Why does $\sum\limits_{n=0}^{\infty}c_nz^n=0 \implies c_n=0, \forall n$? In so many arguments for solutions to ordinary differential equations via the Frobenius Method do I see this argument - that if an infinite polynomial with constant coefficients is identically zero over some dense domain, the coefficients must all be identically zero.
$\sum\limits_{n=0}^{\infty}c_nx^n=0 \implies c_n=0, \forall n$
I know how to show that this is true for any finite polynomial, but I don't know if its as simple as, let's say, saying that the number of zeros here is $\aleph_0$, by construction, and so the number of zeros can't possibly cover the entire domain of interest. I also don't know if I could just say that, by analytic continuation (assuming this power series/infinite polynomial is analytic), it must be identically zero. Could you please lend me some insight to this?
 A: If the power series converges at some $a \neq 0$, then it is absolutely convergent in $|x| <|a|$. In particular, it is continuous. Now as $f(x) = \sum c_nx^n$ is zero on a dense subset of $|x|<|a|$, the continuity of $f$ implies that $f$ is identically zero in $|x|<|a|$. 
Now we argue that $c_n = 0$ for all $n$. First of all, put $x = 0$ into the power seres $\sum c_n x^n$ gives $c_0 = 0$. For $c_1$, as 
$$f(x) = \sum_{n=1} c_nx^n = c_1 x + c_2 x^2 + c_3 x^3 + \cdots \Rightarrow f'(x) = c_1 + 2 c_2 x + 3c_3 x^2 + \cdots $$
but also $f'$ is identically zero on $|x|<|a|$ as $f$ is identically zero. Plug in $x = 0$ again shows $c_1 = 0$. Now differentiate again (and again) evaluate at $0$ would imply that $c_n = 0$ for all $n$. 
A: Assume that your series converge for some open disk near 0, then it is uniform converge on any closed disk contained in this disk and hence analytic. By identity theorem, as zeros of it has a limit point in the disk, it must be identically zero.
Now consider circular path $C$ centered at $0$ contained in the disk, then $2\pi ic_k=\sum^\infty_{n=-k-1}c_{n+k+1}\int_Cz^n\mathrm{d}z=\int_C\sum^\infty_{n=0}c_nz^n/z^{k+1}\mathrm{d}z=0$ for every $k$, by existence of primitive of $z^n$ for $n\neq-1$ on $C$. This implies $c_k=0$ for every $k$.
