Show that $f(x)=0$ if there is zero arbitrarily close to $0$ for $f(x)=\sum_{n=0}^{\infty}a_nx^n$ Suppose that,
$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$
converges on $(-R,R)$ for some $R>0$. If we let $(x_n)$ be a sequence in $(-R,R)$ with $x_n\ne 0$ but $\lim x_n=0$. If $f(x_n)=0$ for all $n\in N$, I need to show that $f(x)=0$ for all $x\in (-R,R)$.
My thinking is that I could use Rolle's Theorem. This theorem states that if we let $f$ be continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. So if $f(a)=f(b)$ then there is at lest one point $c$ in $(a,b)$ where $f'(c)=0$
Can I use this logic? and if so how?
 A: If $f$ is not identically zero, there must be some minimal $N$ such that $a_N\neq0$. This implies
$$f(x)=x^N\sum_{n=0}^\infty a_{n+N}x^n.$$
In particular, we have $\lim_{x\to0}\frac{f(x)}{x^N}=a_N$. Since $x_k\neq0$ and $f(x_k)=0$, this implies
$$0=\frac{f(x_k)}{x_k^N}\to a_N\qquad\text{as }k\to\infty,$$
that is, $a_N=0$ which is a contradiction. This completes the proof.
A: Notation: $f^{(j)}$ denotes the $j$th derivative of $f$, with $f^{(0)}=f$.
(I). Suppose $\{n:x_n>0\}$ is infinite. Take a strictly decreasing positive sub-sequence $(y_{0,n})_n$ of $(x_n)_n$. 
Let $S(j)$ be the sentence: There exists a strictly decreasing sequence $(y_{j,n})_n$ in $(0,R)$, converging to $0,$ with $f^{(j)}(y_{j,n})=0$ for all $n.$ 
We have $S(0).$
If $S(j)$ then  by Rolle's Theorem, for each $n$ there exists $y_{j+1,n}\in (y_{j,n+1},y_{j,n})$ with $f^{(j+1)}(y_{j+1,n})=0.$ So $S(j)\implies S(j+1).$
So by induction on $j$ we have $S(j)$ for all $j\geq 0.$ 
Each $f^{(j)}$ is differentiable and therefore continuous, so for each $j$ we have $$j!a_j=f^{(j)}(0)=\lim_{n\to \infty}f^{(j)}(y_{j,n})=\lim_{n\to \infty}0=0.$$
(II). If $\{n:x_n>0\}$ is finite then $\{n:x_n<0\}$ is infinite and we can take a strictly increasing negative sub-sequence $(y_{0,n})_n$ of $(x_n)_n$ and proceed simiarly to the method of (I)...... Alternatively, we can let $g(x)=f(-x)$ and apply (I) to $g,$ and the last line is $(-1)^j\cdot j!a_j=0$.
