Show that $b_n=0$

Question

Let $\Omega$ be a region in $\Bbb C$ .Let $\{a_n\}$ be a sequence of non-zero elements in $\Omega$ such that $a_n\to 0$ as $n\to \infty$. Let $\{b_n\}$ be a sequence of complex numbers such that $\lim_{n\to \infty}\dfrac{b_n}{a_n^k}$=0 for every non-negative integer $k$ .Suppose that $f:\Omega \to \Bbb C$ is an entire function such that $f(a_n)=b_n \forall n$.

Show that $b_n=0$

As $\{a_n\}\to 0\implies f(a_n)\to 0$ as $f$ is continuous. Thus $b_n\to 0$. But I am required to show that $b_n=0$ which I am not getting how to show.

Please give some hints.

Answer 1

Hint:

Since $f$ is entire $f(z) = \sum_{k=0}^\infty c_k z^k$. Show $c_k = 0$ for all $k$.

We have by continuity $\lim_{z \to 0}f(z) = c_0 = 0.$

Then

$$b_n = f(a_n) = c_1a_n + \sum_{k=2}^\infty c_ka_n^k \\ \implies c_1 = \lim_{n \to \infty} \frac{b_n}{a_n} - \lim_{n \to \infty} \sum_{k=2}^\infty c_ka_n^{k-1} = 0. $$

Proceed inductively to show $c_k = 0$ for $k \geqslant 2$.

Answer 2

Your argument that $f(0)=0$ doesn't seem right. I'd argue this way: By continuity and the hypothesis $b_n/a_n\to 0,$ we have $$f(0)= \lim_{n\to \infty} f(a_n) = \lim_{n\to \infty} b_n = \lim_{n\to \infty} [(b_n/a_n)\cdot a_n] = 0\cdot 0=0.$$