Here's the question:
Let $f(x)=\sum a_n(x-x_0)$ on $(x_0-r,x_0+r)$ for some $r>0$. Show that if there is a sequence $(s_n)$ of distinct roots of $f$ (i.e. $f(s_n)=0$ for all $n$) such that $\lim_{n\to \infty}s_n=x_0$ then $a_n=0$ for all $n$.
I have attempted a proof of my own and I think it's correct but I'm looking for simpler proofs. Here's my proof.:
Suppose $a_n\ne0$ for some $n$. Let $k$ be the least positive integer such that $a_k\ne 0$. Since $(s_n)$ converges to $x_0$, we can find a monotone subsequence $(s_{n_k})$ which converges to $x_0$. Assume wlog that $(s_{n_k})$ is strictly increasing sequence. Now by the Rolle's Theorem, for all $k$, there exists $t_k\in(s_{n_k},s_{n_{k+1}})$ such that $f'(t_k)=0$. It is easy to see that $(t_n) \to x_0$. Again, by Rolle's Theorem, for all $n$, there exists $u_n\in (t_n, t_{n+1})$ such that $f''(u_n)=0$. And also, $(u_n)\to x_0$. By induction, we can obtain a sequence $(b_n)\to x_0$ such that $f^{(k)}(b_n)=0$ for all $n$.
Now, by the continuity of $f^{(k)}$ on the domain, we have that $\lim_{n\to\infty}f^{(k)}(b_n)=f^{(k)}(x_0)$, i.e., $0=k!a_k$ thus, $a_k =0$. This contradicts our assumption.
Is this proof correct? I'm looking for alternative proofs.