On the Sums of Reciprocal Generalized Fibonacci Numbers

Kantaphon Kuhapatanakul^{1} Department of Mathematics Faculty of Science Kasetsart University Bangkok 10900 Thailand

                       fscikpkk@ku.ac.th

Abstract In this article we consider the infinite sums of reciprocal generalized Fibonacci numbers and the infinite sums of reciprocal generalized Fibonacci sums. Applying the floor function to the reciprocals of these sums, our results generalize some identities of Holliday and Komatsu and extend some results of Liu and Zhao.

1 Introduction

Let a, b be two positive integers and c non-negative integer. Define the generalized Fibonacci numbers V_{n}(c; a, b), briefly V_{n}, by the following relation

V_{0} = c, V_{1} = 1 and V_{n+1} = aV_{n} + bV_{n−1} (n ≥ 1).

Here note that V_{n}(0; 1, 1) = F_{n} are the Fibonacci numbers, V_{n}(2; 1, 1) = L_{n} are the Lucas numbers and V_{n}(0; 2, 1) = P_{n} are the Pell numbers.

Ohtsuka and Nakamura [2] derived a formula for infinite sums of reciprocal Fibonacci numbers, as follows,

⌊(∑_{k=n}^{∞} \frac{1}{F_{k}} )^{−1}⌋ = \begin{cases}F_{n−2}, & if n is even and n ≥ 2; \\ F_{n−2} − 1, & if n is odd and n ≥ 1, \end{cases}  (1)

1Supported by the Kasetsart University Research and Development Institute (KURDI), Thailand.

where ⌊·⌋ is the floor function.

Wenpeng and Tingting [4] gave analogue of the identity (1) for the Pell numbers:

⌊(∑_{k=n}^{∞} \frac{1}{P_{k}} )^{−1}⌋ = \begin{cases}P_{n} − P_{n−1}, & if n is even and n ≥ 2; \\ P_{n} − P_{n−1} − 1, & if n is odd and n ≥ 1. \end{cases}  (2)

Holliday and Komatsu [1] generalized (1) and (2) to the generalized Fibonacci numbers V_{n}(0; a, 1), briefly u_{n}, and V_{n}(c; 1, 1) for c ≥ 1, briefly G_{n}. They showed that

⌊(∑_{k=n}^{∞} \frac{1}{u_{k}} )^{−1}⌋ = \begin{cases}u_{n} − u_{n−1}, & if n is even and n ≥ 2; \\ u_{n} − u_{n−1} − 1, & if n is odd and n ≥ 1, \end{cases}  (3)

and

⌊(∑_{k=n}^{∞} \frac{1}{G_{k}} )^{−1}⌋ = \begin{cases}G_{n−2}, & if n is odd and n ≥ n_{1}; \\ G_{n−2} − 1, & if n is even and n ≥ n_{2}, \end{cases}

where n_{1} and n_{2} are determined depending only on the value of c. For example, if G_{n} = L_{n} or c = 2, then n_{1} = 2 and n_{2} = 3.

By the same proof as the one for (3), it is easy to verify that the identity (3) still holds for V_{n}(0; a, b) in place of V_{n}(0; a, 1), provided that 1 ≤ b ≤ a.

In this paper, we first give the analogue of the identity (3) for the alternating sums of reciprocals of V_{n}(0; a, b) and prove the following result in the next section.

Theorem 1. Let U_{n} := V_{n}(0; a, b) with 1 ≤ b ≤ a. Then

⌊(∑_{k=n}^{∞} \frac{(−1)^{k}}{U_{k}} )^{−1}⌋ = (−1)^{n} (U_{n} + U_{n−1}) − 1 (n ≥ 1).

We have a following corollary.

Corollary 2. For a positive integer n, we have

(1) ⌊(∑_{k=n}^{∞} \frac{(−1)^{k}}{F_{k}} )^{−1}⌋ = (−1)^{n}F_{n+1} − 1.

(2) ⌊(∑_{k=n}^{∞} \frac{(−1)^{k}}{P_{k}} )^{−1}⌋ = (−1)^{n} (P_{n} + P_{n−1}) − 1.

