Finite and Infinite Sums Involving Reciprocals of Products of Gibonacci Numbers

1. Motivation and Preliminaries

Sums and series of reciprocals of recurrence sequences is a topic of great interest in recent research articles. In [13] the authors consider a very general sums of the form

$$\sum _{n=0}^{m-1}{\displaystyle \frac{B^{f\left(n\right)}{u}_{\Delta f\left(n\right)}}{w_{f\left(n\right)}{w}_{f(n+1)}}},$$

where $w_n$ is an arbitrary second order recurrence sequence (Horadam sequence), B is some parameter related to that sequence and $\Delta$ is the difference operator. In the paper from 2023, Adegoke et al. [3] study three-parameter sums and series of the form

$$\sum _{i=1}^N{\displaystyle \frac{q^{m(i-k)}}{w_{m(i-k)+n}{w}_{m(i+k)+n}}}and\sum _{i=1}^N{\displaystyle \frac{q^{m(2i-k)}}{w_{m(2i-k)}{w}_{m(2i+k)}}},$$

where q is a parameter, and $m,n$ and k are integers. Similar sums are the subject of interest in [1,9,10,14]. Extraordinary work on reciprocal sums with three and more (generalized) Fibonacci factors has been done by Melham. In a series of papers starting in 2000 he gives closed forms for many types of these sums (see [15,16,17,18,19,20,21,22]). But their history can be tracked back further to André-Jeannin [4], Melham and Shannon [23], Rabinowitz [28] and even to Hoggart and Bicknell [11], Bruckman and Good [8] and Brother Brousseau back in 1967.

Brousseau in his articles [6,7] initiated a trend that is currently named after him “Brousseau sums” and is related to finding various sums with inverses of Fibonacci-like numbers and their (weighted) products. In particular, Brousseau applied a telescoping summation method (referenced in this article as (T1)) in [7] to obtain many Fibonacci and Lucas identities, which are, among others:

$$\begin{array}{cccc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{F_k}{F_{k+1}{F}_{k+2}}}& =1,\hfill & \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{F_{k+1}}{F_k{F}_{k+3}}}& ={\displaystyle \frac{5}{4}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{F_{4k+3}}{F_{2k}{F}_{2k+1}{F}_{2k+2}{F}_{2k+3}}}& ={\displaystyle \frac{1}{2}},\hfill & \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{F_{2k+5}}{F_k{F}_{k+1}{F}_{k+2}{F}_{k+3}{F}_{k+4}{F}_{k+5}}}& ={\displaystyle \frac{1}{15}}.\hfill \end{array}$$

For example, the first identity on the right hand side is a consequence of the following relation:

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{F_n}}-{\displaystyle \frac{1}{F_{n+3}}}& ={\displaystyle \frac{F_{n+3}-F_n}{F_n{F}_{n+3}}}={\displaystyle \frac{2F_{n+1}}{F_n{F}_{n+3}}},\hfill \end{array}$$

and thus applying the telescoping principle we arrive at

$$\sum _{k=1}^{\infty }{\displaystyle \frac{F_{n+1}}{F_n{F}_{n+3}}}={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{F_1}}+{\displaystyle \frac{1}{F_2}}+{\displaystyle \frac{1}{F_3}}\right)={\displaystyle \frac{5}{4}}.$$

Notice that the only place where initial values matter is at the very end of the computation, therefore the reasoning can be easily generalized to any Gibonacci sequence. This idea will be heavily used in our article.

The idea for writing this article has a second source of motivation. In the section Advanced Problems and Solutions of the journal The Fibonacci Quarterly Ohtsuka [25] challenges the readers to show that

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{F_k{F}_{k+4}}}\left({\displaystyle \frac{1}{F_{k+1}}}+{\displaystyle \frac{1}{F_{k+2}}}-{\displaystyle \frac{1}{F_{k+3}}}\right)& ={\displaystyle \frac{1}{3}},\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{F_k{F}_{k+4}}}\left({\displaystyle \frac{1}{F_{k+1}}}+{\displaystyle \frac{1}{F_{k+2}}}-{\displaystyle \frac{1}{F_{k+3}}}\right)& =-{\displaystyle \frac{1}{6}},\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{F_k{F}_{k+1}{F}_{k+2}{F}_{k+3}{F}_{k+4}}}\left({\displaystyle \frac{1}{F_{k+1}}}+{\displaystyle \frac{1}{F_{k+2}}}-{\displaystyle \frac{1}{F_{k+3}}}\right)& ={\displaystyle \frac{1}{24}}.\hfill \end{array}$$

(3)

The problem was solved by Bataille [5] (among others). A Lucas version of the problem also exists [27]:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{3}{L_k{L}_{k+4}}}\left({\displaystyle \frac{1}{L_{k+1}}}+{\displaystyle \frac{1}{L_{k+2}}}-{\displaystyle \frac{1}{L_{k+3}}}\right)& ={\displaystyle \frac{1}{7}},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k}3}{L_k{L}_{k+4}}}\left({\displaystyle \frac{1}{L_{k+1}}}+{\displaystyle \frac{1}{L_{k+2}}}-{\displaystyle \frac{1}{L_{k+3}}}\right)& =-{\displaystyle \frac{3}{28}},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{3}{L_k{L}_{k+1}{L}_{k+2}{L}_{k+3}{L}_{k+4}}}\left({\displaystyle \frac{1}{L_{k+1}}}+{\displaystyle \frac{1}{L_{k+2}}}-{\displaystyle \frac{1}{L_{k+3}}}\right)& ={\displaystyle \frac{1}{672}}.\hfill \end{array}$$

(6)

As it turns out, these sums can be easily generalized to the Gibonacci sequence as well and furthermore, these identities have given rise to a search for similar identities. In addition, we have found other problem proposals that allow for such generalizations.

In the next sections, we find the aforementioned identities, but we also go back to some classic problems, such as summation of

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+m}}}$$

considered by André-Jeannin [4], and later generalized by Melham and Shannon [23] and Melham [18,19,22].

Throughout the article we use the standard notation for the Fibonacci numbers $F_n$, the Lucas numbers $L_n$ and the Gibonacci numbers $G_n$. Recall that the Gibonacci sequences has the same recurrence relation as the Fibonacci sequence but starts with arbitrary initial values, i.e.,

$$G_k=G_{k-1}+G_{k-2},(k\ge 2),$$

with $G_0$ and $G_1$ arbitrary numbers (usually integers) not both zero. When $G_0=0$ and $G_1=1$ then $G_n=F_n$, and when $G_0=2$ and $G_1=1$ then $G_n=L_n$, respectively. The sequence obeys the generalized Binet formula

$$G_n=A{\alpha }^n+B{\beta }^n,$$

where $\alpha =\frac{1+\sqrt{5}}{2}$, $\beta =\frac{1-\sqrt{5}}{2}$ are the roots of the characteristic equation $x^2=x+1$, and $A=\frac{G_1-G_0\beta }{\alpha -\beta }$ and $B=\frac{G_0\alpha -G_1}{\alpha -\beta }$. To avoid indeterminates, we assume that $G_n\ne 0$ for all n.

We recall the following following classic telescoping summation formulas:

$$\begin{array}{cc}\hfill \sum _{j=1}^n(f_{j+1}-f_j)& =f_{n+1}-f_n,\hfill \\ \hfill \sum _{j=1}^n{(-1)}^{j-1}(f_{j+1}+f_j)& ={(-1)}^{n+1}{f}_{n+1}+f_1,\hfill \\ \hfill \sum _{j=1}^n{d}^{n-j}{c}^{j-1}(c{f}_{j+1}-d{f}_j)& =c^n{f}_{n+1}-d^n{f}_1,\hfill \end{array}$$

where we assume that $f_j$, d and c are real numbers with $cd\ne 0$. Throughout the article, we will mainly use (T1) and a variation of (T3).

We note that the range of summation in our article is from $k=1$. This is just a cosmetic choice and any formula obtained for the Gibonacci sequence in range from $k=1$ can be adjusted accordingly to any other range from $k=\ell$ where ℓ is any non-negative integer.

1.1. Gibonacci identities

We conclude with a collection of Gibonacci identities which will be used later in various forms. Two classical identities for Gibonacci numbers are

$$\begin{array}{cc}\hfill G_{k+m}& =F_m{G}_{k+1}+F_{m-1}{G}_k,\hfill \\ \hfill G_{k-m}& ={(-1)}^m(F_{m+1}{G}_k-F_m{G}_{k+1}).\hfill \end{array}$$

(7)

The first identity is Equation (8) in Vajda’s book [29]. Two other general identities from Vajda’s book are

$$\begin{array}{cc}\hfill G_{k+m}+{(-1)}^m{G}_{k-m}& =L_m{G}_k,\hfill \\ \hfill G_{k+m}-{(-1)}^m{G}_{k-m}& =F_m(G_{k-1}+G_{k+1}),\hfill \end{array}$$

(8)

which are Equations (10a) and (10b), respectively. These can be turned into identities involving fractions:

$$\begin{array}{cc}\hfill {\displaystyle \frac{{(-1)}^r}{G_{k+2r}}}+{\displaystyle \frac{1}{G_k}}& ={\displaystyle \frac{L_r{G}_{k+r}}{G_k{G}_{k+2r}}},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k}}-{\displaystyle \frac{F_{r-1}}{G_{k+r}}}& ={\displaystyle \frac{F_r{G}_{k+1}}{G_k{G}_{k+r}}}.\hfill \end{array}$$

(10)

The first one is a restatement of Equation (10a) in Vajda’s book, which we include for further reference:

$$G_{k+2m}+{(-1)}^m{G}_k=L_m{G}_{k+m},$$

(11)

while the second one is a restatement of (7). We also have a special case of (8): for odd m we have

$$G_{k+2m}=F_m(G_{k+m-1}+G_{k+m+1)})-G_k.$$

(12)

If we now change indices in (11) and rearrange, we get

$$G_{nk}=L_n{G}_{nk-n}+{(-1)}^{n+1}{G}_{nk-2n}.$$

(13)

An identity of Howard [12] can be written as

$${\displaystyle \frac{F_{m-r}}{G_{k+r}}}-{(-1)}^r{\displaystyle \frac{F_m}{G_k}}={(-1)}^{r-1}{\displaystyle \frac{F_r{G}_{k+m}}{G_k{G}_{k+r}}}.$$

(14)

Still another classical identity, the Catalan identity, reads as

$$G_{k-r}{G}_{k+r}-G_k^2={(-1)}^{k-r}(G_0{G}_2-G_1^2)F_r^2,r\ge 0,$$

of which the Cassini identity

$$G_{k-1}{G}_{k+1}-G_k^2={(-1)}^{k+1}(G_0{G}_2-G_1^2)$$

(15)

is a special case.

Finally, we note the following limit properties.

Lemma 1.

