Let $a$ be an algebraic number such that $|a|<1$ and $(d_k)_{k\in\mathbb N}$ be a sequence of positive integers. Assume that $\sum_{k\ge0}\frac{a^k}{d_k}$ is an algebraic number. Denote by $a_1,\cdots,a_n$ the conjugates of $a$ over $\mathbb Q$. Suppose that for every $1\le i\le n$ $|a_i|<1$. Is the series $\sum_{k\ge0}\frac{a_i^k}{d_k}$ an algebraic number for every $1\le i\le n$?

  • $\begingroup$ After your edit, the question makes even less sense. It's now unclear to me what you exactly want to ask. So I deleted my answer, which says nothing but the sum is $\frac 1{1 - a_i}$ hence algebraic. Edit: alright this last edit finally makes sense. $\endgroup$
    – WhatsUp
    Dec 13, 2019 at 3:12
  • $\begingroup$ I made edit to my answer and now it gives you a counterexample. $\endgroup$
    – WhatsUp
    Dec 13, 2019 at 4:09

2 Answers 2


No, not necessarily.

We first prove a lemma:

For every real number $x$ such that $0 < x < 2$, there exists a squence of positive integers $(d_k)_{k\geq0}$ such that $\sum_{k\geq 0}\frac{2^{-k}}{d_k} = x$.

Proof: we define sequences $(x_k)$ and $(d_k)$ recursively by: $$x_0 = x, d_k = \lfloor \frac 1 {x_k}\rfloor + 1, x_{k + 1} = 2(x_k - 1/d_k).$$ To verify that this is well-defined, we show that $0< x_k < 2$ implies $0 < x_{k + 1} < 2$.

Since we have $\frac 1 {x_k} < \lfloor \frac 1 {x_k} \rfloor + 1 = d_k$, it is clear that $x_{k + 1} > 0$.

To show that $x_{k + 1} < 2$, we separate into two cases.

  • If $x_k \leq 1$, then we have $x_{k + 1} < 2x_k \leq 2$.

  • If $x_k > 1$, then we have $d_k = 1$ and hence $x_{k + 1} = 2(x_k - 1) < 2$.

Thus both sequences are well-defined and every $d_k$ is a positive integer.

By induction on $k$, it is easy to show that $x - \sum_{k = 0}^{n - 1}\frac{2^{-k}}{d_k} = 2^{-n}x_n$.

Taking limit $n\rightarrow \infty$, we have $x = \sum_{k\geq 0}\frac{2^{-k}}{d_k}$ as desired.

Now it's easy to give counterexamples to your question.

We choose $a = 1/\sqrt 2$ so that its conjugate is $-a$.

Choose a transcendental number $x$ with $0 < x < \sqrt 2$.

By the lemma, there exist sequences of positive integers $(u_k)$ and $(v_k)$, such that $\sum_{k\geq 0} \frac{2^{-k}}{u_k} = x$ and $\sum_{k\geq 0} \frac{2^{-k}}{v_k} = \sqrt 2 x$.

We define a sequence $(d_k)$ such that $d_{2k} = u_k$ and $d_{2k + 1} = v_k$.

We then calculate: $$\sum_{k \geq 0}\frac{(\pm a)^k}{d_k} = \sum_{k \geq 0}\frac{2^{-k}}{u_k} \pm \frac{1}{\sqrt 2}\sum_{k \geq 0}\frac{2^{-k}}{v_k} = x \pm x.$$

Hence for $a$, the sum is $0$, which is algebraic, while for $-a$, the sum is $2x$, which is transcendental.

  • $\begingroup$ Very impressive. Thanks!! $\endgroup$
    – joaopa
    Dec 13, 2019 at 4:50

Let $S$ be the set of sequences $d_k\ge 4^k$ such that $\sum_k \frac{(-\sqrt{15})^k}{d_k}=0$. It is uncountably infinite thus for most of those $\sum_k \frac{(\sqrt{15})^k}{d_k}$ is transcendental.

It is hard to give a concrete example because there is a natural norm on $\overline{\Bbb{Q}}$ which is $$\|x\|=\sup_{\sigma\in Gal(\overline{\Bbb{Q}}/\Bbb{Q})} |\sigma(x)|$$ If $(x_n)$ is Cauchy for $\|.\|$ then for all $\sigma$, $(\sigma(x_n))$ is Cauchy in $\Bbb{C}$. This way we can define $$\sigma(\exp(\sqrt{2})) = \exp(\sigma(\sqrt{2}))=\exp(-\sqrt{2})$$ even if $\sigma$ is not continuous $\Bbb{Q}[\sqrt{2}]\to \Bbb{Q}[\sqrt{2}]$. The completion of $\overline{\Bbb{Q}}$ for $\|.\|$ is weird, it is not the complex numbers, ie. in there $\exp(\sqrt{2})$ is not a complex number, it is the limit of the exponential defining series, not the same.

The common functions such as $\exp,\log$, the algebraic functions satisfy their properties in $\overline{\Bbb{Q}},\|.\|$ not only in $\Bbb{C}$, thus we can't obtain a counter-example to your problem from them. My dream is to use $\overline{\Bbb{Q}},\|.\|$ to connect the properties (the non-trivial zeros..) of $L(s,\chi^\sigma)$ to those of $L(s,\chi)$.

  • $\begingroup$ This $\|\cdot\|$ is not a norm in the usual sense, as it is not multiplicative. The multiplicative norms are of course well classified. $\endgroup$
    – WhatsUp
    Dec 13, 2019 at 14:17
  • $\begingroup$ In particular, you will not be able to show that the completion is a field. In fact it's easy to see what it looks like. Let's consider the restriction to $\Bbb Q[\sqrt 2]$. Then the completion of $\Bbb Q[\sqrt 2]$ w.r.t. this norm is nothing but $\Bbb Q[\sqrt 2]\otimes_{\Bbb Q }\Bbb R$, which as a ring is isomorphic to $\Bbb R\times \Bbb R$, hence not a field. This example of course generalizes to arbitrary number field $F$, whose completion becomes $F\otimes_{\Bbb Q }\Bbb R$, which is the product of $F_v$ for all archimedean places $v$ of $F$. $\endgroup$
    – WhatsUp
    Dec 13, 2019 at 18:21
  • $\begingroup$ Therefore the completion of $\overline{ \Bbb Q}$ w.r.t. this norm is something like direct product (or sum?) of all $\overline {\Bbb Q}_v$, where $v$ ranges through all archimedean places of $\overline {\Bbb Q}$. The topology is perhaps something like weak topology, which I don't understand well. And the action of any $\sigma\in Aut(\overline {\Bbb Q})$ is simply permuting the components, hence not very interesting. I don't think this norm and the completion can be seriously useful. $\endgroup$
    – WhatsUp
    Dec 13, 2019 at 18:30

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