Transcendental series of reciprocals of Fibonacci and Lucas numbers

Let $F_{1}=1,F_{2}=1,\ldots$ be the Fibonacci sequence and let $L_{1}=1$, $L_{n}=F_{n-1}+F_{n}$ for $n\geq 2$ be the Lucas sequence. In the chapter “Irrationality and Transcendence” of their book [EG80, p. 64–65], starting from the Millin series:

Erdös and Graham asked the following:

The transcendence counterparts of this and many questions in [EG80, Chapter 7] were implicit throughout the chapter, hence its title. Indeed, this topic inspired intense research activities most of which involved the so called Mahler’s method. As a consequence of our main result (see Theorem 1.3), we resolve the transcendence version of Question 1.1 even when one is allowed to randomly mix the Fibonacci and Lucas numbers:

Although Fibonacci and Lucas numbers have been discovered for hundreds of years, some of their basic properties have been established only recently thanks to powerful modern methods, for example [BMS06, Ste13]. The key ingredient of the proof of our main theorems is a new application of the Subspace Theorem in treating transcendence of series in which it is hard to control the denominators of the partial sums. Before providing more details about the method, let us provide a very brief and incomplete survey of known results on irrationality and transcendence of sums like $\displaystyle\sum_{k=1}^{\infty}\frac{1}{F_{n_{k}}}$. In a nutshell, we can divide previous results in two groups.

The first group treats series in which the denominators of the partial sums are very small compared to the reciprocals of the error terms. This includes work of Mignotte on the transcendence of $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!F_{2^{k}}}$ and $\displaystyle\sum_{k=1}^{\infty}\frac{2+(-1)^{k}}{F_{2^{k}}}$ [Mig71, Mig77]. In fact, these results predate Erdös-Graham question. As another example, consider $s:=\displaystyle\sum_{k=0}^{\infty}\frac{1}{F_{2^{k}+1}}$. For every sufficiently large integer $N$, thanks to divisibility properties of the Fibonacci sequence and the fact that $2^{k}+1$ divides $2^{3k}+1$, the denominator of $\displaystyle\sum_{k=1}^{N}\frac{1}{F_{2^{k}+1}}$ is at most $\displaystyle\prod_{k=\lfloor N/3\rfloor}^{N}F_{2^{k}+1}$ which is $o(F_{2^{N+1}+1})$, hence $s$ must be irrational. We refer the readers to [Bad93] and the references there for similar results. In fact, if $n_{1}<n_{2}<\ldots$ satisfies $\displaystyle\frac{n_{k+1}}{n_{k}}\geq c>2$ then $\displaystyle\sum_{k=1}^{\infty}\frac{1}{F_{n_{k}}}$ is irrational since $F_{n_{1}}\cdots F_{n_{N}}=o(F_{n_{N+1}})$ as $N\to\infty$. Likewise, if $\displaystyle\frac{n_{k+1}}{n_{k}}\geq c>3$ then $\displaystyle\sum_{k=1}^{\infty}\frac{1}{F_{n_{k}}}$ is transcendental by applying Roth’s theorem and the fact that $F_{n_{1}}\cdots F_{n_{N}}=O\left(F_{n_{N+1}}^{(1/2)-\epsilon}\right)$ for an appropriate $\epsilon>0$. However, replacing those easy bounds $2$ and $3$ respectively by smaller numbers for irrationality and transcendence problems appears to be a very difficult task.

The second group constitutes the majority of results in this topic. In 1975, Mahler [Mah75] reproved Mignotte’s result using the method he had invented nearly 50 years earlier (see Nishioka’s notes [Nis96] for an introduction to Mahler’s method). This method is applicable when the sequence $n_{k}$ has the special form $n_{k}=d^{k}+r$ where $d,r\in{\mathbb{Z}}$ with $d\geq 2$. We refer the readers to [BT94, DKT02, KKS09] and the references there for further details. In fact, before this paper, there has not been one result establishing the transcendence of $\displaystyle\sum_{k=1}^{\infty}\frac{1}{F_{n_{k}}}$ for an arbitrary sequence $n_{k}$ with $\displaystyle\frac{n_{k+1}}{n_{k}}\geq c$ for any single value $c<3$ (as explained above, $c>3$ is the easy bound due to an immediate application of Roth’s theorem).