If m is a non-negative integer, then

$$\underset{k\to \infty }{lim}{\displaystyle \frac{G_k}{G_{k+m}}}={\displaystyle \frac{1}{{\alpha }^m}}={(-1)}^m{\beta }^m={(-1)}^m(\beta F_m+F_{m-1})$$

and

$$\underset{k\to \infty }{lim}{\displaystyle \frac{F_k}{G_{k+m}}}={\displaystyle \frac{1}{(G_1-G_0\beta ){\alpha }^m}}={\displaystyle \frac{{(-1)}^{m-1}}{{\epsilon }_G}}(G_{m+1}-\alpha G_m),$$

(16)

where here and throughout this paper ${\epsilon }_G=G_0^2-G_1^2+G_1{G}_0$.

Proof. 

These results are consequences of the Binet formula. Note that in obtaining the second equality in (16), we used

$${\displaystyle \frac{1}{{\alpha }^m}}={(-1)}^m{\beta }^m={(-1)}^m(\beta F_m+F_{m-1})$$

and (7). □

For convenience and to shorten the formulas, we will commonly write $G_k\cdots G_{k+n}$ instead of $G_k·G_{k+1}\cdots G_{k+n}$.

2. The Generalization of (1)–(6) and Other Identities

Our first set of results is the direct generalization of the identities (1)–(6) to the Gibonacci case. Thus, we shall utilize elementary identities. A general approach or the structure of some identities will be discussed in the next section.

Theorem 1.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)={\displaystyle \frac{1}{G_1{G}_2{G}_4}}.$$

(17)

Proof. 

Notice that

$$G_{k+3}(G_{k+3}+G_k)=G_{k+3}·2G_{k+2}=G_{k+2}(G_{k+1}+G_{k+4}),$$

so

$$G_{k+3}^2-G_{k+1}{G}_{k+2}=G_{k+2}{G}_{k+4}-G_k{G}_{k+3}$$

(18)

and thus

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k{G}_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_k{G}_{k+4}}}·{\displaystyle \frac{G_{k+3}^2-G_{k+1}{G}_{k+2}}{G_{k+1}{G}_{k+2}{G}_{k+3}}}\hfill \\ \hfill & ={\displaystyle \frac{G_{k+2}{G}_{k+4}-G_k{G}_{k+3}}{G_k\cdots G_{k+4}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+3}}}-{\displaystyle \frac{1}{G_{k+1}{G}_{k+2}{G}_{k+4}}}.\hfill \end{array}$$

Hence the desired sum telescopes and we have via (T1)

$$\underset{n\to \infty }{lim}\left({\displaystyle \frac{1}{G_1{G}_2{G}_4}}-{\displaystyle \frac{1}{G_{n+1}{G}_{n+2}{G}_{n+4}}}\right)={\displaystyle \frac{1}{G_1{G}_2{G}_4}},$$

which gives (17). □

The next result is the following alternating version of (17).

Theorem 2.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{G_k{G}_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_3}{G_1{G}_4}}-{\displaystyle \frac{1}{G_3}}-{\displaystyle \frac{1}{G_4}}\right).$$

(19)

Proof. 

We use the computation used to prove (17) and the Cassini identity for the Gibonacci numbers (15). This implies

$$\begin{array}{cc}\hfill {\displaystyle \frac{{(-1)}^k}{G_k{G}_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{{(-1)}^k}{G_k{G}_{k+1}{G}_{k+3}}}+{\displaystyle \frac{{(-1)}^{k+1}}{G_{k+1}{G}_{k+2}{G}_{k+4}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_{k+1}{G}_{k+3}-G_{k+2}^2}{G_k{G}_{k+1}{G}_{k+3}}}+{\displaystyle \frac{G_{k+2}{G}_{k+4}-G_{k+3}^2}{G_{k+1}{G}_{k+2}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{1}{G_k}}-{\displaystyle \frac{G_{k+2}^2}{G_k{G}_{k+1}{G}_{k+3}}}+{\displaystyle \frac{1}{G_{k+1}}}-{\displaystyle \frac{G_{k+3}^2}{G_{k+1}{G}_{k+2}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_{k+2}}{G_k{G}_{k+1}}}-{\displaystyle \frac{G_{k+2}^2}{G_k{G}_{k+1}{G}_{k+3}}}-{\displaystyle \frac{G_{k+3}^2}{G_{k+1}{G}_{k+2}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_{k+2}{G}_{k+3}-G_{k+2}^2}{G_k{G}_{k+1}{G}_{k+3}}}-{\displaystyle \frac{G_{k+3}{G}_{k+2}+G_{k+3}{G}_{k+1}}{G_{k+1}{G}_{k+2}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_{k+2}}{G_k{G}_{k+3}}}-{\displaystyle \frac{G_{k+3}}{G_{k+1}{G}_{k+4}}}-{\displaystyle \frac{G_{k+3}}{G_{k+2}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_{k+2}}{G_k{G}_{k+3}}}-{\displaystyle \frac{G_{k+3}}{G_{k+1}{G}_{k+4}}}-\left({\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+4}}}\right)\right).\hfill \end{array}$$

Applying the telescoping formula (T1) we complete the proof of (19). □

The final generalized sum is as follows.

Theorem 3.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}{G}_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)={\displaystyle \frac{1}{2G_1{G}_2^2{G}_3^3{G}_4}}.$$

(20)

Proof. 

We use (18) so that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k\cdots G_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{G_{k+2}{G}_{k+4}-G_k{G}_{k+3}}{G_k{G}_{k+1}^2{G}_{k+2}^2{G}_{k+3}^2{G}_{k+4}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{G_k{G}_{k+1}^2{G}_{k+2}{G}_{k+3}^2}}-{\displaystyle \frac{1}{G_{k+1}^2{G}_{k+2}^2{G}_{k+3}{G}_{k+4}}}.\hfill \end{array}$$

Letting

$$A\left(k\right)={\displaystyle \frac{1}{G_k{G}_{k+1}^2{G}_{k+2}{G}_{k+3}^2}},B\left(k\right)={\displaystyle \frac{1}{G_k^2{G}_{k+1}^2{G}_{k+2}{G}_{k+3}}}$$

we have

$${\displaystyle \frac{1}{G_k\cdots G_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)=\left(A\left(k\right)-B\left(k\right)\right)+\left(B\left(k\right)-B(k+1)\right).$$

Further notice that

$$G_{k+3}^2-G_k^2=(G_{k+3}-G_k)(G_{k+3}+G_k)=4G_{k+1}{G}_{k+2}=-2(G_k-G_{k+3})G_{k+2},$$

therefore

$$\begin{array}{cc}\hfill A\left(k\right)-B\left(k\right)& ={\displaystyle \frac{G_k-G_{k+3}}{G_k^2{G}_{k+1}^2{G}_{k+2}{G}_{k+3}^2}}\hfill \\ \hfill & =-{\displaystyle \frac{1}{2}}·{\displaystyle \frac{G_{k+3}^2-G_k^2}{{(G_k\cdots G_{k+3})}^2}}\hfill \\ \hfill & =-{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{G_k^2{G}_{k+1}^2{G}_{k+2}^2}}-{\displaystyle \frac{1}{G_{k+1}^2{G}_{k+2}^2{G}_{k+3}^2}}\right).\hfill \end{array}$$

Finally, we have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}{G}_{k+4}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& \\ \hfill & =\underset{n\to \infty }{lim}\left(-{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{G_1^2{G}_2^2{G}_3^2}}-{\displaystyle \frac{1}{G_{n+1}^2{G}_{n+2}^2{G}_{n+3}^2}}\right)+B\left(1\right)-B(n+1)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2{G}_2^2{G}_3{G}_4}}-{\displaystyle \frac{1}{2G_1^2{G}_2^2{G}_3^2}}\hfill \\ \hfill & ={\displaystyle \frac{2G_3{G}_4-G_4^2}{2{(G_1\cdots G_4)}^2}}={\displaystyle \frac{G_1{G}_4}{2{(G_1\cdots G_4)}^2}}={\displaystyle \frac{1}{2G_1{G}_2^2{G}_3^3{G}_4}}\hfill \end{array}$$

and we get (20). □

Continuing the current trend of generalizations, we shall recall the following identities obtained by Ohtsuka in 2018, proposed as Problem H-818 in The Fibonacci Quarterly, and solved in [24]:

$$\begin{array}{cc}\hfill S=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{F_k{F}_{k+1}{F}_{k+2}{F}_{k+4}}}& ={\displaystyle \frac{38-15\sqrt{5}}{36}},\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill T=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{F_k{F}_{k+2}{F}_{k+3}{F}_{k+4}}}& ={\displaystyle \frac{-32+15\sqrt{5}}{36}}.\hfill \end{array}$$

(22)

We now evaluate the Gibonacci case of these sums.

Theorem 4.

The following summation hold:

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}={\displaystyle \frac{1}{2(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right).$$

Consequently, the following identities hold:

$$\begin{array}{cc}\hfill S_G=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+4}}}& \\ \hfill & ={\displaystyle \frac{1}{6(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right)+{\displaystyle \frac{1}{3G_1{G}_2{G}_3{G}_4}},\hfill \\ \hfill T_G=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2}{G}_{k+3}{G}_{k+4}}}& \\ \hfill & =-{\displaystyle \frac{1}{6(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right)+{\displaystyle \frac{2}{3G_1{G}_2{G}_3{G}_4}}.\hfill \end{array}$$

Proof. 

Adding the two series gives

$$\begin{array}{cc}\hfill S_G+T_G& =\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+1}+G_{k+3}}{G_k\cdots G_{k+4}}}=\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+4}-G_k}{G_k\cdots G_{k+4}}}\hfill \\ \hfill & =\sum _{k=1}^{\infty }\left({\displaystyle \frac{1}{G_k\cdots G_{k+3}}}-{\displaystyle \frac{1}{G_{k+1}\cdots G_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_4}}.\hfill \end{array}$$

Subtracting, on the other hand, gives,

$$\begin{array}{cc}\hfill S_G-T_G& =\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+3}-G_{k+1}}{G_k\cdots G_{k+4}}}=\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+2}}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}{G}_{k+4}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}·\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+4}+G_k}{G_k\cdots G_{k+4}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}·\sum _{k=1}^{\infty }\left({\displaystyle \frac{1}{G_{k+1}\cdots G_{k+4}}}+{\displaystyle \frac{1}{G_k\cdots G_{k+3}}}\right).\hfill \end{array}$$

We now apply identity (23) from [19] to obtain the following finite case of our sum:

$$2(G_1^2-G_0^2-G_0{G}_1)·G_1·\sum _{k=1}^{n-1}{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}={\displaystyle \frac{G_3+G_5}{G_2{G}_3}}-\left({\displaystyle \frac{F_{n-1}}{G_n}}+{\displaystyle \frac{3F_n}{G_{n+1}}}+{\displaystyle \frac{F_{n+1}}{G_{n+2}}}\right).$$

Using (16) we obtain

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}={\displaystyle \frac{1}{2(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right).$$

(23)

This implies

$$S_G-T_G={\displaystyle \frac{1}{3(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right)-{\displaystyle \frac{1}{3G_1{G}_2{G}_3{G}_4}}.$$

Since $S_G+T_G$ and $S_G-T_G$ are now explicitly given, solving for $S_G$ and $T_G$ gives the final result. □

We note that setting $G_n=F_n$ in Theorem 4 restores (21) and (22). If we now let $G_n=L_n$, then we obtain a Lucas version of the identities:

$$\begin{array}{cc}\hfill S_L& =\sum _{k=1}^{\infty }{\displaystyle \frac{1}{L_k{L}_{k+1}{L}_{k+2}{L}_{k+4}}}={\displaystyle \frac{115-42\sqrt{5}}{2520}},\hfill \\ \hfill T_L& =\sum _{k=1}^{\infty }{\displaystyle \frac{1}{L_k{L}_{k+2}{L}_{k+3}{L}_{k+4}}}={\displaystyle \frac{-85+42\sqrt{5}}{2520}}.\hfill \end{array}$$

The next three corollaries are immediate consequences of Theorem 4.

