Infinite series of $1/9^{F_n}$ yields algebraic number?

Question

This document contains a problem that reads as follows:

Let $F_n$ be the nth Fibonacci number defined by $F_1 = 1$ and $F_2 = 1$ and for all $n\geq 3$, $F_n = F_{n-1} + F_{n-2}$. Prove that $$\sum_{n=1}^\infty \frac{1}{9^{F_{n+2}}}$$ is an irrational number but not a transcendental number.

I fiddled with some algebraic manipulations on the value of this series hoping to find some nontrivial polynomial identity, with no luck so far.

Is it true that the value of this series is algebraic? I find it a little unbelievable, but it is usually much harder to prove something transcendental than to prove it algebraic.

Answer 1

Aas the comments indicate, the problem is wrong and this constant (let us call it $\alpha$) is transcendental. I would suppose that one can give an elementary proof (by supposing a minimal polynomial and arguing in base 9), but this seems rather messy. There are certain papers that prove similar results with elementary methods for the Kempner number $\sum_{n=0}^\infty 2^{-2^n}$ (Knight [1] has an elementary proof, but that paper is not freely available. Adamczewski gives their proof in Section 2 of [2]).

There are various deep mathemical results that imply that your number $\alpha$ is transcendental. Let me use one due to Bugeaud and (once more) Adamczewski [3], which is also one of the strongest results we have for the basis expansions of algebraic numbers:

Theorem. Let $\beta$ be an irrational real algebraic number and $b \ge 2$. Then, write $p(k)$ for the factor complexity of $\beta$ in base $b$ (that is, the number of different strings $a_1a_2\cdots a_k \in \{0,\dots, b-1\}^k$ that occur in the base-$b$ expansion of $\beta$). Then, $$ \lim_{k\to+\infty} \frac{p(k)}{k} = +\infty\,. $$ Now fix $b = 9$ and let us try to bound $p(k)$ for your constant $\alpha$. Thus, in base $9$, $$\alpha = 0.\alpha_1\alpha_2\alpha_3\cdots = 0.011010010000100000001000\cdots\,.$$ We want to prove that $p(k) \le 3k$ for all $k \ge 1$.

To see this, fix $k$ and we will count the number of $m$ such that $\alpha_m\cdots\alpha_{m+k-1}$ contains two non-zero digits. Then, for some natural number $n$, $m \le F_n < F_{n+1} \le m + k - 1$. Thus, $m \le F_n$ and $k > F_{n+1} - F_n = F_{n-1} > \frac{1}{2}{F_n}$ and so $m \le 2k - 1$. Thus, the only words $k$ in the base-$9$ expansion of $\alpha$ that contain two non-zero digits are $\alpha_1\cdots \alpha_{k-1},\dots,\alpha_{2k-1}\cdots\alpha_{3k-2}$. Hence, as there are at most $k+1$ possible combinations to have a single non-zero digit (which has to be $1$) or no non-zero digit, we indeed have that $p(k) \le 2k-1 + k+1 = 3k$.

Only, $\frac{p(k)}{k} \le \frac{3k}{k} = 3$ and thus $\lim_{k\to\infty} \frac{p(k)}{k}$ is not infinity. Hence, by the theorem, $\alpha$ is not an irrational real algebraic number. As noted in the comments, the base-$9$ expansion of $\alpha$ is not periodic and so $\alpha$ is irrational. Thus, $\alpha$ is transcendental.

[1] Knight, M. J. "An “oceans of zeros” proof that a certain non-Liouville number is transcendental." The American Mathematical Monthly 98.10 (1991): 947-949.

[2] Adamczewski, Boris. "The Many Faces of the Kempner Number." Journal of Integer Sequences 16.2 (2013): 3. https://adamczewski.perso.math.cnrs.fr/Kempner.pdf

[3] Adamczewski, Boris, and Yann Bugeaud. "On the complexity of algebraic numbers I. Expansions in integer bases." Annals of Mathematics (2007): 547-565. https://arxiv.org/pdf/math/0511674