From now on, let $\alpha\neq\pm 1$ be a real quadratic unit; this means ${\mathbb{Q}}(\alpha)$ is real quadratic and $\alpha\neq\pm 1$ a unit in the ring of algebraic integers. Let $\sigma$ be the non-trivial automorphism of ${\mathbb{Q}}(\alpha)$ and let $\beta=\sigma(\alpha)$. Without loss of generality, we assume $|\beta|<1<|\alpha|$. Let $H$ and $h$ respectively denote the absolute multiplicative and logarithmic Weil height on $\bar{{\mathbb{Q}}}$, our main result is the following:

There have been two general transcendence results using the Subspace Theorem recently and both involve values of a power series $\sum d_{n}z^{n}$ at an algebraic number $z_{0}$. One is a result of Adamczewski-Bugeaud [AB07] extending an earlier work of Troi-Zannier [TZ99]. In their work, the coefficients of $d_{n}$’s form an automatic sequence and $z_{0}$ is the reciprocal of a Pisot number. The authors rely on the repeating pattern of automatic sequences to apply the Subspace Theorem using linear forms in three variables. The other is a result of Corvaja-Zannier [CZ02] treating the case that $\sum d_{n}z^{n}$ is lacunary with positive real coefficients and $z_{0}\in(0,1)$. The problem considered here is different from all the above. While it is true that one can express $\displaystyle\sum_{k=1}^{\infty}\frac{1}{F_{n_{k}}}$ as the value of a power series at $1/\alpha$ with $\alpha=\displaystyle\frac{1+\sqrt{5}}{2}$, neither the coefficients are automatic nor the series is lacunary for an arbitrary choice of $n_{k}$ with $n_{k+1}/n_{k}\geq c>2$.

The more subtle difference and key reason for the difficulty in settling our current problem are as follows. Let us consider the example $s=\displaystyle\sum_{k=1}^{\infty}\frac{1}{F_{n_{k}}}$ and $\alpha=\displaystyle\frac{1+\sqrt{5}}{2}$ again. Let $s_{N}=\displaystyle\sum_{k=1}^{N}\frac{1}{F_{n_{k}}}$ be the sequence of partial sums so that $|s-s_{N-1}|=O(|\alpha|^{-n_{N}})$. Now the usual idea is to fix a large integer $P$, then truncate each

where $s_{N,i}$ is a finite sum of units for $1\leq i\leq P-1$. Then we have:

and after assuming that $s$ is algebraic, one might attempt to apply the Subspace Theorem to this equation for $s$, $s_{N-P}$ and the individual units in each $s_{N,i}$. The difference compared to work of Adamczewski-Bugeaud or Corvaja-Zannier is that while terms in their application of the Subspace Theorem are $S$-integers for an appropriate choice of a finite set of places $S$ in an appropriate number field, here we cannot find such an $S$ so that the term $s_{N-P}$ above is an $S$-integer for infinitely many $N$. For this reason, in our situation, when applying the Subspace Theorem we may have the contribution $H(s_{N-M})^{D}$ where $D$ is the number of terms. With just the constraint $c>2$, it is entirely possible for the above contribution to offset the error term $O(|\alpha|^{-n_{N}})$ and one fails to apply the Subspace Theorem. Therefore new ideas are needed to overcome this crucial issue. Moreover, after one applies the Subspace Theorem, it remains a highly nontrivial task to arrive at the desired conclusion from the resulting linear relation. This paper promotes the innovation that one should start with a certain “minimal expression” before applying the Subspace Theorem in order to maximize the benefit of the resulting linear relation. We refer the readers to the discussion right after Proposition 3.6 for more details.

Acknowledgements. The author wishes to thank Professors Yann Bugeaud and Maurice Mignotte for useful comments. The author is partially supported by an NSERC Discovery Grant and a CRC II Research Stipend from the Government of Canada.

The Subspace Theorem is one of the milestones of diophantine geometry in the last 50 years. The first version was obtained by Schmidt [Sch70] and further versions were obtained by Schlickewei and Evertse [Sch92, Eve96, ES02]. This section follows the exposition in the book of Bombieri-Gubler [BG06].