Corollary 1.

We have

$$\begin{array}{cc}\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k-1}}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}{G}_{k+4}}}\hfill & \\ & \hfill =-{\displaystyle \frac{2}{3(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right)+{\displaystyle \frac{5}{3G_1{G}_2{G}_3{G}_4}}.\end{array}$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{F_{k-1}}{F_k{F}_{k+1}{F}_{k+2}{F}_{k+3}{F}_{k+4}}}={\displaystyle \frac{30\sqrt{5}-67}{18}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{L_{k-1}}{L_k{L}_{k+1}{L}_{k+2}{L}_{k+3}{L}_{k+4}}}={\displaystyle \frac{84\sqrt{5}-185}{1260}}.$$

Proof. 

Calculate $3T_G-S_G$ and use

$$G_{k+3}-3G_{k+1}=-G_{k-1}.$$

□

Corollary 2.

We have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{G_{k-1}}{G_{k+1}{G}_{k+2}{G}_{k+3}{G}_{k+4}{G}_{k+5}}}& ={\displaystyle \frac{7}{6(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right)\hfill \\ \hfill & -{\displaystyle \frac{8}{3G_1{G}_2{G}_3{G}_4}}-{\displaystyle \frac{G_{-1}}{G_1{G}_2{G}_3{G}_4{G}_5}}.\hfill \end{array}$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{F_{k-1}}{F_{k+1}{F}_{k+2}{F}_{k+3}{F}_{k+4}{F}_{k+5}}}={\displaystyle \frac{1174-525\sqrt{5}}{180}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{L_{k-1}}{L_{k+1}{L}_{k+2}{L}_{k+3}{L}_{k+4}{L}_{k+5}}}={\displaystyle \frac{7235-3234\sqrt{5}}{27720}}.$$

Proof. 

Calculate $2S_G-5T_G$ and use

$$2G_{k+3}-5G_{k+1}=G_{k-2}.$$

□

Corollary 3.

We have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+1}+G_{k-1}}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}{G}_{k+4}}}& =-{\displaystyle \frac{5}{6(G_1^2-G_0^2-G_0{G}_1)}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right)\hfill \\ \hfill & +{\displaystyle \frac{7}{3G_1{G}_2{G}_3{G}_4}}.\hfill \end{array}$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{L_k}{F_k{F}_{k+1}{F}_{k+2}{F}_{k+3}{F}_{k+4}}}={\displaystyle \frac{75\sqrt{5}-166}{36}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{F_k}{L_k{L}_{k+1}{L}_{k+2}{L}_{k+3}{L}_{k+4}}}={\displaystyle \frac{6\sqrt{5}-13}{360}}.$$

Proof. 

Calculate $4T_G-S_G$. □

3. Some Mixed Sums of Type (1)

In this section we are interested in the following kind of sums:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{3}{G_{k+1}}}-{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{4}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_2{G}_3{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left(-{\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{2}{G_{k+2}}}+{\displaystyle \frac{3}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_3{G}_4{G}_5}},\hfill \end{array}$$

and similar ones (the two above will be justified later). Note that each term in the sum is the product of:

• the product of the inverse of certain Gibonacci numbers,
• the linear combination of the inverse of certain Gibonacci numbers,

• and each Gibonacci number appears at most once in each term. In this section, we show several identities of that kind.

Theorem 5.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{k+1}{G}_{k+4}}}\left({\displaystyle \frac{3}{G_k}}-{\displaystyle \frac{2}{G_{k+2}}}-{\displaystyle \frac{3}{G_{k+3}}}\right)={\displaystyle \frac{2}{G_1{G}_3{G}_4}}.$$

Proof. 

The proof follows the flow of the proof of Theorem 1. First, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{3}{G_k}}-{\displaystyle \frac{2}{G_{k+2}}}-{\displaystyle \frac{3}{G_{k+3}}}& ={\displaystyle \frac{3G_{k+2}{G}_{k+3}-2G_k{G}_{k+3}-3G_k{G}_{k+2}}{G_k{G}_{k+2}{G}_{k+3}}}\hfill \\ \hfill & ={\displaystyle \frac{2G_{k+1}{G}_{k+4}-2G_k{G}_{k+2}}{G_k{G}_{k+2}{G}_{k+3}}},\hfill \end{array}$$

where we used the identity

$$3G_{k+2}{G}_{k+3}-2G_k{G}_{k+3}-3G_k{G}_{k+2}=2G_{k+1}{G}_{k+4}-2G_k{G}_{k+2}.$$

This identity is justified by writing in the equivalent form

$$\begin{array}{c}\hfill 3G_{k+2}(G_{k+3}-G_k)=2G_{k+1}{G}_{k+4}-2G_k(G_{k+2}-G_{k+3})\end{array}$$

and then noting that each of the sides equals $6G_{k+1}{G}_{k+2}$. Note that we have used $G_k+G_{k+4}=3G_{k+2}$ and $G_{k+3}-G_k=2G_{k+1}$. □

Theorem 6.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{k+1}{G}_{k+4}}}\left({\displaystyle \frac{3}{G_k}}-{\displaystyle \frac{2}{G_{k+2}}}+{\displaystyle \frac{1}{G_{k+3}}}\right)={\displaystyle \frac{2}{G_1{G}_2{G}_4}}.$$

Proof. 

This time we use

$${\displaystyle \frac{3}{G_k}}-{\displaystyle \frac{2}{G_{k+2}}}+{\displaystyle \frac{1}{G_{k+3}}}={\displaystyle \frac{3G_{k+2}{G}_{k+3}-2G_k{G}_{k+3}+G_k{G}_{k+2}}{G_k{G}_{k+2}{G}_{k+3}}}$$

and apply the identity

$$3G_{k+2}{G}_{k+3}-2G_k{G}_{k+3}+G_k{G}_{k+2}=G_{k+3}{G}_{k+4}-G_k{G}_{k+1},$$

which can be verified by simple manipulation based on the recurrence relation. This implies

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{k+1}{G}_{k+4}}}\left({\displaystyle \frac{3}{G_k}}-{\displaystyle \frac{2}{G_{k+2}}}+{\displaystyle \frac{1}{G_{k+3}}}\right)& =\sum _{k=1}^{\infty }\left({\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}}}-{\displaystyle \frac{1}{G_{k+2}{G}_{k+3}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1{G}_2{G}_3}}+{\displaystyle \frac{1}{G_2{G}_3{G}_4}}\hfill \\ \hfill & ={\displaystyle \frac{G_1+G_4}{G_1\cdots G_4}}={\displaystyle \frac{2G_3}{G_1\cdots G_4}}={\displaystyle \frac{2}{G_1{G}_2{G}_4}}.\hfill \end{array}$$

□

We have discovered many more identities of the form described at the beginning of this section. The following set shares the same indices in the corresponding places of the left-hand side of the identity, but the coefficients differ.

Corollary 4.

We have the following identities:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{6}{G_{k+1}}}+{\displaystyle \frac{3}{G_{k+2}}}-{\displaystyle \frac{3}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_4}}+{\displaystyle \frac{1}{G_2{G}_3{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{3}{G_{k+1}}}-{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{4}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_2{G}_3{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left(-{\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{2}{G_{k+2}}}+{\displaystyle \frac{3}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_3{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{2}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{3}{G_{k+2}}}+{\displaystyle \frac{2}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{3}{G_{k+1}}}+{\displaystyle \frac{4}{G_{k+2}}}+{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_4}}.\hfill \end{array}$$

Proof. 

All identities can be verified with simple Gibonacci identities and with the recurrence relation (similarly to other identities from this section). In particular, it can be verified that

$$\begin{array}{cc}\hfill {\displaystyle \frac{6}{G_{k+1}}}+{\displaystyle \frac{3}{G_{k+2}}}-{\displaystyle \frac{3}{G_{k+3}}}& ={\displaystyle \frac{G_{k+4}{G}_{k+5}-G_k{G}_{k+1}}{G_{k+1}{G}_{k+2}{G}_{k+3}}},\hfill \\ \hfill {\displaystyle \frac{3}{G_{k+1}}}-{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{4}{G_{k+3}}}& ={\displaystyle \frac{G_{k+1}{G}_{k+5}-G_k{G}_{k+1}}{G_{k+1}{G}_{k+2}{G}_{k+3}}},\hfill \\ \hfill -{\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{2}{G_{k+2}}}+{\displaystyle \frac{3}{G_{k+3}}}& ={\displaystyle \frac{G_{k+1}{G}_{k+5}-G_k{G}_{k+2}}{G_{k+1}{G}_{k+2}{G}_{k+3}}},\hfill \\ \hfill {\displaystyle \frac{2}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}& ={\displaystyle \frac{G_{k+2}{G}_{k+5}-G_k{G}_{k+3}}{G_{k+1}{G}_{k+2}{G}_{k+3}}},\hfill \\ \hfill {\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{3}{G_{k+2}}}+{\displaystyle \frac{2}{G_{k+3}}}& ={\displaystyle \frac{G_{k+3}{G}_{k+5}-G_k{G}_{k+4}}{G_{k+1}{G}_{k+2}{G}_{k+3}}},\hfill \\ \hfill {\displaystyle \frac{3}{G_{k+1}}}+{\displaystyle \frac{4}{G_{k+2}}}+{\displaystyle \frac{1}{G_{k+3}}}& ={\displaystyle \frac{G_{k+4}{G}_{k+5}-G_k{G}_{k+5}}{G_{k+1}{G}_{k+2}{G}_{k+3}}}.\hfill \end{array}$$

Then, for example,

$${\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}·{\displaystyle \frac{G_{k+4}{G}_{k+5}-G_k{G}_{k+1}}{G_{k+1}{G}_{k+2}{G}_{k+3}}}={\displaystyle \frac{1}{G_k\cdots G_{k+3}}}-{\displaystyle \frac{1}{G_{k+2}\cdots G_{k+5}}}$$

from which the first identity follows. □

Remark 1.

