}F_{k+2}^{2}} = 1, ∑_{k=1}^{∞} \frac{L_{k+2}}{F_{k}^{2}F_{k+1}^{2}F_{k+2}F_{k+3}^{2}F_{k+4}^{2}} = \frac{1}{108} . \end{matrix}  (100)

\begin{matrix}At (m, n, q) = (3, 2, 2) : \\ ∑_{k=1}^{∞} & \frac{L_{2k+12}}{F_{2k}^{2}F_{2k+4}^{2}F_{2k+8}^{2}F_{2k+12}F_{2k+16}^{2}F_{2k+20}^{2}F_{2k+24}^{2}} \\ = \frac{636693716175181614930457}{1701394375843622618689225675379000792710492054565683200} . \end{matrix}  (101)

Proof of Theorem 36

By making appropriate choices for the indices u and v in the identity (12a), it is straightforward to establish the following identity:

\frac{F_{mnq}F_{2nk+mnq}}{F_{nk}^{2}F_{nk+mnq}^{2}} = \frac{1}{F_{nk}^{2}} − \frac{(−1)^{mnq}}{F_{nk+mnq}^{2}} ,  (102)

so that

\frac{F_{mnq}F_{2nk+mnq}}{F_{nk}^{2}F_{nk+mnq}^{2}} = \frac{1}{F_{nk}^{2}} − \frac{1}{F_{nk+mnq}^{2}} , mnq even  (103)

and

\frac{F_{mnq}F_{2nk+mnq}}{F_{nk}^{2}F_{nk+mnq}^{2}} = \frac{1}{F_{nk}^{2}} + \frac{1}{F_{nk+mnq}^{2}} , mnq odd .  (104)

From (103), with f(k) = 1/F_{k}^{2} in Lemma 1, we have the finite summation identity

\begin{aligned}∑_{k=1}^{N} \frac{F_{2nk+mnq}}{∏_{j=0}^{m} F_{nk+jnq}^{2}} & = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{nk+jnq}^{2}} \\ & − \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{nk+nN+jnq}^{2}} , mnq even . \end{aligned}  (105)

As N approaches infinity, we have Theorem 36, while specifically requiring m to be even gives Corollary 37.

Theorem 38. If m, n and q are integers such that q is even \underline{or} mnq is odd, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}F_{2nk+mnq}}{∏_{j=0}^{m} F_{nk+jnq}^{2}} ) = \frac{1}{F_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+jnq}^{2}} ) .

Proof. The statement of the theorem follows from (102) and identity (18).

In particular, we have

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+nq}}{F_{nk}^{2}F_{nk+nq}^{2}} = \frac{1}{F_{nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{F_{nk}^{2}} , q even or nq odd ,  (106)

which generalizes Brousseau’s result ([2], Formula (6)), also derived by Melham [8] as a special case of a more general result. Brousseau’s formula (6) corresponds to n = 1 in (106). Another Brousseau’s result ([2], Formula (15)) is also contained in the identity (106) above at n = 3, q = 1.

More examples from Theorem 38:

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+3nq}}{F_{nk}^{2}F_{nk+nq}^{2}F_{nk+2nq}^{2}F_{nk+3nq}^{2}} = \frac{1}{F_{3nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{F_{nk}^{2}F_{nk+nq}^{2}F_{nk+2nq}^{2}} , q even or nq odd . (107)

Corollary 39. If q is a positive \underline{even} integer, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}L_{nk+mnq}}{F_{nk+mnq} ∏_{j=0}^{m−1} F_{nk+jnq}^{2} ∏_{j=m+1}^{2m} F_{nk+jnq}^{2}} ) = \frac{1}{F_{2mnq}} ∑_{k=1}^{q} (−1)^{k−1} ( ∏_{j=0}^{2m−1} \frac{1}{F_{nk+jnq}^{2}} ) .

In particular:

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{k+q}}{F_{k}^{2}F_{k+q}F_{k+2q}^{2}} = \frac{1}{F_{2q}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{F_{k}^{2}F_{k+q}^{2}} , q even .  (108)

We note that Theorem 36, Theorem 38, Corollary 37 and Corollary 39 correspond to setting p = 0 in Theorem 48 and Theorem 49 of section 2.12.

