Limit of $\sum_{i=1}^k a_iz_i^n$ as $n\to\infty$ where $|z_i| = 1$ Let $z_1, ..., z_k \in S^1\setminus\{1\}$, $a_1, ..., a_k\in\mathbb{C}\setminus \{0\}$ with $z_i\neq z_j$ for $i\neq j$, and consider the sequence of the form $b_n = \sum_{i=1}^k a_iz_i^n$. Is it true that $\lim_{n\to\infty} b_n$ can never exist?
I think that if we assume $z_1, ..., z_k$ are all roots of unity, then $b_n$ must be non-constant and periodic, so the limit doesn't exist.
If $z$ isn't a root of unity, it's known that $\{z^n : n\in\mathbb{N}\}$ is dense in $S^1$, so I think that in this case the limit also won't exist, but I'm not sure if the density of two such terms won't "cancel out".
 A: Wlog $k \ge 2$ and we first assume $a=a_1+..+a_k \ne 0$ and show that $b_n$ cannot converge and then we reduce the case $a=0$ to the previous case in $k-1$ variables so completing the problem.
So let $a \ne 0, b_n \to b$
Let $g_m(z)=1-\Pi_{q=1}^k(1-z\bar z_q^m)=\sum_{r=1}^kc_{rm}z^r$ and note that $g_m(z_q^m)=1, q=1,..k$, hence multiplying with $a_q$ and summing $q=1,...k$ one gets that $\sum_{r=1}^kc_{rm}b_{rm}=a$
Let $c_m=\sum_{r=1}^kc_{rm}$, so $c_mb-a=\sum_{r=1}^kc_{rm}(b-b_{rm})$ and since obviously $|c_{rm}| \le 2^k$ as they are $r$ symmetric sums in $\bar z_q^m$ and $b_{rm} \to b, m \to \infty$ for each $r=1,..k$, one gets that $c_m \to a/b, m \to \infty$ (and of course $b \ne 0$)
But now $c_m=g_m(1)=1-\Pi_{q=1}^k(1-\bar z_q^m)$ so one gets that $d_m=\Pi_{q=1}^k(1-\bar z_q^m)$ converges and that is easily shown to be impossible as in the one dimensional case (if $y_q=\bar z_q$, the image of $(y_1^m, y_2^m,..y_k^m)$ in the $k$-torus depends of the equivalence classes under the relation $y_q$~$y_r$ iff $y_q/y_r$ is a root of unity, so one either has the case where all $y_q$ are roots of unity and the image hence $d_m$ is periodic, or one gets density on each equivalence class that is not $1$ independently of the other and periodicity again for the roots of unity, and $d_m$ cannot converge in this case either -see below for more detail ***)
Assume now $a=0$ so $c_mb \to 0$, but by the above $c_m$ cannot converge so $b=0$; and let $y_q=z_q\bar z_k, q=1,..k-1$. Then with $f_n=\sum_{q=1}^{k-1}a_qy_q^n$, one has $f_n+a_k=f_n+a_kz_k^n\bar z_k^n=b_n\bar z_k^n \to 0$ as $b_n \to 0$, hence $f_n \to -a_k$ and $a_1+a_2+..a_{k-1}=-a_k \ne 0$ so we are in the previous case for the $k-1$ circle numbers $y_q$ and we showed that was impossible, hence we are done!
Note - the algebraic manipulations and ideas above are adapted from Turan's famous method of power sums where various interesting things about the sequence $b_n$ (in the more general case $\min |z_q|=1$) are proved that way
*** as per comments request, we will elaborate a little on the fact that if $|y_q|=1, y_q \ne 1, q=1,..,k$, then $d_m=\Pi_{q=1}^k(1-y_q^m)$ cannot converge; first we note that either $y_1^m$ is periodic or dense (while $|\Pi_{q=2}^k(1-y_q^m)| \le 2^{k-1}$), so choosing an appropriate subsequence for which $y_1^m \to 1$ (or of course $y^m=1$ in the periodic case), if $d_m$ converges, then $d_m \to 0$.
Then if there are roots of unity among the $y_q$, picking $N$ the lcm of all their orders and considering only $mN+1$ powers, those terms become constant ($y_q^n=1, n|N$ then $y_q^{mN+1}=y_q$) with the corresponding product constant non zero, while the other terms are of the type $y_q(y_q^N)^k$ so we apply uniform distribution results to $y_q^N$ instead and let's rename them $w_q, q=1,..l \le k$.
Assuming $l \ge 1$ (so there is at least one nonroot of unity among the $y_q$ as otherwise we are done) the group generated by $w_q$ is finitely generated free abelian, so it has a bunch of generators (which we can rename) $w_1,..w_r, r \le l$ and the others are of the type $w_q=w_1^{n_1}..w_r^{n_r}$ for fixed numbers $n_1,..n_r$ depending on $q$; now by general theory and actually not hard to prove with Weyl criterion, $(w_1^m,..w_r^m)$ are dense in the $r$ Torus so we can pick any combination of $r$ unit circle numbers we want and approximate it well infinitely many times, and then we can clearly do this with $y_1w_1^m,...,y_rw_1^m$ instead by rotating back.
So (with the usual principal argument in $[-\pi,\pi]$) we can choose to approximate well infinitely many times, such $r$ circle numbers, $\alpha_1,..\alpha_r$ with the property $|\arg \alpha_q|>\delta>0$ and also $|\arg {y_q(y_1^{-1}\alpha_1^{n_1}..y_r^{-1}\alpha_r^{n_r}})|>\delta$ for the particular $q$ for which $w_q=w_1^{n_1}..w_r^{n_r}$; hence approximating as above well enough we get that all $|\arg(y_q^{mN+1}|>\delta/2$ on that particular subsequence, so $1-y_q^{mN+1}$ stays away from $1$ on that particular subsequence of $m$'s and $d_m$ cannot converge to zero either.
A: Here is a proof when $a_i>0$ : suppose that $(b_n)$ converges and let $\sigma_1:\mathbb{N}\rightarrow\mathbb{N}$ be such that $(z_1^{\sigma_1(n)})_{n\geqslant 0}$ converges to a limit point $\ell_1$ of $(z_1^n)_{n\geq 0}$, then we can construct $\sigma:\mathbb{N}\rightarrow\mathbb{N}$ such that $(z_i^{\sigma(n)})_{n\geqslant 0}$ converges to some $\ell_i$ (this is the same limit as before when $i=1$). We can suppose without loss of generality that $(\sigma(n+1)-\sigma(n))_{n\geqslant 0}$ is strictly increasing so that $\lim\limits_{n\rightarrow +\infty}b_{\sigma(n+1)-\sigma(n)}=\sum_{i=1}^k a_i$. Since $(b_n)$ converges, then $\lim\limits_{n\rightarrow +\infty}b_n=\sum_{i=1}^k a_i$ and therefore
$$ \sum_{i=1}^k a_i\ell_i=\lim\limits_{n\rightarrow +\infty}b_{\sigma(n)}=\sum_{i=1}^k a_i \implies \sum_{i=1}^k a_i(l_i-1) = 0$$
Since $\text{Im}(l_i-1) \leq 0$ and $l_i\in S^1$ for all $i$, we need to have $l_1 = ... = l_k = 1$, so that $1$ is the only limit point of $(z_1^n)_{n\geqslant 0}$, therefore $\lim\limits_{n\rightarrow +\infty}z_1^n=1$ and $z_1=\frac{z_1^{n+1}}{z_1^n}\underset{n\rightarrow +\infty}{\longrightarrow}1$ which is a contradiction.
