If $\sum_k |a_k|<\infty$ and $z_{n}\to z$, then $\lim_{n}\sum_{k}a_{k}z_{n}^{k}=\sum_{k}a_{k}z^{k}$?

Question

Let $(a_{k})_{k\in\mathbb{Z}}\in\ell^{1}(\mathbb{Z})$ and let $(z_{n})_{n\in\mathbb{N}}$ be a sequence in $S^{1}:=\{z\in\mathbb{C}:|z|=1\}$ that converges to $z$. Is it then true that $$\lim_{n\to\infty}\sum_{k\in\mathbb{Z}}a_{k}z_{n}^{k}=\sum_{k\in\mathbb{Z}}a_{k}z^{k}?$$ I was hoping to use compactness of $S^{1}$ and some uniform convergence argument. However, I'm not even sure whether the sum converges uniformly or not.

By the way, I'm trying to avoid measure theoreretic arguments.

Answer 1

$$ \Big|\sum_{k=1}^\infty a_kz^k_n-\sum_{k=1}^\infty a_kz^n\Big|\le \Big|\sum_{k=1}^N a_kz^k_n-\sum_{k=1}^N a_kz^n\Big| +\Big|\sum_{k=N+1}^\infty a_kz^k_n-\sum_{k=N+1}^\infty a_kz^n\Big|\\ \le \Big|\sum_{k=1}^N a_kz^k_n-\sum_{k=1}^N a_kz^n\Big|+\Big|\sum_{k=N+1}^\infty a_kz^k_n\Big|+\Big|\sum_{k=N+1}^\infty a_kz^n\Big| \\ \le \Big|\sum_{k=1}^N a_k(z^k_n-z^k)\Big|+ 2\sum_{k=N+1}^\infty |a_k|<\frac{\varepsilon}{2}+ \frac{\varepsilon}{2} $$ To achieve the above, first find $N\in\mathbb N$, such that $$ \sum_{k=N+1}^\infty |a_n|<\frac{\varepsilon}{4}, $$ and then find $n_0\in\mathbb N$, such that $$ n\ge n_0\quad\Longrightarrow\quad |a_k||z^k_n-z^k|<\frac{\varepsilon}{2N}, \quad \text{for all $k=1,\ldots,N$.} $$

Answer 2

By the way I can only use measure theoretic argument, sorry about that: $|a_{k}z_{n}^{k}|=|a_{k}|$ and $(a_{k})\in l^{1}$, so by Lebesgue Dominated Convergence Theorem the limit and the sum can be swiped.