Theorem 40. If m, n and q are positive integers, then

∑_{k=1}^{∞} (\frac{(−1)^{nk−1}F_{2nk+mnq} ∏_{j=1}^{m−1} L_{nk+jnq}^{2}}{∏_{j=0}^{m} F_{nk+jnq}^{2}} ) = \frac{5^{m}q}{4F_{mnq}} − \frac{1}{4F_{mnq}} ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{L_{nk+jnq}^{2}}{F_{nk+jnq}^{2}} ) .

In particular

∑_{k=1}^{∞} \frac{(−1)^{nk−1}F_{2nk+nq}}{F_{nk}^{2}F_{nk+nq}^{2}} = \frac{5q}{4F_{nq}} − \frac{1}{4F_{nq}} ∑_{k=1}^{q} \frac{L_{nk}^{2}}{F_{nk}^{2}} .  (109)

Theorem 41. If n and q are positive integers such that n is odd and q is even, then

∑_{k=1}^{∞} (\frac{F_{2nk+mnq} ∏_{j=1}^{m−1} L_{nk+jnq}^{2}}{∏_{j=0}^{m} F_{nk+jnq}^{2}} ) = \frac{1}{4F_{mnq}} ∑_{k=1}^{q} ((−1)^{k} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}^{2}}{F_{nk+jnq}^{2}} ) .

Multiplying identity (8a) and identity (8b) and choosing u and v judiciously, it is easy to establish that the following identity holds for positive integers m, n, q and k:

(−1)^{nk−1} \frac{4F_{mnq}F_{2nk+mnq}}{F_{nk}^{2}F_{nk+mnq}^{2}} = \frac{L_{nk+mnq}^{2}}{F_{nk+mnq}^{2}} − \frac{L_{nk}^{2}}{F_{nk}^{2}} .  (110)

From identity (110) and f(k) = L_{k}^{2}/F_{k}^{2} in Lemma 1 we have the finite summation formula:

\begin{matrix}4F_{mnq} ∑_{k=1}^{N} \frac{(−1)^{nk−1}F_{2nk+mnq} ∏_{j=1}^{m−1} L_{nk+jnq}^{2}}{∏_{j=0}^{m} F_{nk+jnq}^{2}} \\ = ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{L_{nk+nN+jnq}^{2}}{F_{nk+nN+jnq}^{2}} − ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{L_{nk+jnq}^{2}}{F_{nk+jnq}^{2}} , \end{matrix}  (111)

from which Theorem 40 follows in the limit N approaches infinity. Theorem 41 follows from identity (18).

2.11 Sums with L_{nk}^{2}L_{nk+nq}^{2} ⋯ L_{nk+mnq−nq}^{2}L_{nk+mnq}L_{nk+mnq+nq}^{2} ⋯ L_{nk+2mnq}^{2} or L_{nk}^{2}L_{nk+nq}^{2} ⋯ L_{nk+mnq}^{2} in the denominator

The theorems in this section are the Lucas versions of those of the previous section. We omit their proofs. The basic identity is

\frac{5F_{mnq}F_{2nk+mnq}}{L_{nk}^{2}L_{nk+mnq}^{2}} = \frac{1}{L_{nk}^{2}} − \frac{(−1)^{mnq}}{L_{nk+mnq}^{2}} ,  (112)

which follows from the identity (12b).

Theorem 42. If m, n and q are positive integers such that mnq is even, then

∑_{k=1}^{∞} (\frac{F_{2nk+mnq}}{∏_{j=0}^{m} L_{nk+jnq}^{2}} ) = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{1}{L_{nk+jnq}^{2}} ) .

Explicitly,

∑_{k=1}^{∞} \frac{F_{2nk+mnq}}{L_{nk}^{2}L_{nk+nq}^{2} ⋯ L_{nk+mnq}^{2}} = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} \frac{1}{L_{nk}^{2}L_{nk+nq}^{2} ⋯ L_{nk+(m−1)nq}^{2}} .

Corollary 43. If m, n and q are positive integers, then

∑_{k=1}^{∞} (\frac{F_{nk+mnq}}{L_{nk+mnq} ∏_{j=0}^{m−1} L_{nk+jnq}^{2} ∏_{j=m+1}^{2m} L_{nk+jnq}^{2}} ) = \frac{1}{5F_{2mnq}} ∑_{k=1}^{q} ( ∏_{j=0}^{2m−1} \frac{1}{L_{nk+jnq}^{2}} ) .

Theorem 44. If m, n and q are positive integers such that q is even \underline{or} mnq is odd, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}F_{2nk+mnq}}{∏_{j=0}^{m} L_{nk+jnq}^{2}} ) = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+jnq}^{2}} ) .

In particular, we have

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+nq}}{L_{nk}^{2}L_{nk+nq}^{2}} = \frac{1}{5F_{nq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{L_{nk}^{2}} , q even or nq odd ,  (113)

Corollary 45. If q is a positive \underline{even} integer, then

∑_{k=1}^{∞} (\frac{(−1)^{k−1}F_{nk+mnq}}{L_{nk+mnq} ∏_{j=0}^{m−1} L_{nk+jnq}^{2} ∏_{j=m+1}^{2m} L_{nk+jnq}^{2}} ) = \frac{1}{5F_{2mnq}} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{2m−1} L_{nk+jnq}^{2}} ) .

In particular:

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{k+q}}{L_{k}^{2}L_{k+q}L_{k+2q}^{2}} = \frac{1}{5F_{2q}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{L_{k}^{2}L_{k+q}^{2}} , q even .  (114)

Theorem 46. If m, n and q are positive integers, then

∑_{k=1}^{∞} (\frac{(−1)^{nk−1}F_{2nk+mnq} ∏_{j=1}^{m−1} F_{nk+jnq}^{2}}{∏_{j=0}^{m} L_{nk+jnq}^{2}} ) = \frac{1}{4F_{mnq}} ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{F_{nk+jnq}^{2}}{L_{nk+jnq}^{2}} ) − \frac{1}{4F_{mnq}} \frac{q}{5^{m}} .

In particular

∑_{k=1}^{∞} \frac{(−1)^{nk−1}F_{2nk+nq}}{L_{nk}^{2}L_{nk+nq}^{2}} = \frac{1}{4F_{nq}} ∑_{k=1}^{q} \frac{F_{nk}^{2}}{L_{nk}^{2}} − \frac{q}{20F_{nq}} .  (115)

Theorem 47. If n and q are positive integers such that n is odd and q is even, then

∑_{k=1}^{∞} (\frac{F_{2nk+mnq} ∏_{j=1}^{m−1} F_{nk+jnq}^{2}}{∏_{j=0}^{m} L_{nk+jnq}^{2}} ) = \frac{1}{4F_{mnq}} ∑_{k=1}^{q} ((−1)^{k−1} ∏_{j=0}^{m−1} \frac{F_{nk+jnq}^{2}}{L_{nk+jnq}^{2}} ) .

In particular

∑_{k=1}^{∞} \frac{F_{2nk+nq}}{L_{nk}^{2}L_{nk+nq}^{2}} = \frac{1}{4F_{nq}} ∑_{k=1}^{q} (−1)^{k−1} \frac{F_{nk}^{2}}{L_{nk}^{2}} , n odd, q even .  (116)

2.12 Sums with F_{nk}F_{nk+np}F_{nk+nq}F_{nk+nq+np}F_{nk+2nq}F_{nk+2nq+np}⋯ F_{nk+mnq}F_{nk+mnq+np} in the denominator

Theorem 48. If m, n, q are positive integers such that mnq is even; and p is a non-negative integer, then

∑_{k=1}^{∞} \frac{F_{2nk+mnq+np}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+np}} = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} F_{nk+jnq}F_{nk+jnq+np}} .

Theorem 49. If m, n, q are positive integers such that mnq is odd or q is even; and p is a non-negative integer, then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq+np}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+np}} = \frac{1}{F_{mnq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} F_{nk+jnq}F_{nk+jnq+np}} .