We note that in Corollary 4 the second and the sixth generate all of the cases. This is because

$$\begin{array}{cc}\mathrm{span}\left\{\right(6,3,-3),(3,-1,-4),(-1,2,3),(2,1,-1),(1,3,2),(3,4,1\left)\right\}\hfill & \\ & \hfill =\mathrm{span}\left\{\right(3,-1,-4),(3,4,1\left)\right\}.\end{array}$$

Despite our best efforts, we were unable to find the exceptional identity with (rational) coefficients $A,B$ and C, that is, the identity which left-hand side is

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{A}{G_{k+1}}}+{\displaystyle \frac{B}{G_{k+2}}}+{\displaystyle \frac{C}{G_{k+3}}}\right),$$

such that

$$\mathrm{span}\{(3,-1,-4),(3,4,1),(A,B,C)\}={\mathbb{R}}^3.$$

This would imply the closed form for all identities with arbitrary coefficients $A,B$ and C.

4. Miscellaneous Identities

Theorem 7.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2}{G}_{k+3}}}={\displaystyle \frac{1}{2G_1{G}_2{G}_3}}.$$

(24)

Proof. 

Use

$${\displaystyle \frac{1}{G_k{G}_{k+2}{G}_{k+3}}}={\displaystyle \frac{1}{2}}{\displaystyle \frac{G_{k+3}-G_k}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}$$

from which (T1) can be applied. □

Theorem 8.

For all $m\ge 0$ we have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k\cdots G_{k+2m}{G}_{k+2m+2}\cdots G_{k+4m+2}}}={\displaystyle \frac{1}{L_{2m+1}{G}_1\cdots G_{4m+2}}}.$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2}}}={\displaystyle \frac{1}{G_1{G}_2}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+4}{G}_{k+5}{G}_{k+6}}}={\displaystyle \frac{1}{4G_1\cdots G_6}}.$$

Proof. 

We use (11). Take an odd m and make the replacement $m\to 2m+1$ in (11) to get

$$L_{2m+1}{G}_{k+2m+1}=G_{k+4m+2}-G_k.$$

Applying (T1) gives for a fixed $n>1$:

$$\begin{array}{cc}\hfill \sum _{k=1}^n{\displaystyle \frac{L_{2m+1}{G}_{k+2m+1}}{G_k\cdots G_{k+2m}{G}_{k+2m+1}{G}_{k+2m+2}\cdots G_{k+4m+2}}}& =\sum _{k=1}^n{\displaystyle \frac{G_{k+4m+2}-G_k}{G_k\cdots G_{k+4m+2}}}\hfill \\ \hfill & =\sum _{k=1}^n\left({\displaystyle \frac{1}{G_k\cdots G_{k+4m+1}}}-{\displaystyle \frac{1}{G_{k+1}\cdots G_{k+4m+2}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1\cdots G_{4m+2}}}-{\displaystyle \frac{1}{G_{n+1}\cdots G_{n+4m+2}}}.\hfill \end{array}$$

□

The next theorem showcases a simple example of an infinite family of identities with similar structure.

Theorem 9.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{k+2}\cdots G_{k+m}}}\left({\displaystyle \frac{F_m}{G_k}}+{\displaystyle \frac{F_{m-1}-1}{G_{k+1}}}\right)={\displaystyle \frac{1}{G_1\cdots G_m}}.$$

(25)

Proof. 

Use

$${\displaystyle \frac{F_m}{G_k}}+{\displaystyle \frac{F_{m-1}-1}{G_{k+1}}}={\displaystyle \frac{G_{k+1}{F}_m-G_k+F_{m-1}{G}_k}{G_k{G}_{k+1}}}={\displaystyle \frac{G_{k+m}-G_k}{G_k{G}_{k+1}}}$$

and apply (T1). □

The the next relation is an alternating version of (25) in case $m=4$.

Theorem 10.

We have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{G_{k+2}{G}_{k+3}{G}_{k+4}}}\left({\displaystyle \frac{3}{G_k}}+{\displaystyle \frac{1}{G_{k+1}}}\right)& ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{1}{G_1{G}_2}}+{\displaystyle \frac{1}{G_1{G}_4}}-{\displaystyle \frac{1}{G_3{G}_4}}-{\displaystyle \frac{1}{2G_2{G}_3}}\right)\hfill \\ \hfill & -{\displaystyle \frac{1}{2(G_1^2-G_0{G}_2)}}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}}}.\hfill \end{array}$$

Proof. 

Start with

$${\displaystyle \frac{3}{G_k}}+{\displaystyle \frac{1}{G_{k+1}}}={\displaystyle \frac{3G_{k+1}+G_k}{G_k{G}_{k+1}}}={\displaystyle \frac{G_{k+4}-G_k}{G_k{G}_{k+1}}}$$

and

$${\displaystyle \frac{{(-1)}^k}{G_{k+2}{G}_{k+3}{G}_{k+4}}}\left({\displaystyle \frac{3}{G_k}}+{\displaystyle \frac{1}{G_{k+1}}}\right)={\displaystyle \frac{{(-1)}^k}{G_k\cdots G_{k+3}}}+{\displaystyle \frac{{(-1)}^{k+1}}{G_{k+1}\cdots G_{k+4}}}.$$

We now apply the Cassini identity to have:

$$\begin{array}{cc}\hfill {\displaystyle \frac{{(-1)}^k}{G_k\cdots G_{k+3}}}+{\displaystyle \frac{{(-1)}^{k+1}}{G_{k+1}\cdots G_{k+4}}}& \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{G_{k+1}{G}_{k+3}-G_{k+2}^2}{G_k\cdots G_{k+3}}}+{\displaystyle \frac{G_{k+2}{G}_{k+4}-G_{k+3}^2}{G_{k+1}\cdots G_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{1}{G_k{G}_{k+2}}}+{\displaystyle \frac{G_{k+1}-G_{k+3}}{G_k{G}_{k+1}{G}_{k+3}}}+{\displaystyle \frac{1}{G_{k+1}{G}_{k+3}}}-{\displaystyle \frac{G_{k+1}+G_{k+2}}{G_{k+1}{G}_{k+2}{G}_{k+4}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{1}{G_k{G}_{k+2}}}+{\displaystyle \frac{1}{G_k{G}_{k+3}}}-{\displaystyle \frac{1}{G_k{G}_{k+1}}}+{\displaystyle \frac{1}{G_k{G}_{k+3}}}-{\displaystyle \frac{1}{G_{k+2}{G}_{k+4}}}-{\displaystyle \frac{1}{G_{k+1}{G}_{k+4}}}\right).\hfill \end{array}$$

From that we get, using (T1),

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{G_{k+2}{G}_{k+3}{G}_{k+4}}}\left({\displaystyle \frac{3}{G_k}}+{\displaystyle \frac{1}{G_{k+1}}}\right)& ={\displaystyle \frac{1}{G_1^2-G_0{G}_2}}\left({\displaystyle \frac{1}{G_1{G}_2}}+{\displaystyle \frac{1}{G_1{G}_4}}-{\displaystyle \frac{1}{G_3{G}_4}}-{\displaystyle \frac{1}{2G_2{G}_3}}\right)\hfill \\ \hfill & -{\displaystyle \frac{1}{2(G_1^2-G_0{G}_2)}}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}}},\hfill \end{array}$$

as

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2}}}={\displaystyle \frac{1}{G_1{G}_2}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+3}}}={\displaystyle \frac{1}{2}}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}}}-{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{G_2{G}_3}}.$$

□

One of the original Brousseau sums is [7, Identity (9)]:

$$\sum _{k=1}^{\infty }{\displaystyle \frac{F_{k+3}}{F_k{F}_{k+2}{F}_{k+4}{F}_{k+6}}}={\displaystyle \frac{17}{480}}.$$

(26)

This sum is a special case of the following sum with the structure similar to the previous sums from this section.

Theorem 11.

Let $p,q,m$ be odd positive integers. Then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{G_{km+pqm}}{G_{km}{G}_{km+2qm}\cdots G_{km+2pqm}}}={\displaystyle \frac{1}{L_{pqm}}}\sum _{j=1}^{2q}{\displaystyle \frac{1}{G_{jm}{G}_{jm+2qm}\cdots G_{jm+2(p-1)qm}}}.$$

Proof. 

Write (with the use of (11))

$$\begin{array}{cc}\hfill {\displaystyle \frac{L_{pqm}{G}_{km+pqm}}{G_{km}{G}_{km+2qm}\cdots G_{km+2pqm}}}& ={\displaystyle \frac{G_{km+2pqm}-G_{km}}{G_{km}{G}_{km+2qm}\cdots G_{km+2pqm}}},\hfill \end{array}$$

simplify and apply (T1). Note that the first term in the telescoping sum is

$${\displaystyle \frac{1}{G_m{G}_{m+2qm}\cdots G_{m+2(p-1)qm}}}-{\displaystyle \frac{1}{G_{m+2qm}{G}_{m+3qm}\cdots G_{m+2pqm}}},$$

which justifies the sum obtained in the final form. □

If we now set $m=q=1$ and $p=3$ in Theorem 11, we restore (26).

Theorem 12.

Let $m>0$ be odd. Then we have

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k-1}}{G_{mk}{G}_{m(k+3)}}}\left({\displaystyle \frac{G_{m(k+1)}}{G_{m(k+2)}}}+{\displaystyle \frac{G_{m(k+2)}}{G_{m(k+1)}}}\right)={\displaystyle \frac{{(-1)}^{n+1}}{G_{m(n+1)}{G}_{m(n+3)}}}+{\displaystyle \frac{1}{G_m{G}_{3m}}}$$

and in the limiting case we also have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k-1}}{G_{mk}{G}_{m(k+3)}}}\left({\displaystyle \frac{G_{m(k+1)}}{G_{m(k+2)}}}+{\displaystyle \frac{G_{m(k+2)}}{G_{m(k+1)}}}\right)={\displaystyle \frac{1}{G_m{G}_{3m}}}.$$

Proof. 