Let $M_{{\mathbb{Q}}}=M_{{\mathbb{Q}}}^{\infty}\cup M_{{\mathbb{Q}}}^{0}$ where $M_{{\mathbb{Q}}}^{0}$ is the set of $p$-adic valuations and $M_{{\mathbb{Q}}}^{\infty}$ is the singleton consisting of the usual archimedean valuation. More generally, for every number field $K$, write $M_{K}=M_{K}^{\infty}\cup M_{K}^{0}$ where $M_{K}^{\infty}$ is the set of archimedean places and $M_{K}^{0}$ is the set of finite places. Throughout this paper, we fix an embedding of $\bar{{\mathbb{Q}}}$ into ${\mathbb{C}}$ and let $|\cdot|$ denote the usual absolute value on ${\mathbb{C}}$. Hence for a number field $K$, the set $M_{K}^{\infty}$ corresponds to the set of real embeddings and pairs of complex-conjugate embeddings of $K$ into ${\mathbb{C}}$. For every $w\in M_{K}$, let $K_{w}$ denote the completion of $K$ with respect to $w$ and denote $d(w/v)=[K_{w}:{\mathbb{Q}}_{v}]$ where $v$ is the restriction of $w$ to ${\mathbb{Q}}$. Following [BG06, Chapter 1], for every $w\in M_{K}$ restricting to $v$ on ${\mathbb{Q}}$, we normalize $|\cdot|_{w}$ as follows:

Let $m\in{\mathbb{N}}$, for every vector ${\mathbf{u}}=(u_{0},\ldots,u_{m})\in K^{m+1}\setminus\{\mathbf{0}\}$ and $w\in M_{K}$, let $|{\mathbf{u}}|_{w}:=\displaystyle\max_{0\leq i\leq m}|u_{i}|_{w}$. For $P\in{\mathbb{P}}^{m}(\bar{{\mathbb{Q}}})$, let $K$ be a number field such that $P$ has a representative ${\mathbf{u}}\in K^{m+1}\setminus\{\mathbf{0}\}$ and define:

It is an easy fact that this is independent of the choice of ${\mathbf{u}}$ and the number field $K$. Then we define $h(P)=\log(H(P))$. For $\alpha\in\bar{{\mathbb{Q}}}$, write $H(\alpha)=H([\alpha:1])$ and $h(\alpha)=\log(H(\alpha))$. Later on, we will use the classical version of Roth’s theorem [BG06, Chapter 6] to give a weak upper bound on $n_{k+1}/n_{k}$:

Let $m\in{\mathbb{N}}$, for every vector ${\mathbf{x}}=(x_{0},\ldots,x_{m})\in K^{m+1}\setminus\{\mathbf{0}\}$, let $\tilde{{\mathbf{x}}}$ denote the corresponding point in ${\mathbb{P}}^{m}(K)$. For every $w\in M_{K}$, denote $|{\mathbf{x}}|_{w}:=\displaystyle\max_{0\leq i\leq m}|x_{i}|_{w}$. We have:

Throughout this section, we assume the notation in the statement of Theorem 1.3 and put $A_{k}=a_{n_{k}}$, $B_{k}=b_{n_{k}}$, $C_{k}=c_{n_{k}}$, $U_{k}=u_{n_{k}}$ for every $k$ to simplify the notation. From now on, assume that $s:=\displaystyle\sum_{k=1}^{\infty}\frac{C_{k}}{U_{k}}$ is algebraic and let $K$ be the Galois closure of ${\mathbb{Q}}(\alpha,s)$. For $m\geq 1$, let $s_{m}=\displaystyle\sum_{k=1}^{m}\frac{C_{k}}{U_{k}}$ be the sequence of partial sums. We will repeatedly use the following observation: if $(t_{n})_{n}$ is a sequence in $K^{*}$ such that $h(t_{n})/n\rightarrow 0$ as $n\to\infty$ then $|t_{n}|_{v}=e^{o(n)}$ as $n\to\infty$ for every $v\in M_{K}$; this means for every $\epsilon>0$, we have $e^{-\epsilon n}<|t_{n}|_{v}<e^{\epsilon n}$ for all sufficiently large $n$.

We are given that $C_{k}\neq 0$ for every $k$. We may assume that $A_{k}B_{k}\neq 0$ for every $k$, as follows. Let $\sigma$ denote the nontrivial automorphism of ${\mathbb{Q}}(\alpha)$. Suppose that $A_{k}=0$ then $U_{k}=-B_{k}\beta^{n_{k}}\in{\mathbb{Q}}^{*}$. Applying $\sigma$ gives $U_{k}=-B_{k}\beta^{n_{k}}=-\sigma(B_{k})/\alpha^{n_{k}}$, therefore $\sigma(B_{k})/B_{k}=(\alpha/\beta)^{n_{k}}$. Since $|\sigma(B_{k})/B_{k}|=e^{o(n_{k})}$, we conclude that $A_{k}=0$ is possible for only finitely many $k$. A similar conclusion holds for $B_{k}=0$ too. By ignoring the first finitely many $n_{k}$’s, we may assume $A_{k}B_{k}\neq 0$ for every $k$.