In 2012, Liu and Zhao [3] showed the formulas for the infinite sums of reciprocal hyperfibonacci numbers and hyperlucas numbers as:

⌊(∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} F_{i}} )^{−1}⌋ = F_{n} − 1 (n ≥ 3),  (4)
⌊(∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} L_{i}} )^{−1}⌋ = L_{n} − 1 (n ≥ 4).  (5)

Especially, they also gave a following general result for u_{n} = V_{n}(0; a, 1) as:

⌊(∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} u_{i}} )^{−1}⌋ = u_{n} − 1 (n ≥ 3).

Next, we will extend the above results on the Fibonacci and Lucas numbers and V_{n}(0; a, 1) to the generalized Fibonacci numbers V_{n}(c; 1, 1) and V_{n}(0; a, b) and prove the following results in the next section. Theorem 3. Let G_{n} = V_{n}(c; 1, 1) for c ≥ 1. We have

⌊(∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} G_{i}} )^{−1}⌋ = G_{n} − 1 (n ≥ n_{0}),

where n_{0} is determined depending only on the value of c.

For example, we can determine n_{0} for a fixed c as follows:

\begin{matrix}c & 1 & 2 & 3-5 & 6-13 & 14-34 & 35-89 & 90-233 & 234-610 & 611-1597 & 1598-4181 \\ n_{0} & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 \end{matrix}

Put c = 1, 2 in Theorem 3, we can, respectively, deduce the identities (4) and (5). Theorem 4. Let a ≥ b ≥ 1 and U_{n} = V_{n}(0; a, b). We have

⌊(∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} U_{i}} )^{−1}⌋ = U_{n} − 1 (n ≥ N_{a}),

where N_{a} = 3 for a = 1 and N_{a} = 2 for a ≥ 2.

We have the following corollary. Corollary 5. We have

⌊(∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} P_{i}} )^{−1}⌋ = P_{n} − 1 (n ≥ 2).

2 Proofs of Theorems

We begin with some identities of the generalized Fibonacci numbers V_{n}(c; a, b) whose induction proofs are omitted. Lemma 6. Let a, b be positive integers and c non-negative integer. Then for n ≥ 1 we have (1) V_{n+1}V_{n−1} − V_{n}^{2} = (−1)^{n}b^{n−1}(1 − ac − bc^{2}).

(2) ∑_{i=0}^{n} V_{i} = \frac{1}{a + b − 1} (V_{n+1} + bV_{n} + ac − c − 1).

Proof of Theorem 1 : By Lemma 6, we have

U_{n+1}U_{n−1} − U_{n}^{2} = (−1)^{n}b^{n−1}

and

∑_{i=0}^{n} U_{i} = \frac{1}{a + b − 1} (U_{n+1} + bU_{n} − 1) .

Since a ≥ b ≥ 1, we have U_{n+1} − U_{n−1} > a^{n−1} ≥ b^{n−1}, so

\frac{(−1)^{n}}{U_{n} + U_{n−1} − (−1)^{n}} − \frac{(−1)^{n+1}}{U_{n+1} + U_{n} − (−1)^{n+1}} − \frac{(−1)^{n}}{U_{n}}

= \frac{(−1)^{n+1}U_{n−1} + 1}{U_{n} (U_{n} + U_{n−1} − (−1)^{n})} − \frac{(−1)^{n+1}}{U_{n+1} + U_{n} − (−1)^{n+1}}

= \frac{(−1)^{n+1}U_{n−1}U_{n+1} + (−1)^{n}U_{n}^{2} + U_{n+1} − U_{n−1} + (−1)^{n}}{U_{n} (U_{n} + U_{n−1} − (−1)^{n}) (U_{n+1} + U_{n} − (−1)^{n+1})}

= \frac{(−1)^{n−1}b^{n−1} + U_{n+1} − U_{n−1} + (−1)^{n}}{U_{n} (U_{n} + U_{n−1} − (−1)^{n}) (U_{n+1} + U_{n} − (−1)^{n+1})}

> 0.

By applying the above inequality repeatedly, we obtain

\frac{1}{(−1)^{n}(U_{n} + U_{n−1}) − 1} > \frac{(−1)^{n}}{U_{n}} + \frac{1}{(−1)^{n+1}(U_{n+1} + U_{n}) − 1}

> \frac{(−1)^{n}}{U_{n}} + \frac{(−1)^{n+1}}{U_{n+1}} + \frac{1}{(−1)^{n+2}(U_{n+2} + U_{n+1}) − 1}

⋮

> \frac{(−1)^{n}}{U_{n}} + \frac{(−1)^{n+1}}{U_{n+1}} + \frac{(−1)^{n+2}}{U_{n+2}} + · · · .

Therefore,

∑_{k=n}^{∞} \frac{(−1)^{k}}{U_{k}} < \frac{1}{(−1)^{n}(U_{n} + U_{n−1}) − 1}.  (6)

Similarly, we have

\frac{(−1)^{n}}{U_{n}} − \frac{(−1)^{n}}{U_{n} + U_{n−1}} + \frac{(−1)^{n+1}}{U_{n+1} + U_{n}} = (−1)^{n} (\frac{U_{n−1}}{U_{n}(U_{n} + U_{n−1})} + \frac{1}{U_{n+1} + U_{n}} )

= (−1)^{n} (\frac{U_{n−1}U_{n+1} − U_{n}^{2}}{U_{n}(U_{n} + U_{n−1})(U_{n+1} + U_{n})})

= \frac{b^{n−1}}{U_{n}(U_{n} + U_{n−1})(U_{n+1} + U_{n})}

> 0,

so

\frac{(−1)^{n}}{U_{n} + U_{n−1}} < \frac{(−1)^{n}}{U_{n}} + \frac{(−1)^{n+1}}{U_{n+1} + U_{n}} .

Repeating the above inequality, we obtain

∑_{k=n}^{∞} \frac{(−1)^{k}}{U_{k}} > \frac{(−1)^{n}}{U_{n} + U_{n−1}} .  (7)

Combining the (6) and (7), we get

\frac{1}{(−1)^{n}(U_{n} + U_{n−1})} < ∑_{k=n}^{∞} \frac{(−1)^{k}}{U_{k}} < \frac{1}{(−1)^{n}(U_{n} + U_{n−1}) − 1},

it is equivalent to

⌊(∑_{k=n}^{∞} \frac{(−1)^{k}}{U_{k}} )^{−1}⌋ = (−1)^{n}(U_{n} + U_{n−1}) − 1.

Proof of Theorem 3 : By Lemma 6, we have

G_{n+1}G_{n−1} − G_{n}^{2} = (−1)^{n}(1 − c − c^{2})

and

∑_{i=0}^{n} G_{n} = G_{n+2} − 1.

Suppose c ≥ 1, we have

\frac{1}{G_{n} − 1} − \frac{1}{G_{n+1} − 1} − \frac{1}{∑_{i=0}^{n} G_{i}} = \frac{1}{G_{n} − 1} − \frac{1}{G_{n+1} − 1} − \frac{1}{G_{n+2} − 1}

= \frac{G_{n+1}}{(G_{n} − 1)(G_{n+2} − 1)} − \frac{1}{G_{n+1} − 1}

= \frac{2G_{n} − 1 + (−1)^{n+1}(c^{2} + c − 1)}{(G_{n} − 1)(G_{n+1} − 1)(G_{n+2} − 1)}.

Since G_{n} is monotone increasing with n, we can take n so large that 2G_{n} ≥ (−1)^{n}(c^{2} + c) for a fixed c. Hence, the numerator of the right-hand side of the above identity is positive if n ≥ N_{1} for some positive integer N_{1}, so we get

\frac{1}{G_{n} − 1} ≥ \frac{1}{∑_{i=0}^{n} G_{i}} + \frac{1}{G_{n+1} − 1}

≥ \frac{1}{∑_{i=0}^{n} G_{i}} + \frac{1}{∑_{i=0}^{n+1} G_{i}} + \frac{1}{G_{n+2} − 1}

≥ \frac{1}{∑_{i=0}^{n} G_{i}} + \frac{1}{∑_{i=0}^{n+1} G_{i}} + \frac{1}{∑_{i=0}^{n+2} G_{i}} + · · · .