The identity follows from

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_{mk}{G}_{m(k+3)}}}\left({\displaystyle \frac{G_{m(k+1)}}{G_{m(k+2)}}}+{\displaystyle \frac{G_{m(k+2)}}{G_{m(k+1)}}}\right)& ={\displaystyle \frac{G_{m(k+1)}^2+G_{m(k+2)}^2}{G_{mk}{G}_{m(k+1)}{G}_{m(k+2)}{G}_{m(k+3)}}}\hfill \end{array}$$

and further noting that, with the aid of Identity (13),

$$\begin{array}{cc}\hfill G_{m(k+1)}^2+G_{m(k+2)}^2& ={(-1)}^{m+1}{G}_{m(k+1)}^2+G_{m(k+2)}(L_m{G}_{m(k+1)}+G_{mk})\hfill \\ \hfill & =G_{m(k+1)}(L_m{G}_{m(k+2)}+{(-1)}^{m+1}{G}_{m(k+1)})+G_{mk}{G}_{m(k+2)}\hfill \\ \hfill & =G_{m(k+1)}{G}_{m(k+3)}+G_{mk}{G}_{m(k+2)}.\hfill \end{array}$$

We can now apply (T2) with $f_k={\displaystyle \frac{1}{G_{mk}{G}_{m(k+2)}}}$. □

5. Series with Higher Powers

This part of our article is inspired by another problem proposal by Ohtsuka from 2024. In Problem H-938 [26] the readers are asked to prove the following relation:

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{F_k^2{F}_{k+1}^2}}=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{F_k^2{F}_{k+2}^2}}+{\displaystyle \frac{13-5\sqrt{5}}{2}}.$$

(27)

First, we derive a Gibonacci generalization of this statement. Second, we study similar series those closed forms involve squares of certain Gibonacci numbers.

Theorem 13.

We have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k^2{G}_{k+1}^2}}& =\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k^2{G}_{k+2}^2}}+{\displaystyle \frac{G_3^2-G_2^2-G_2}{G_1{G}_2^2{G}_3^2}}\hfill \\ \hfill & +{\displaystyle \frac{1}{G_1^2-G_0^2-G_0{G}_1}}\left({\displaystyle \frac{G_3+G_5}{G_1{G}_2{G}_3}}-{\displaystyle \frac{5}{G_1^2\alpha +G_0{G}_1}}\right).\hfill \end{array}$$

(28)

Proof. 

Let $Q_G$ denote the difference of the sums in question, i.e.,

$$Q_G=\sum _{k=1}^{\infty }\left({\displaystyle \frac{1}{G_k^2{G}_{k+1}^2}}-{\displaystyle \frac{1}{G_k^2{G}_{k+2}^2}}\right).$$

We have

$$\begin{array}{cc}\hfill Q_G& =\sum _{k=1}^{\infty }{\displaystyle \frac{G_{k+2}^2-G_{k+1}^2}{G_k^2{G}_{k+1}^2{G}_{k+2}^2}}\hfill \\ \hfill & =\sum _{k=1}^{\infty }{\displaystyle \frac{G_k{G}_{k+1}+G_k{G}_{k+2}}{G_k^2{G}_{k+1}^2{G}_{k+2}^2}}\hfill \\ \hfill & =\sum _{k=1}^{\infty }\left({\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}^2}}+{\displaystyle \frac{1}{G_k{G}_{k+1}^2{G}_{k+2}}}\right).\hfill \end{array}$$

Now, as

$${\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}^2}}={\displaystyle \frac{G_{k+3}}{G_k{G}_{k+1}{G}_{k+2}^2{G}_{k+3}}}={\displaystyle \frac{1}{G_k{G}_{k+2}^2{G}_{k+3}}}+{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}$$

and

$${\displaystyle \frac{1}{G_k{G}_{k+1}^2{G}_{k+2}}}={\displaystyle \frac{G_{k+1}-G_k}{G_{k-1}{G}_k{G}_{k+1}^2{G}_{k+2}}}={\displaystyle \frac{1}{G_{k-1}{G}_k{G}_{k+1}{G}_{k+2}}}-{\displaystyle \frac{1}{G_{k-1}{G}_{k+1}^2{G}_{k+2}}},$$

we can write

$$\begin{array}{cc}\hfill Q_G& ={\displaystyle \frac{1}{G_1^2{G}_2^2}}-{\displaystyle \frac{1}{G_1^2{G}_3^2}}+\sum _{k=2}^{\infty }\left({\displaystyle \frac{1}{G_k{G}_{k+2}^2{G}_{k+3}}}-{\displaystyle \frac{1}{G_{k-1}{G}_{k+1}^2{G}_{k+2}}}\right)\hfill \\ \hfill & +\sum _{k=2}^{\infty }\left({\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}+{\displaystyle \frac{1}{G_{k-1}{G}_k{G}_{k+1}{G}_{k+2}}}\right).\hfill \end{array}$$

We apply the telescoping principle (T1) and end up with

$$Q_G=={\displaystyle \frac{G_3^2-G_2^2}{G_1{G}_2^2{G}_3^2}}-{\displaystyle \frac{1}{G_1{G}_3^2{G}_4}}-{\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_4}}+2\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+3}}}.$$

The statement follows from (23) and some basic simplifications. □

When $G_n=F_n$ then we get (27) from (28). When $G_n=L_n$ then (28) yields

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{L_k^2{L}_{k+1}^2}}=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{L_k^2{L}_{k+2}^2}}+{\displaystyle \frac{25-9\sqrt{5}}{90}}.$$

Theorem 14.

We have for all $m,q>0$:

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}}}\left({\displaystyle \frac{1}{G_{km}}}+{\displaystyle \frac{{(-1)}^{m+1}}{G_{km+2qm}}}\right)={\displaystyle \frac{1}{L_{qm}}}\sum _{j=1}^q{\displaystyle \frac{1}{G_{jm}^2{G}_{jm+qm}^2}}.$$

(29)

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}}}\left({\displaystyle \frac{1}{G_k}}+{\displaystyle \frac{1}{G_{k+2}}}\right)={\displaystyle \frac{1}{G_1^2{G}_2^2}}$$

(30)

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{2k}{G}_{2(k+1)}{G}_{2(k+2)}}}\left({\displaystyle \frac{1}{G_{2k}}}-{\displaystyle \frac{1}{G_{2(k+2)}}}\right)={\displaystyle \frac{1}{3G_2^2{G}_4^2}}.$$

Proof. 

Notice that by (11):

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}}}\left({\displaystyle \frac{1}{G_{km}}}+{\displaystyle \frac{{(-1)}^{m+1}}{G_{km+2qm}}}\right)& ={\displaystyle \frac{(G_{km+2qm}+{(-1)}^{m+1}{G}_{km})G_{km+qm}}{G_{km}^2{G}_{km+qm}^2{G}_{km+2qm}^2}}\hfill \\ \hfill & ={\displaystyle \frac{(G_{km+2qm}+{(-1)}^{m+1}{G}_{km})(G_{km+2qm}+{(-1)}^m{G}_{km})}{L_{qm}{G}_{km}^2{G}_{km+qm}^2{G}_{km+2qm}^2}}\hfill \\ \hfill & ={\displaystyle \frac{G_{km+2qm}^2-G_{km}^2}{L_{qm}{G}_{km}^2{G}_{km+qm}^2{G}_{km+2qm}^2}}\hfill \\ \hfill & ={\displaystyle \frac{1}{L_{qm}}}\left({\displaystyle \frac{1}{G_{km}^2{G}_{km+qm}^2}}-{\displaystyle \frac{1}{G_{km+qm}^2{G}_{km+2qm}^2}}\right).\hfill \end{array}$$

Use (T1). □

Theorem 15.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k^2{G}_{k+1}\cdots G_{k+3}{G}_{k+4}^2}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+3}}}\right)={\displaystyle \frac{1}{3{(G_1\cdots G_4)}^2}}.$$

(31)

Proof. 

Use $G_{k+4}+G_k=3G_{k+2}$ and $G_{k+4}-G_k=G_{k+1}+G_{k+3}$ to obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k^2{G}_{k+1}\cdots G_{k+3}{G}_{k+4}^2}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{G_{k+1}{G}_{k+2}+G_{k+2}{G}_{k+3}}{{(G_k\cdots G_{k+4})}^2}}\hfill \\ \hfill & ={\displaystyle \frac{G_{k+4}^2-G_k^2}{3{(G_k\cdots G_{k+4})}^2}}.\hfill \end{array}$$

□

A more general result, similar to (31) can also be obtained.

Theorem 16.

For odd $m>0$ we have

$$\begin{array}{cc}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k^2\cdots G_{k+m-2}^2{G}_{k+m-1}{G}_{k+m}{G}_{k+m+1}{G}_{k+m+2}^2\cdots G_{k+2m}^2}}\left({\displaystyle \frac{1}{G_{k+m-1}}}+{\displaystyle \frac{1}{G_{k+m+1}}}\right)\hfill & \\ & \hfill ={\displaystyle \frac{1}{F_{2m}{(G_1\cdots G_{2m})}^2}},\end{array}$$

(32)

where the case $m=1$ is understood as

$$\sum _{k=1}^{+\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}}}\left({\displaystyle \frac{1}{G_k}}+{\displaystyle \frac{1}{G_{k+2}}}\right)={\displaystyle \frac{1}{G_1^1{G}_2^2}}.$$

(33)

Proof. 

Rewrite

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k^2\cdots G_{k+m-2}^2{G}_{k+m-1}{G}_{k+m}{G}_{k+m+1}{G}_{k+m+2}^2\cdots G_{k+2m}^2}}\left({\displaystyle \frac{1}{G_{k+m-1}}}+{\displaystyle \frac{1}{G_{k+m+1}}}\right)& \\ \hfill & ={\displaystyle \frac{G_{k+m}(G_{k+m-1}+G_{k+m+1})}{{(G_k\cdots G_{k+2m})}^2}}\hfill \\ \hfill & ={\displaystyle \frac{1}{F_m}}{\displaystyle \frac{1}{L_m}}{\displaystyle \frac{(G_{k+2m}-G_k)(G_{k+2m}+G_k)}{{(G_k\cdots G_{k+2m})}^2}}\hfill \\ \hfill & ={\displaystyle \frac{G_{k+2m}^2-G_k^2}{F_{2m}{(G_k\cdots G_{k+2m})}^2}}.\hfill \end{array}$$

Note that we have used $G_{k+2m}-G_k=L_m{G}_{k+m}$ (cf. Identity (11)) as well as $G_{k+2m}+G_k=F_m(G_{k+m-1}+G_{k+m+1})$ for odd m (cf. Identity (12)). □

When $G_n=F_n$ or $G_n=L_n$, then we get from (33) the sums

$$\sum _{k=1}^{+\infty }{\displaystyle \frac{1}{F_k{F}_{k+1}{F}_{k+2}}}\left({\displaystyle \frac{1}{F_k}}+{\displaystyle \frac{1}{F_{k+2}}}\right)=1,\sum _{k=1}^{+\infty }{\displaystyle \frac{1}{L_k{L}_{k+1}{L}_{k+2}}}\left({\displaystyle \frac{1}{L_k}}+{\displaystyle \frac{1}{L_{k+2}}}\right)={\displaystyle \frac{1}{4}}$$

and we get from (32) the sum

$$\sum _{k=1}^{+\infty }{\displaystyle \frac{1}{F_k^2{F}_{k+1}^2{F}_{k+2}{F}_{k+3}{F}_{k+4}{F}_{k+5}^2{F}_{k+6}^2}}\left({\displaystyle \frac{1}{F_{k+2}}}+{\displaystyle \frac{1}{F_{k+4}}}\right)={\displaystyle \frac{1}{460800}}.$$

Theorem 17.

