Prove if a power series is zero on an interval then all coefficients are zero Suppose the power series $P(x) =  \sum_{n=1}^\infty b_n x^n$  converges for $|x| \leq 1$ and that for some $0<c\le1$ it is given that $$P(x)=0 \quad \forall x \;\text{such that}\;|x| < c$$ Show that $b_n = 0$ for all $n \geq 1$.
What I was able to do was:
$$\lim_{n\to\infty} |b_n x^n|^{1/n} < 1$$ (by the root theorem for convergence) 
and as $P(x)=0$ for $|x| < c$, we have:
$$\sum_{n=1}^\infty b_n x^n = 0$$ 
Now how to get what is asked in the question from these known facts? I am not able to proceed beyond this. Thanks.
 A: A power series has the same radius of convergence as the series of its derivatives. In particular, we can take the derivative term by term for $|x|<c$. We have $P^{(k)}(0)=0$, and translating this in term of $\sum_k a_kx^k$, this means that $a_k=0$.
A: Assume otherwise and let $n$ be moinimal with $b_n\ne 0$.
Then $Q(x)=\sum_{k=0}^\infty b_{n+k}x^k$ converges for $|x|<1$ and we have $P(x)=x^nQ(x)$ and hence also $Q(x)=0$ for $0<|x|<c$. by continuity, also $Q(0)=0$, i.e. $b_n=0$ - contradiction.
A: 
Let $f(x) = \sum_{k=0}^\infty c_k x^k$ be a power series with radius of convergence $R \gt 0$ and $f(x)=0$ for all $|x| \lt S$ where $0 \lt S \lt R$, then $c_k=0$ for all $k$.

Consider the $n$-th derivative at the point $x=0$. We get
$$f^{(n)}(0) = \sum_{k=n}^\infty \frac{k!}{(k-n)!}c_k0^{k-n} = n!c_{n}$$
since all summands cancel except of the first. Keep in mind that we defined for power series that $0^0=1$ (but only for power series!).
Since we know that $f(x)=0$ on $(-S,S)$, we know that its derivative must also be zero on the same interval. Therefore it follows that $c_n=0$ for all $n$.
But since each coefficient is zero, the whole power series is the zero function and in fact zero on all of $\Bbb R$.