Proof of Theorem 48 and Theorem 49

By dividing through the identity (13a) by F_{u}F_{u+p}F_{u+v}F_{u+v+p} and setting u = nk and v = mnq, the following identity is established

\frac{F_{mnq}F_{2nk+mnq+np}}{F_{nk}F_{nk+np}F_{nk+mnq}F_{nk+mnq+np}} = \frac{1}{F_{nk}F_{nk+np}} + \frac{(−1)^{mnq+1}}{F_{nk+mnq}F_{nk+mnq+np}} ,  (117)

so that

\frac{F_{mnq}F_{2nk+mnq+np}}{F_{nk}F_{nk+np}F_{nk+mnq}F_{nk+mnq+np}} ; = \frac{1}{F_{nk}F_{nk+np}} − \frac{1}{F_{nk+mnq}F_{nk+mnq+np}} , mnq even  (118)

and

\frac{F_{mnq}F_{2nk+mnq+np}}{F_{nk}F_{nk+np}F_{nk+mnq}F_{nk+mnq+np}} ; = \frac{1}{F_{nk}F_{nk+np}} + \frac{1}{F_{nk+mnq}F_{nk+mnq+np}} , mnq odd .  (119)

From (118) and f(k) = 1/(F_{k}F_{k+np}) in Lemma 15 we obtain the finite summation identity

\begin{matrix}F_{mnq} ∑_{k=1}^{N} (\frac{F_{2nk+mnq+np}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+np}} ) \\ = ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{m−1} F_{nk+jnq}F_{nk+jnq+np}} ) \\ − ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{m−1} F_{nk+nN+jnq}F_{nk+nN+jnq+np}} ) , \end{matrix}  (120)

from which Theorem 48 follows in the limit as N approaches infinity. The statement of Theorem 49 follows from (117) and identity (18).

2.13 Sums with L_{nk}L_{nk+np}L_{nk+nq}L_{nk+nq+np}L_{nk+2nq}L_{nk+2nq+np}⋯ L_{nk+mnq}L_{nk+mnq+np} in the denominator

The results in this section are the Lucas versions of the results in the preceding section. The basic identities are

\frac{5F_{mnq}F_{2nk+mnq+np}}{∏_{j=0}^{m} L_{nk+jnq}L_{nk+jnq+np}} = \frac{1}{∏_{j=0}^{m−1} L_{nk+jnq}L_{nk+jnq+np}} + \frac{(−1)^{mnq+1}}{∏_{j=1}^{m} L_{nk+jnq}L_{nk+jnq+np}} , (121)

and, if mnq is even

\begin{matrix}5F_{mnq} ∑_{k=1}^{N} (\frac{F_{2nk+mnq+np}}{∏_{j=0}^{m} L_{nk+jnq}L_{nk+jnq+np}} ) \\ = ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{m−1} L_{nk+jnq}L_{nk+jnq+np}} ) \\ − ∑_{k=1}^{q} (\frac{1}{∏_{j=0}^{m−1} L_{nk+nN+jnq}L_{nk+nN+jnq+np}} ) , \end{matrix}  (122)

while if mnq is odd or q is even:

\begin{matrix}5F_{mnq} ∑_{k=1}^{N} (\frac{(−1)^{k−1}F_{2nk+mnq+np}}{∏_{j=0}^{m} L_{nk+jnq}L_{nk+jnq+np}} ) \\ = ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+jnq}L_{nk+jnq+np}} ) \\ + (−1)^{N−1} ∑_{k=1}^{q} (\frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+nN+jnq}L_{nk+nN+jnq+np}} ) . \end{matrix}  (123)

Theorem 50. If m, n, q are positive integers such that mnq is even; and p is a non-negative integer, then

∑_{k=1}^{∞} \frac{F_{2nk+mnq+np}}{∏_{j=0}^{m} L_{nk+jnq}L_{nk+jnq+np}} = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} \frac{1}{∏_{j=0}^{m−1} L_{nk+jnq}L_{nk+jnq+np}} .

Theorem 51. If m, n, q are positive integers such that mnq is odd or q is even; and p is a non-negative integer, then

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq+np}}{∏_{j=0}^{m} L_{nk+jnq}L_{nk+jnq+np}} = \frac{1}{5F_{mnq}} ∑_{k=1}^{q} \frac{(−1)^{k−1}}{∏_{j=0}^{m−1} L_{nk+jnq}L_{nk+jnq+np}} .