We have

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k\cdots G_{k+3}}}\left({\displaystyle \frac{G_{k+3}}{G_{k+2}}}+{\displaystyle \frac{G_k}{G_{k+1}}}\right)={\displaystyle \frac{1}{G_1{G}_2^2{G}_3}}.$$

Proof. 

We calculate

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k\cdots G_{k+3}}}\left({\displaystyle \frac{G_{k+3}}{G_{k+2}}}+{\displaystyle \frac{G_k}{G_{k+1}}}\right)& ={\displaystyle \frac{G_{k+1}{G}_{k+3}+G_k{G}_{k+2}}{G_k{G}_{k+1}^2{G}_{k+2}^2{G}_{k+3}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}^2}}+{\displaystyle \frac{1}{G_{k+1}^2{G}_{k+2}{G}_{k+3}}}.\hfill \end{array}$$

Set

$$\begin{array}{ccc}\hfill A\left(k\right)={\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}^2}},& \hfill & \hfill B\left(k\right)={\displaystyle \frac{1}{G_k^2{G}_{k+1}{G}_{k+2}}}.\end{array}$$

Then,

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_k\cdots G_{k+3}}}\left({\displaystyle \frac{G_{k+3}}{G_{k+2}}}+{\displaystyle \frac{G_k}{G_{k+1}}}\right)& =A\left(k\right)+B(k+1)\hfill \\ \hfill & =\left(A\left(k\right)+B\left(k\right)\right)+\left(B(k+1)-B\left(k\right)\right).\hfill \end{array}$$

Thus, using (30) (where the series involving $A\left(k\right)+B\left(k\right)$ is computed) we obtain

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k\cdots G_{k+3}}}\left({\displaystyle \frac{G_{k+3}}{G_{k+2}}}+{\displaystyle \frac{G_k}{G_{k+1}}}\right)& =\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}}}\left({\displaystyle \frac{1}{G_k}}+{\displaystyle \frac{1}{G_{k+2}}}\right)+\sum _{k=1}^{\infty }\left(B(k+1)-B\left(k\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{G_1^2{G}_2^2}}-B\left(1\right)={\displaystyle \frac{1}{G_1^2{G}_2^2}}-{\displaystyle \frac{1}{G_1^2{G}_2{G}_3}}\hfill \\ & ={\displaystyle \frac{1}{G_1{G}_2^2{G}_3}}.\hfill \end{array}$$

The proof is completed. □

This is a generalization of the previous result.

Theorem 18.

We have for all odd $m>0$

$$\begin{array}{cc}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}{G}_{km+3qm}}}\left({\displaystyle \frac{G_{km+3qm}}{G_{km+2qm}}}+{\displaystyle \frac{{(-1)}^{m+1}{G}_{km}}{G_{km+qm}}}\right)\hfill & \\ & \hfill ={\displaystyle \frac{1}{L_{qm}}}\sum _{k=1}^q{\displaystyle \frac{1}{G_{jm}{G}_{jm+qm}^2{G}_{jm+2qm}}}\end{array}$$

(34)

and for all even $m>0$

$$\begin{array}{cc}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}{G}_{km+3qm}}}\left({\displaystyle \frac{G_{km+3qm}}{G_{km+2qm}}}+{\displaystyle \frac{{(-1)}^{m+1}{G}_{km}}{G_{km+qm}}}\right)\hfill & \\ & \hfill ={\displaystyle \frac{1}{L_{qm}}}\sum _{j=1}^q{\displaystyle \frac{G_{jm+2qm}+L_{qm}{G}_{jm+qm}}{G_{jm}^2{G}_{jm+qm}^2{G}_{jm+2qm}}}.\end{array}$$

(35)

Proof. 

Calculate

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}{G}_{km+3qm}}}\left({\displaystyle \frac{G_{km+3qm}}{G_{km+2qm}}}+{\displaystyle \frac{{(-1)}^{m+1}{G}_{km}}{G_{km+qm}}}\right)& \\ \hfill & ={\displaystyle \frac{G_{km+qm}{G}_{km+3qm}+{(-1)}^{m+1}{G}_{km+2qm}{G}_{km}}{G_{km}{G}_{km+qm}^2{G}_{km+2qm}^2{G}_{km+3qm}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}^2}}+{\displaystyle \frac{{(-1)}^{m+1}}{G_{km+qm}^2{G}_{km+2qm}{G}_{km+3qm}}}.\hfill \end{array}$$

Let

$$A\left(k\right)={\displaystyle \frac{1}{G_{km}{G}_{km+qm}{G}_{km+2qm}^2}},B\left(k\right)={\displaystyle \frac{{(-1)}^{m+1}}{G_{km}^2{G}_{km+qm}{G}_{km+2qm}}}.$$

Then each term of the desired sum equals

$$A\left(k\right)+B(k+q)=\left(A\right(k)+B(k\left)\right)+\left(B\right(k+q)-B(k\left)\right).$$

We now use (29) to compute ${\sum }_{k=1}^{\infty }(A\left(k\right)+B\left(k\right))$ and we notice that

$$\sum _{k=1}^{\infty }(B(k+q)-B\left(k\right))=-\sum _{j=1}^{q}B\left(j\right),$$

thus the desired sum equal, with the aid of (11),

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{L_{qm}}}\sum _{k=1}^q{\displaystyle \frac{1}{G_{jm}^2{G}_{jm+qm}^2}}-\sum _{j=1}^q{\displaystyle \frac{{(-1)}^{m+1}}{G_{jm}^2{G}_{jm+qm}{G}_{jm+2qm}}}& ={\displaystyle \frac{1}{L_{qm}}}\sum _{j=1}^q{\displaystyle \frac{G_{jm+2qm}+{(-1)}^m{L}_{qm}{G}_{jm+qm}}{G_{jm}^2{G}_{jm+qm}^2{G}_{jm+2qm}}}.\hfill \end{array}$$

If m is odd, then the numerator simplifies to $G_{jm}$ via (11), which concludes (34). Otherwise, for even m we have (35). □

6. A New Look at Brousseau Sums

Particular cases of the sums discussed in this section appeared in [4, Corollary 1], but the final form is different than ours.

Theorem 19.

If m and q are positive integers, then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{mk}}{G_{mk}{G}_{mk+mq}}}={\displaystyle \frac{1}{{\epsilon }_G{F}_{mq}}}\left(q\alpha -\sum _{k=1}^q{\displaystyle \frac{G_{mk+1}}{G_{mk}}}\right).$$

(36)

Proof. 

It is known that [2]

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{mk}}{G_{mk}{G}_{mk+mq}}}={\displaystyle \frac{F_{mn}}{F_{mq}}}\sum _{k=1}^q{\displaystyle \frac{{(-1)}^{mk}}{G_{mk}{G}_{mk+mn}}},m,q\in {\mathbb{Z}}^{+}.$$

(37)

Therefore,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{mk}}{G_{mk}{G}_{mk+mq}}}={\displaystyle \frac{1}{F_{mq}}}\sum _{k=1}^q{\displaystyle \frac{{(-1)}^{mk}}{G_{mk}}}\underset{n\to \infty }{lim}{\displaystyle \frac{F_{mn}}{G_{mk+mn}}},$$

from which (36) follows on account of (16). □

Theorem 20.

If m and q are positive integers, then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_{mk}{G}_{mk+2mq}}}={\displaystyle \frac{1}{{\epsilon }_G{F}_{2mq}}}\sum _{k=1}^{2q}{(-1)}^{mk-1}{\displaystyle \frac{G_{mk+1}}{G_{mk}}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k(m-1)}}{G_{mk}{G}_{mk+2mq}}}={\displaystyle \frac{1}{{\epsilon }_G{F}_{2mq}}}\sum _{k=1}^{2q}{(-1)}^{k-1}{\displaystyle \frac{G_{mk+1}}{G_{mk}}}.$$

Proof. 

Similar to the proof of Theorem 19. We use [2]:

$$\sum _{k=1}^{2n}{\displaystyle \frac{{(\pm 1)}^{k(m-1)}}{G_{mk}{G}_{mk+2mq}}}={\displaystyle \frac{F_{2mn}}{F_{2mq}}}\sum _{k=1}^{2q}{\displaystyle \frac{{(\pm 1)}^{k(m-1)}}{G_{mk}{G}_{mk+2mn}}}.$$

(38)

□

Theorem 21.

If m and n are non-negative integers, then

$$\sum _{k=1}^{2n}{\displaystyle \frac{1}{G_k{G}_{k+2m}}}={\displaystyle \frac{F_{2n}}{F_{2m}}}\sum _{k=1}^{2m}{\displaystyle \frac{1}{G_k{G}_{k+2n}}}$$

(39)

and

$$\begin{array}{cc}\hfill \sum _{k=1}^{2n}{\displaystyle \frac{1}{G_k{G}_{k+2m+1}}}& ={\displaystyle \frac{1}{F_{2m+1}}}\sum _{k=1}^{2n}{\displaystyle \frac{1}{G_k{G}_{k+1}}}-{\displaystyle \frac{F_{2n}}{F_{2m+1}}}\sum _{k=1}^{2m}{\displaystyle \frac{1}{G_k{G}_{k+2n}}}-{\displaystyle \frac{F_{2m}}{F_{2m+1}{G}_{2n+1}{G}_{2n+2m+1}}}\hfill \\ \hfill & +{\displaystyle \frac{F_{2m}}{F_{2m+1}{G}_1{G}_{2m+1}}}.\hfill \end{array}$$

(40)

Proof. 

Set $m=1$ in (38) and write m for q to obtain (39).

Write $2m+1$ for r in (10):

$${\displaystyle \frac{1}{G_k}}-{\displaystyle \frac{F_{2m}}{G_{k+2m+1}}}={\displaystyle \frac{F_{2m+1}{G}_{k+1}}{G_k{G}_{k+2m+1}}},$$

and arrange as

$${\displaystyle \frac{F_{2m+1}}{G_k{G}_{k+2m+1}}}={\displaystyle \frac{1}{G_k{G}_{k+1}}}-{\displaystyle \frac{F_{2m}}{G_{k+1}{G}_{k+2m+1}}}.$$

Now sum, to obtain, after a shift of summation index,

$$\begin{array}{cc}\hfill \sum _{k=1}^n{\displaystyle \frac{1}{G_k{G}_{k+2m+1}}}& ={\displaystyle \frac{1}{F_{2m+1}}}\sum _{k=1}^n{\displaystyle \frac{1}{G_k{G}_{k+1}}}-{\displaystyle \frac{F_{2m}}{F_{2m+1}}}\sum _{k=1}^n{\displaystyle \frac{1}{G_k{G}_{k+2m}}}-{\displaystyle \frac{F_{2m}}{F_{2m+1}{G}_{n+1}{G}_{n+2m+1}}}\hfill \\ \hfill & +{\displaystyle \frac{F_{2m}}{F_{2m+1}{G}_1{G}_{2m+1}}}.\hfill \end{array}$$

Write $2n$ for n and use (39) to re-write the second term on the right hand side; this gives (40). □

As a corollary, we obtain the infinite series version of the above theorem.

