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$a_n\neq 0$, $b_n$ be complex null sequences, such that $\lim_{n}\frac{b_n}{a_n^k}=0\forall k\in\mathbb{N}$, suppose $f$ is analytic in domain $U$ which contains $0$ and all $a_n$, we need to show $f(a_n)=b_n=0\forall n$

well, by Identity Theroem of analytic function, zero set of $f$ has limit point in the domain of $f$ hence $f\equiv 0$ but how to show $f(a_n)=b_n=0\forall n$ ? Hint please. $|\frac{b_n}{a_n^k}|<\epsilon$ then?

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My interpretation of the question:

We have two sequences $(a_n), (b_n)$ which both converge to $0$. Also, $a_n\ne 0$ for all $n$, and for every $k$ we have $$ \lim_{n\to\infty }\frac{b_n}{a_n^k}\to 0 \tag{1}$$ Goal: if $f$ is analytic in a neighborhood of $0$ and $f(a_n)=b_n$ for all $n$, then $f\equiv0$ (and in particular, this means all $b_n$ had to be equal to $0$).

Idea of proof: suppose $f$ is not identically zero. Consider its Taylor expansion at $0$. Let $c_kz^k$ be its first nonzero term, that is, $f(z)=c_kz^k+\dots $. Obtain a constradiction with (1).

Spoiler included in case you get stuck again.

Further hint: since $f(z)/z^k\to c_k$ as $z\to 0$, it follows that $\lim_{n\to \infty} \frac{f(a_n)}{a_n^k} = c_k $, a contradiction.

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