Thus,

∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} G_{i}} ≤ \frac{1}{G_{n} − 1} (n ≥ N_{1}).  (8)

On the other hand, we have

\frac{1}{∑_{i=0}^{n} G_{i}} − \frac{1}{G_{n}} + \frac{1}{G_{n+1}} = \frac{1}{G_{n+2} − 1} − \frac{1}{G_{n}} + \frac{1}{G_{n+1}}

= \frac{1 − G_{n+1}}{G_{n}(G_{n+2} − 1)} + \frac{1}{G_{n+1}}

= \frac{G_{n−1} + (−1)^{n}(c^{2} + c − 1)}{G_{n}G_{n+1}(G_{n+2} − 1)}.

Similarly, we can take n so large that G_{n−1} + (−1)^{n}(c^{2} + c − 1) > 0 for a fixed c. Hence, the numerator of the right-hand side of the above identity is positive if n ≥ N_{2} for some positive integer N_{2}, so we get

\frac{1}{G_{n}} < \frac{1}{∑_{i=0}^{n} G_{i}} + \frac{1}{G_{n+1}}

< \frac{1}{∑_{i=0}^{n} G_{i}} + \frac{1}{∑_{i=0}^{n+1} G_{i}} + \frac{1}{G_{n+2}}

< \frac{1}{∑_{i=0}^{n} G_{i}} + \frac{1}{∑_{i=0}^{n+1} G_{i}} + \frac{1}{∑_{i=0}^{n+2} G_{i}} + · · · .

Thus,

∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} G_{i}} > \frac{1}{G_{n}} (n ≥ N_{2}).  (9)

Combining the two inequalities (8) and (9) we obtain

\frac{1}{G_{n}} < ∑_{k=n}^{∞} \frac{1}{∑_{i=0}^{k} G_{i}} ≤ \frac{1}{G_{n} − 1},

where n ≥ n_{0} = max{N_{1}, N_{2}}, which completes the proof.

Proof of Theorem 4 : The case a = 1 has already been proved, it suffices to show the case a ≥ 2. Defining S_{n} = ∑_{i=0}^{n} U_{i}. We have

\frac{1}{S_{n}} − \frac{1}{U_{n}} + \frac{1}{U_{n+1}} = \frac{1}{U_{n+1}} − \frac{S_{n−1}}{U_{n}S_{n}}

= \frac{U_{n}S_{n} − U_{n+1}S_{n−1}}{U_{n}U_{n+1}S_{n}}

= \frac{U_{n+1} − U_{n} + (−1)^{n+1}}{aU_{n}U_{n+1}S_{n}} > 0,

and for n ≥ 2

\frac{1}{U_{n} − 1} − \frac{1}{U_{n+1} − 1} − \frac{1}{S_{n}} = \frac{S_{n−1} + 1}{(U_{n} − 1)S_{n}} − \frac{1}{U_{n+1} − 1}

= \frac{U_{n+1}S_{n−1} − U_{n}S_{n} + aS_{n}}{(U_{n} − 1)(U_{n+1} − 1)S_{n}}

= \frac{2U_{n} + (−1)^{n} − 1}{a(U_{n} − 1)(U_{n+1} − 1)S_{n}} > 0.

Then, we obtain

\frac{1}{U_{n}} < ∑_{k=n}^{∞} \frac{1}{S_{k}} ≤ \frac{1}{U_{n} − 1} (n ≥ 2),

which completes the proof.

3 Acknowledgement

The author is grateful to the anonymous referee for his/her helpful comments.

References

[1] S. H. Holliday and T. Komatsu, On the sum of reciprocal generalized Fibonacci numbers, Integers 11A (2011).

[2] H. Ohtsuka and S. Nakamura, On the sum of reciprocal sums of Fibonacci numbers, Fibonacci Quart. 46/47 (2008/2009), 153–159.

[3] R. Liu and F. Z. Zhao, On the sums of reciprocal hyperfibonacci numbers and hyperlucas numbers, J. Integer Seq. 15 (2012), Article 12.4.5.

[4] Z. Wenpeng and W. Tingting, The infinite sum of reciprocal Pell numbers, Appl. Math. Comput. 218 (2012), 6164–6167.

2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39. Keywords: second-order recurrence, Fibonacci number, reciprocal sum.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A000071}, \underline{A000129}, and \underline{A001610}.)

Received April 26 2013; revised version received July 3 2013. Published in Journal of Integer Sequences, July 30 2013.

Return to Journal of Integer Sequences home page.