Corollary 5.

If m is a non-negative integer, then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2m}}}={\displaystyle \frac{1}{{\epsilon }_G{F}_{2m}}}\sum _{k=1}^{2m}{(-1)}^{k-1}{\displaystyle \frac{G_{k+1}}{G_k}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2m+1}}}={\displaystyle \frac{1}{F_{2m+1}}}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}}}-{\displaystyle \frac{1}{{\epsilon }_G{F}_{2m+1}}}\sum _{k=1}^{2m}{(-1)}^{k-1}{\displaystyle \frac{G_{k+1}}{G_k}}+{\displaystyle \frac{F_{2m}}{F_{2m+1}{G}_1{G}_{2m+1}}}.$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+2}}}={\displaystyle \frac{1}{G_1{G}_2}}$$

and

$$\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+3}}}={\displaystyle \frac{1}{2}}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{G_k{G}_{k+1}}}-{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{G_2{G}_3}}.$$

Theorem 22.

If p is a positive integer and m and r are non-negative integers such that m is greater than r, then,

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{pk}}{G_{pk+pr}{G}_{pk+pm}}}={(-1)}^{pr}{\displaystyle \frac{F_{pn}}{F_{pm-pr}}}\sum _{k=r+1}^m{\displaystyle \frac{{(-1)}^{pk}}{G_{pk}{G}_{pk+pn}}}.$$

In particular,

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^k}{G_{k+r}{G}_{k+m}}}={(-1)}^r{\displaystyle \frac{F_n}{F_{m-r}}}\sum _{k=r+1}^m{\displaystyle \frac{{(-1)}^k}{G_k{G}_{k+n}}}.$$

Proof. 

Making the replacements $k\to pk$, $m\to pm$ and $r\to pr$ in (14), we have

$${\displaystyle \frac{F_{pm-pr}}{G_{pk+pr}{G}_{pk+pm}}}={(-1)}^{pr}{\displaystyle \frac{F_{pm}}{G_{pk}{G}_{pk+pm}}}-{(-1)}^{pr}{\displaystyle \frac{F_{pr}}{G_{pk}{G}_{pk+pr}}}.$$

Now multiply each term by ${(-1)}^{pk}$ and sum from $k=1$ to $k=n$, making use of (37). □

Corollary 6.

If p is a positive integer and m and r are non-negative integers such that m is greater than r, then,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{pk}}{G_{pk+pr}{G}_{pk+pm}}}={\displaystyle \frac{{(-1)}^{pr}}{{\epsilon }_G{F}_{pm-pr}}}\left((m-r)\alpha -\sum _{k=r+1}^m{\displaystyle \frac{G_{pk+1}}{G_{pk}}}\right).$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{G_{k+r}{G}_{k+m}}}={\displaystyle \frac{{(-1)}^r}{{\epsilon }_G{F}_{m-r}}}\left((m-r)\alpha -\sum _{k=r+1}^m{\displaystyle \frac{G_{k+1}}{G_k}}\right).$$

Remark 2.

It should be noted that our results cannot be compared tothe sums obtained by Farhi in [9]. Our results concern arbitrary Gibonacci sequences, whereas his results are related to Lucas sequences of the first and second kinds.

7. Sums with Products of Gibonacci Numbers in the Denominator

The identity (T3) can be generalized to the following, which can be easily proven, for example, by direct computation.

Lemma 2.

If $g\left(k\right)={\left(g_k\right)}_{k\in {\mathbb{Z}}^{+}}$ is a sequence of complex numbers, n and r are integers and c and d are any numbers, then

$$\begin{array}{cc}\hfill & \sum _{k=1}^n{d}^{n-k}{c}^{k-1}{g}_{k+1}{g}_{k+2}\cdots g_{k+r-1}(c{g}_{k+r}-d{g}_k)\hfill \\ \hfill & =c^n{g}_{n+1}{g}_{n+2}\cdots g_{n+r}-d^n{g}_1{g}_2\cdots g_r.\hfill \end{array}$$

(41)

The Lemma turns out to be extremely useful in finding a variety of identities of our interest. The remainder of this section showcases many of its applications

Theorem 23.

If r is a positive integer, then

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k(r+1)}}{G_k{G}_{k+1}\cdots G_{k+r-1}{G}_{k+r+1}\cdots G_{k+2r}}}={\displaystyle \frac{{(-1)}^{rn+n+r}}{L_r{G}_{n+1}{G}_{n+2}\cdots G_{n+2r}}}-{\displaystyle \frac{{(-1)}^r}{L_r{G}_1{G}_2\cdots G_{2r}}}.$$

(42)

In particular,

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^k}{G_k{G}_{k+1}{G}_{k+3}{G}_{k+4}}}={\displaystyle \frac{{(-1)}^n}{3G_{n+1}{G}_{n+2}{G}_{n+3}{G}_{n+4}}}-{\displaystyle \frac{1}{3G_1{G}_2{G}_3{G}_4}}$$

and

$$\sum _{k=1}^n{\displaystyle \frac{1}{G_k{G}_{k+1}{G}_{k+2}{G}_{k+4}{G}_{k+5}{G}_{k+6}}}={\displaystyle \frac{1}{4G_1{G}_2{G}_3{G}_4{G}_5{G}_6}}-{\displaystyle \frac{1}{4G_{n+1}{G}_{n+2}{G}_{n+3}{G}_{n+4}{G}_{k+5}{G}_{k+6}}}.$$

Proof. 

With (9) in mind, identity (42) follows upon use of $g\left(k\right)=1/G_k$, $c={(-1)}^r$ and $d=-1$ in (41). □

Corollary 7.

If r is a positive integer, then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k(r+1)}}{G_k{G}_{k+1}\cdots G_{k+r-1}{G}_{k+r+1}\cdots G_{k+2r}}}={\displaystyle \frac{{(-1)}^{r-1}}{L_r{G}_1{G}_2\cdots G_{2r}}}.$$

Theorem 24.

If m and r are positive integers such that $r>m$, then

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k(m+1)}{\left(F_{r-m}/F_m\right)}^k}{G_k{G}_{k+1}\cdots G_{k+m-1}{G}_{k+m+1}\cdots G_{k+r-1}{G}_{k+r}}}={\displaystyle \frac{{(-1)}^{nm+n+m}{\left(F_{r-m}/F_m\right)}^n{F}_{r-m}}{F_r{G}_{n+1}{G}_{n+2}\cdots G_{n+r}}}-{\displaystyle \frac{{(-1)}^m{F}_{r-m}}{F_r{G}_1{G}_2\cdots G_r}}.$$

(43)

In particular, at $m=2$, $r=3$, we find

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k-1}}{G_k{G}_{k+1}{G}_{k+3}}}={\displaystyle \frac{{(-1)}^{n-1}}{2G_{n+1}{G}_{n+2}{G}_{n+3}}}+{\displaystyle \frac{1}{2G_1{G}_2{G}_3}}$$

and at $m=2$, $r=4$ we get

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k-1}}{G_k{G}_{k+1}{G}_{k+3}{G}_{k+4}}}={\displaystyle \frac{{(-1)}^{n-1}}{3G_{n+1}{G}_{n+2}{G}_{n+3}{G}_{n+4}}}+{\displaystyle \frac{1}{3G_1{G}_2{G}_3{G}_4}}.$$

Proof. 

Choosing $c=F_{m-r}$ and $d={(-1)}^r{F}_m$ in (41) together with (14) gives (43). □

Remark 3.

Identity (24) is reproduced as the limiting case of choosing $m=1$, $r=3$ in (43).

Corollary 8.

If m and r are positive integers such that $r>m$, then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k(m+1)}{\left(F_{r-m}/F_m\right)}^k}{G_k{G}_{k+1}\cdots G_{k+m-1}{G}_{k+m+1}\cdots G_{k+r-1}{G}_{k+r}}}={\displaystyle \frac{{(-1)}^{m+1}{F}_{r-m}}{F_r{G}_1{G}_2\cdots G_r}}.$$

Setting $r=m+1$ in (43) leads to the next result.

Corollary 9.

If m is a positive integer, then

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k(m+1)}}{F_m^k{G}_k{G}_{k+1}\cdots G_{k+m-1}{G}_{k+m+1}}}={\displaystyle \frac{{(-1)}^{nm+n+m}}{F_m^n{F}_{m+1}{G}_{n+1}{G}_{n+2}\cdots G_{n+m+1}}}-{\displaystyle \frac{{(-1)}^m}{F_{m+1}{G}_1{G}_2\cdots G_{m+1}}},$$

with the limiting value

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k(m+1)}}{F_m^k{G}_k{G}_{k+1}\cdots G_{k+m-1}{G}_{k+m+1}}}={\displaystyle \frac{{(-1)}^{m+1}}{F_{m+1}{G}_1{G}_2\cdots G_{m+1}}}.$$

Theorem 25.

If $r>1$ is a positive integer, then

$$\sum _{k=1}^n{\displaystyle \frac{F_{r-1}^{k-1}}{G_k{G}_{k+2}\cdots G_{k+r}}}={\displaystyle \frac{1}{F_r{G}_1{G}_2\cdots G_r}}-{\displaystyle \frac{F_{r-1}^n}{F_r{G}_{n+1}{G}_{n+2}\cdots G_{n+r}}},$$

with the limiting value

$$\sum _{k=1}^{\infty }{\displaystyle \frac{F_{r-1}^{k-1}}{G_k{G}_{k+2}\cdots G_{k+r}}}={\displaystyle \frac{1}{F_r{G}_1{G}_2\cdots G_r}}.$$

Proof. 

Use $c=F_{r-1}$ and $d=1$ in (41) in conjunction with (10). □

Theorem 26.

