TITLE: Integer-valued power sums
QUESTION [12 upvotes]: 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.

REPLY [5 votes]: 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.