2.14 Evaluation of other sums

Theorem 52. If m, n and q are positive integers, then

∑_{k=1}^{∞} \frac{(−1)^{nk−1}F_{2nk+mnq+2} ∏_{j=1}^{m−1} F_{nk+jnq+1}^{2}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+2}} = \frac{q}{F_{mnq}} − \frac{1}{F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{F_{nk+jnq+1}^{2}}{F_{nk+jnq}F_{nk+jnq+2}} ,

so that

∑_{k=1}^{∞} \frac{F_{2nk+mnq+2} ∏_{j=1}^{m−1} F_{nk+jnq+1}^{2}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+2}} = \frac{1}{F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{F_{nk+jnq+1}^{2}}{F_{nk+jnq}F_{nk+jnq+2}} − \frac{q}{F_{mnq}} , n even (124)

and

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{2nk+mnq+2} ∏_{j=1}^{m−1} F_{nk+jnq+1}^{2}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+2}} = \frac{q}{F_{mnq}} − \frac{1}{F_{mnq}} ∑_{k=1}^{q} ∏_{j=0}^{m−1} \frac{F_{nk+jnq+1}^{2}}{F_{nk+jnq}F_{nk+jnq+2}} , n odd .

(125)

Theorem 53. If m, n and q are positive integers such that n is odd and q is even, then

∑_{k=1}^{∞} (\frac{F_{2nk+mnq+2} ∏_{j=1}^{m−1} F_{nk+jnq+1}^{2}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+2}} ) = \frac{1}{F_{mnq}} ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{(−1)^{k}F_{nk+jnq+1}^{2}}{F_{nk+jnq}F_{nk+jnq+2}} ) .

Proof of Theorem (52) and Theorem (53)

Dividing through the identity (10) by F_{v+1}F_{v−1}F_{u+1}F_{u−1} and setting u = nk + 1 and v = nk + mnq + 1 we obtain the identity

\frac{(−1)^{nk−1}F_{mnq}F_{2nk+mnq+2}}{F_{nk}F_{nk+2}F_{nk+mnq}F_{nk+mnq+2}} = \frac{F_{nk+mnq+1}^{2}}{F_{nk+mnq+2}F_{nk+mnq}} − \frac{F_{nk+1}^{2}}{F_{nk+2}F_{nk}} .  (126)

With f(k) = F_{k+1}^{2}/(F_{k}F_{k+2}) in Lemma 1 and use of the identity (126) we get the finite summation identity

\begin{matrix}F_{mnq} ∑_{k=1}^{N} (\frac{(−1)^{nk−1}F_{2nk+mnq+2} ∏_{j=1}^{m−1} F_{nk+jnq+1}^{2}}{∏_{j=0}^{m} F_{nk+jnq}F_{nk+jnq+2}} ) \\ = ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{F_{nk+nN+jnq+1}^{2}}{F_{nk+nN+jnq}F_{nk+nN+jnq+2}} ) \\ − ∑_{k=1}^{q} (∏_{j=0}^{m−1} \frac{F_{nk+jnq+1}^{2}}{F_{nk+jnq}F_{nk+jnq+2}} ) , \end{matrix}  (127)

from which Theorem (52) follows in the limit as N approaches infinity. Theorem (53) follows from the identity (126) and taking f(k) = F_{k+1}^{2}/(F_{k}F_{k+2}) in identity (18).

Theorem 54.

∑_{k=1}^{∞} \frac{F_{2k+3}}{F_{k}^{4}F_{k+1}^{3}F_{k+2}^{3}F_{k+3}^{4}} = \frac{1}{128}, ∑_{k=1}^{∞} \frac{F_{2k+3}}{L_{k}^{4}L_{k+1}^{3}L_{k+2}^{3}L_{k+3}^{4}} = \frac{1}{829440} .

Theorem 55.

∑_{k=1}^{∞} \frac{F_{3k+1}F_{3k+2}F_{6k+3}}{F_{3k}^{4}F_{3k+3}^{4}} = \frac{1}{128}, ∑_{k=1}^{∞} \frac{L_{3k+1}L_{3k+2}F_{6k+3}}{L_{3k}^{4}L_{3k+3}^{4}} = \frac{1}{10240} .

3 Acknowledgment

The author thanks the anonymous referee whose comments helped improve the presentation.

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2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39. Keywords: Fibonacci-Lucas sum, Fibonacci number, Lucas number, infinite sum, reciprocal sum, telescoping summation.

Received October 27 2017; revised version received February 6 2018. Published in Journal of Integer Sequences, February 8 2018.

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