If n is a non-negative integer, $r>1$ is a positive integer and $G_0\ne 0$, then

$$\sum _{k=1}^n{(-1)}^{k-1}{\displaystyle \frac{F_{k+1}\cdots F_{k+r-1}}{G_k{G}_{k+1}\cdots G_{k+r}}}={\displaystyle \frac{1}{G_0{F}_r}}\left({\displaystyle \frac{F_1{F}_2\cdots F_r}{G_1{G}_2\cdots G_r}}-{\displaystyle \frac{F_{n+1}{F}_{n+2}\cdots F_{n+r}}{G_{n+1}{G}_{n+2}\cdots G_{n+r}}}\right).$$

In particular,

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k-1}}{F_{k+r}{F}_{k+r+1}}}={\displaystyle \frac{1}{F_r}}\left({\displaystyle \frac{1}{F_{r+1}}}-{\displaystyle \frac{F_{n+1}}{F_{n+r+1}}}\right)$$

(44)

and

$$\sum _{k=1}^n{(-1)}^{k-1}{\displaystyle \frac{F_{k+1}}{G_k{G}_{k+1}{G}_{k+2}}}={\displaystyle \frac{1}{G_0}}\left({\displaystyle \frac{1}{G_1{G}_2}}-{\displaystyle \frac{F_{n+1}{F}_{n+2}}{G_{n+1}{G}_{n+2}}}\right).$$

Proof. 

Set $c=1=d$ and $g\left(k\right)=F_k/G_k$ in (41) and use the following identity [29, Identity (21)]:

$${\displaystyle \frac{F_{k+r}}{G_{k+r}}}-{\displaystyle \frac{F_k}{G_k}}={(-1)}^k{\displaystyle \frac{G_0{F}_r}{G_k{G}_{k+r}}}.$$

□

Corollary 10.

If $r>1$ is a positive integer and $G_0\ne 0$, then

$$\sum _{k=1}^{\infty }{(-1)}^{k-1}{\displaystyle \frac{F_{k+1}\cdots F_{k+r-1}}{G_k{G}_{k+1}\cdots G_{k+r}}}={\displaystyle \frac{1}{G_0{F}_r}}\left({\displaystyle \frac{F_1{F}_2\cdots F_r}{G_1{G}_2\cdots G_r}}-{\displaystyle \frac{1}{{(G_1-G_0\beta )}^r}}\right).$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k-1}}{F_{k+r}{F}_{k+r+1}}}={\displaystyle \frac{1}{F_r}}\left({\displaystyle \frac{1}{F_{r+1}}}-{\displaystyle \frac{1}{{\alpha }^r}}\right)$$

(45)

and

$$\sum _{k=1}^{\infty }{(-1)}^{k-1}{\displaystyle \frac{F_{k+1}}{G_k{G}_{k+1}{G}_{k+2}}}={\displaystyle \frac{1}{G_0}}\left({\displaystyle \frac{1}{G_1{G}_2}}-{\displaystyle \frac{1}{{(G_1-G_0\beta )}^2}}\right).$$

Remark 4.

We note that (44) and (45) are not new. These results are special cases of another generalization, namely,

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{m(k+1)}}{F_{km+r}{F}_{km+m+r}}}={\displaystyle \frac{F_{mn}}{F_m{F}_{m+r}{F}_{mn+m+r}}},m\ge 1,r\ge 0,$$

which can be found in [13] or [19]. When $m=1$ then

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k+1}}{F_{k+r}{F}_{k+1+r}}}={\displaystyle \frac{F_n}{F_{r+1}{F}_{n+1+r}}}.$$

The equivalence of (44) and the last identity is readily seen by noting that

$$F_{n+1+r}-F_{r+1}{F}_{n+1}=F_r{F}_n.$$

Theorem 27.

If n is a non-negative integer and $r>1$ is a positive integer, then

$$\sum _{k=1}^n{(-1)}^{k-1}{\displaystyle \frac{G_{k+1}{G}_{k+2}\cdots G_{k+r-1}}{G_{k+r}{G}_{k+r+1}\cdots G_{k+2r}}}={\displaystyle \frac{1}{{\epsilon }_G{F}_r^2}}\left({\displaystyle \frac{G_{n+1}{G}_{n+2}\cdots G_{n+r}}{G_{n+r+1}{G}_{n+r+2}\cdots G_{n+2r}}}-{\displaystyle \frac{G_1{G}_2\cdots G_r}{G_{r+1}{G}_{r+2}\cdots G_{2r}}}\right).$$

In particular,

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k-1}{G}_{k+1}}{G_{k+2}{G}_{k+3}{G}_{k+4}}}={\displaystyle \frac{1}{{\epsilon }_G}}\left({\displaystyle \frac{G_{n+1}{G}_{n+2}}{G_{n+3}{G}_{n+4}}}-{\displaystyle \frac{G_1{G}_2}{G_3{G}_4}}\right).$$

Proof. 

Use $g\left(k\right)=G_k/G_{k+r}$ and $c=1=d$ in (41), noting that (variation on Tagiuri’s identity)

$${\displaystyle \frac{G_{k+r}}{G_{k+2r}}}-{\displaystyle \frac{G_k}{G_{k+r}}}={(-1)}^{k-1}{\displaystyle \frac{F_r^2{\epsilon }_G}{G_{k+r}{G}_{k+2r}}}.$$

□

Corollary 11.

If $r>1$ is a positive integer, then

$$\sum _{k=1}^{\infty }{(-1)}^{k-1}{\displaystyle \frac{G_{k+1}{G}_{k+2}\cdots G_{k+r-1}}{G_{k+r}{G}_{k+r+1}\cdots G_{k+2r}}}={\displaystyle \frac{1}{{\epsilon }_G{F}_r^2}}\left({(-1)}^r\left(\beta F_{r^2}+F_{r^2-1}\right)-{\displaystyle \frac{G_1{G}_2\cdots G_r}{G_{r+1}{G}_{r+2}\cdots G_{2r}}}\right).$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k-1}{G}_{k+1}}{G_{k+2}{G}_{k+3}{G}_{k+4}}}={\displaystyle \frac{1}{{\epsilon }_G}}\left(3\beta +2-{\displaystyle \frac{G_1{G}_2}{G_3{G}_4}}\right).$$

Theorem 28.

If n is a non-negative integer and $r\ge 2$ is an even integer, then

$$\sum _{k=1}^n{\displaystyle \frac{G_{2k+r}}{F_k{G}_k{F}_{k+1}{G}_{k+1}\cdots F_{k+r}{G}_{k+r}}}={\displaystyle \frac{1}{F_r}}\left({\displaystyle \frac{1}{F_1{G}_1\cdots F_r{G}_r}}-{\displaystyle \frac{1}{F_{n+1}{G}_{n+1}\cdots F_{n+r}{G}_{n+r}}}\right).$$

In particular,

$$\sum _{k=1}^n{\displaystyle \frac{G_{2k+2}}{F_k{G}_k{F}_{k+1}{G}_{k+1}{F}_{k+2}{G}_{k+2}}}={\displaystyle \frac{1}{G_1{G}_2}}-{\displaystyle \frac{1}{F_{n+1}{G}_{n+1}{F}_{n+2}{G}_{n+2}}}.$$

Proof. 

Use $g\left(k\right)=1/\left(F_k{G}_k\right)$ and $c=1=d$ in (41), noting that

$${\displaystyle \frac{1}{F_n{G}_n}}-{\displaystyle \frac{1}{F_{n+r}{G}_{n+r}}}={\displaystyle \frac{F_r{G}_{2n+r}}{F_n{G}_n{F}_{n+r}{G}_{n+r}}},$$

which itself is a consequence of

$$F_r{G}_{2n+r}=F_{n+r}{G}_{n+r}-F_n{G}_n,r\mathrm{even}.$$

□

Corollary 12.

If $r\ge 2$ is an even integer, then

$$\sum _{k=1}^{\infty }{\displaystyle \frac{G_{2k+r}}{F_k{G}_k{F}_{k+1}{G}_{k+1}\cdots F_{k+r}{G}_{k+r}}}={\displaystyle \frac{1}{F_r}}{\displaystyle \frac{1}{F_1{G}_1\cdots F_r{G}_r}}.$$

In particular,

$$\sum _{k=1}^{\infty }{\displaystyle \frac{G_{2k+2}}{F_k{G}_k{F}_{k+1}{G}_{k+1}{F}_{k+2}{G}_{k+2}}}={\displaystyle \frac{1}{G_1{G}_2}}.$$

8. Conclusions

It should be clear that the possibilities do not end here and we (or the reader) could derive many more identities based on the telescoping principle. We focused mainly on (T1), (T3) and (41) and these simple rules enabled us to discover a multitude of identities with Gibonacci numbers in the denominator and, in some cases, in the numerator, including the alternating versions of some identities.

In the Introduction we announced that we would not give that much attention to (T2). However, using the ideas introduced in various sections of this article, it possible to utilize that telescoping identity. For example, we can present the following variation of Corollary 4.

Corollary 13.

We have the following identities:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k+1}}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{3}{G_{k+1}}}-{\displaystyle \frac{3}{G_{k+2}}}\right)& ={\displaystyle \frac{1}{G_2{G}_3{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k+1}}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{1}{G_{k+1}}}+{\displaystyle \frac{2}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_3{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k+1}}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{4}{G_{k+1}}}-{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{1}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_4{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k+1}}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{5}{G_{k+1}}}+{\displaystyle \frac{1}{G_{k+2}}}-{\displaystyle \frac{2}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_5}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k+1}}{G_k{G}_{k+4}{G}_{k+5}}}\left({\displaystyle \frac{9}{G_{k+1}}}-{\displaystyle \frac{3}{G_{k+3}}}\right)& ={\displaystyle \frac{1}{G_1{G}_2{G}_3{G}_4}}.\hfill \end{array}$$

Proof. 

Use the following set of identities in order:

$$\begin{array}{cc}\hfill 3G_{k+2}{G}_{k+3}-3G_{k+1}{G}_{k+3}& =G_{k+5}{G}_k+G_{k+1}{G}_k,\hfill \\ \hfill G_{k+2}{G}_{k+3}+2G_{k+1}{G}_{k+3}-G_{k+1}{G}_{k+2}& =G_{k+5}{G}_{k+1}+G_{k+2}{G}_k,\hfill \\ \hfill 4G_{k+2}{G}_{k+3}-G_{k+1}{G}_{k+3}-G_{k+1}{G}_{k+2}& =G_{k+5}{G}_{k+2}+G_{k+3}{G}_k,\hfill \\ \hfill 5G_{k+2}{G}_{k+3}+G_{k+1}{G}_{k+3}-2G_{k+1}{G}_{k+2}& =G_{k+5}{G}_{k+3}+G_{k+4}{G}_k,\hfill \\ \hfill 9G_{k+2}{G}_{k+3}-3G_{k+1}{G}_{k+2}& =G_{k+5}{G}_{k+4}+G_{k+5}{G}_k.\hfill \end{array}$$

Follow the proof of Corollary 4 and apply (T2). □

We observe that the identities appearing in the proof of Corollary 13 exhibit an elegant structure. In each respective case, the right-hand side is $G_{k+5}{G}_{k+i}+G_{k+i+1}{G}_k$ for $i=1,\cdots ,4$. Therefore, it is interesting to find similar patterns that could be used for Gibonacci identities, including patterns involving products of more than two terms.

We encourage readers to explore the potential applications of (T2) and other telescoping principles to summation problems.