Integer-valued power sums

Suppose I have a positive number $d \in \mathbb{R}$ and a sequence of numbers $a_n \in [0,d]$ for $n \in \mathbb{N}$ with the following properties $$\sum_{i=1}^{\infty} a_i^r \in \mathbb{Z}$$ for all $r \in \mathbb{N}$ and $$\sum_{i=1}^{\infty} a_i \leq d \ .$$

Does it follow that only finitely many of the $a_i$ are non-zero?

Note that it does not follow that $a_i \in \mathbb{Z}$ as the sequence $a_1 = 2 + \sqrt{2}$, $a_2 = 2 - \sqrt{2}$, $a_k = 0$ for $k > 2$ with $d=4$ shows. This sequence also shows that the power sums can be unbounded.

• $a_2$ is not in $[0,d]$ in your example. – Brendan McKay Nov 29 '16 at 23:36
• What is the sense of $\sum a_i\leqslant d$ condition? It works for $d=\sum a_i$. – Fedor Petrov Nov 29 '16 at 23:48
• I guess it just means $(a_i)\in\ell^1$. – Fan Zheng Nov 29 '16 at 23:50
• I think Newton's relations only imply that the elementary symmetric functions are all rational. – Jeremy Rouse Nov 30 '16 at 1:04
• The Galois orbit of $a_0$ will be a torsor for a Galois group, and Galois groups are profinite so either finite or uncountable; in particular they're never countably infinite. There are also issues with convergence in this approach. – wrigley Nov 30 '16 at 9:24

The function $$f : z \in \mathbb{C} \longmapsto \sum_{i} \frac{a_i}{1-a_iz}$$ is meromorphic on $\mathbb{C}$ and has integral Taylor coefficients. It follows from a theorem of Borel that such a function must be in $\mathbb{Q}(z)$; see for example Richard Stanley's answer here. In particular $f$ has only finitely many poles ; this implies that only finitely many of the $a_i$'s are nonzero.

• It worth mention that the argument of Borel (archive.org/stream/s2bulletindessci18fran#page/n27/mode/2up) looks to be the same, or at least very similar, as Ilya suggests in his answer to this question – Fedor Petrov Nov 30 '16 at 10:57
• @js21: Isn't there a condition in the paper by Borel about not having singularities except poles in a disc of some radius? How is that satisfied for your $f$? – Ulrich Pennig Dec 1 '16 at 16:14
• @Ulrich Pennig: Such a condition does not appear in Borel's text. He only assumes that $f$ is meromorphic on a disc of radius $>1$. I agree that the sentence "la fonction $A(x)$ n'a pas d'autres singularites que $p$ poles a l'interieur d'un disque de rayon $\rho'>1$" might be misleading : in this context, it should be translated as "the singularities of $A(x)$ in the disc of radius $0$ are at most poles, and there are at most $p$ of them". – js21 Dec 1 '16 at 16:30
• @js21 That is precisely the sentence I stumbled over. Thank you for the clarification. I will have a closer look at the proof of Borel's theorem. – Ulrich Pennig Dec 1 '16 at 16:53

OK. I've finally got the time to correct my deleted answer. If anyone doesn't like complex analysis, this answer only uses real analysis.

Let $f(x)=\prod_{n=1}^\infty (1+a_nx)=\sum_{n=0}^\infty e_nx^n$, where $e_n$ is the elementary symmetric polynomial in $(a_n)$. Then as observed in the comments, $e_n\in\mathbb{Z}/n!$, so if infinitely many $a_n>0$, then $e_n\ge1/n!$, which implies $f(x)\ge e^x$, or $\log f(x)\ge x$ for $x>0$. On the other hand, we can pick $N$ such that $\sum_{n=N}^\infty a_n<1/2$. Then $\sum_{n=N}^\infty \log(1+a_nx)<x/2$. Also $\sum_{n=1}^{N-1} \log(1+a_nx)=o(x)$, so $\log f(x)\le(1/2+o(1))x$, contradiction.

This is not an answer --- but perhaps someone can fill the gap?

Clearly, we have $d>1$. Denote $$n_r=\sum_{i=1}^\infty a_i^r\in\mathbb N.$$ Choose some $k$ such that $$\sum_{i>k}a_i<\frac1{2d}$$ (all $k$ greater than some $k_0$ fit). Set $$P(x)=(x-a_1)\dots(x-a_k)=x^k+p_{k-1}x^{k-1}+\dots+p_0.$$

Now, look at the rows $m_0,\dots,m_{3k}$ of the matrix $M_{k,s}=[n_{s+i+j}]_{0\leq i,j\leq 3k}$ (for some large $s$). We see that for every $i\geq k$ the elements of $$m_i'=m_i+p_{k-1}m_{i-1}+\dots+p_0m_{i-k}$$ are expressed via the sums of large powers of $a_i$, $i>k$, with bounded coefficients. Thus all elements of $m_i'$ are less than $C_k/(2d)^s$ with some constant $C_k$ depending only on $k$. On the other hand, the entries of $m_0,\dots,m_{k-1}$ do not exceed $\alpha_k kd^{s+k}$ for some constant $\alpha_k$ also depending on $k$ only.

Replacing the $m_i$ with $m_i'$ does not chande $\det M_{s,k}$, so $$\det M_{s,k}\leq (3k)!\cdot (\alpha_kkd^{s+k})^k\cdot \left(\frac{C_k}{(2d)^s}\right)^{2k}$$ which tends to $0$ as $s\to\infty$. On the other hand, $\det M_{s,k}$ is integer, so $\det M_{s,k}=0$ for all sufficiently large $s\geq s_k$.

Now I would like to say that this yields $(n_r)$ be a linear recurrent sequence from some moment. Ptifully, this is not true in general, but perhaps some argument may fill this?

Notice that if $(n_r)$ were linear recurrent, the rest is easy. Indeed, we may prove inductively (assuming $a_1\geq a_2\geq \dots$) that all nonzero $a_i$ are the roots of its characteristic polynomial, so they are finitely many.

• Is it possible to adapt the arguments of Borel (see the answer by js21) and get a solution without complex analysis? – Fedor Petrov Nov 30 '16 at 11:02
• Why the implication from vanishing determinant to linear recurrence relation fails? Isn't this Lemma 9 in terrytao.wordpress.com/2014/05/13/…? – Fan Zheng Nov 30 '16 at 18:09