Similarly, from $\displaystyle\sum_{k=m_{1}}^{m_{2}}\frac{C_{k}}{U_{k}}=\frac{C_{m_{1}}}{U_{m_{1}}}+O(C_{m_{1}+1}/U_{m_{1}+1})$ for any $m_{1}<m_{2}\leq\infty$, by ignoring the first finitely many $n_{k}$’s, we may assume:

We start with several easy estimates:

Note that $\alpha\beta=\pm 1$ since they are units. Then we can use the geometric series to express:

which is valid when $k$ is sufficiently large so that $|B_{k}/A_{k}|<|(\alpha/\beta)^{n_{k}}|=|\alpha|^{2n_{k}}.$

Let $P$ be a large positive integers that will be specified later. In the following, $N$ denotes an arbitrarily large positive integer. In the various $O$-notations and $o$-notations, the implied constants might depend on the given data and $P$ but they are independent of $N$. We have:

As mentioned in the Section 1, it is typical in applications of the Subspace Theorem to break $s_{N-1}$ as $s_{N-P}$ and truncate the expression (3) for $k=N-P+1,\ldots,N-1$ to maintain the error term $\alpha^{-n_{N}+o(n_{N})}$.

For $1\leq i\leq P-1$, let

thanks to Proposition 3.4. The explicit upper bound here is not important: the key fact is that these $D_{N,i}$’s can be bounded from above independently of $N$.

At this point, one may attempt to apply the Subspace Theorem using the inequality:

and linear forms in $2+\displaystyle\sum_{i=1}^{P-1}(D_{N,i}+1)$ variables for the terms $s$, $s_{N-P}$, and those in the double sum in a similar manner to [CZ04, p. 180–181] or [KMN19, Proposition 3.4]. However, unlike these previous papers, the term $s_{N-P}$ in our situation is not an $S$-integer (for infinitely many $N$) for any choice of a finite set $S\subset M_{K}$. Because of this, there is a potential contribution of $\displaystyle H(s_{N-P})^{2+\sum(D_{N,i}+1)}$ which could completely offset the error term $|\alpha|^{-n_{N}+o(n_{N})}$.

Our new idea is to consider an extra “buffer zone” by specifying another positive integer $Q<P$, expressing

and rewriting (5) as

We then multiply both sides by $x_{N}$ to get

After that we expand $x_{N}=\displaystyle\prod_{i=1}^{Q}U_{N-P+i}=\displaystyle\prod_{i=1}^{Q}(A_{N-P+i}\alpha^{n_{N-P+i}}-B_{N-P+i}\beta^{n_{N-P+i}})$ as a linear combination of $2^{Q}$ terms, expand $x^{\prime}_{N}$ as a linear combination of $Q2^{Q-1}$ terms. Note that each $x_{N}s$, $x_{N}s_{N-P}$, as well as each individual term in the double sum now consists of $2^{Q}$ many terms. In a typical application of the Subspace Theorem, one is worse off after performing the above steps. Therefore it is amusing that in our current situation, those steps can help reduce the number of terms significantly while the new error $|x_{N}||\alpha|^{-n_{N}+o(n_{N})}$ is not too much larger than the previous $|\alpha|^{-n_{N}+o(n_{N})}$.

Now even if we can apply the Subspace Theorem, there remains one important technical issue to overcome. After expanding $x^{\prime}_{N}$ and $x_{N}$, it might happen that certain terms in the LHS of (7) already satisfied a linear relation and the conclusion of the Subspace Theorem trivially illustrates this fact. For instance, in the double sum in the LHS of (5), if there are two different $(i_{1},j_{1})$ and $(i_{2},j_{2})$ for which $(2j_{1}+1)n_{N-P+i_{1}}$ and $(2j_{2}+1)n_{N-P+i_{2}}$ are close (or even equal) to each other then one should “gather” the two terms corresponding to $(i_{1},j_{1})$ and $(i_{2},j_{2})$ first. So we will also need a way to efficiently “gather similar terms” so that the conclusion of the Subspace Theorem becomes helpful for our purpose. First, we expand $x_{N}$ and $x^{\prime}_{N}$: