$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n$, $\lim a_n = z_0$, then $a_n = 0, \forall n$ I need to show the following:
Consider
$$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n$$
a series with radius of convergence $R>0$. Suppose that there is a non constant sequence $a_n$ such that 
$$\lim a_n = z_0$$
and $f(a_n)=0$, then the series is identically null, that is, $a_n=0, \forall n$
First of all, what does $f(a_n)=0$ means? I think it's a typo. Does somebody know what it should mean? Why this result is important? 
Could somebody help me in how to prove it?
 A: The other answers are correct, but use some sort of theorem. What you need is

The zeros of a non-constant analytic/holomorphic function are isolated
proof : in the neighborhood of $z_0$ a zero of $f$ $f(z) = f(z_0)+C(z-z_0)^n + o(|z-z_0|^n)=C(z-z_0)^n (1+ o(1))$ for some $n \in \mathbb{N}$ and $C\ne 0$, so that $f(z) \ne 0$ on $0<|z-z_0|<\epsilon$

Now if $f$ is analytic around $z_0$ and has a sequence of zeros $f(a_n)=0$ with  $a_n \to z_0$
Then by continuity $f(z_0)=\lim_n f(a_n) = 0$, and if your sequence $a_n$ isn't constant for $n$ large enough, then  $z_0$ is a non-isolated zero of $f$. And by the previous theorem, it means that $f$ is identitcally zero.
A: Is this in a complex analysis course?
The statement $f(a_n) = 0$ means exactly what you think it means.  When you plug in the number $a_n$ you happen to get $0$.
Case 1: Suppose the sequence $(a_n)$ has infinitely many values.  This means that there is an infinite sequence of distinct inputs that are mapped to 0, and these inputs converge to $z_0$.  But $f$ is analytic (has a convergent Taylor series), so that would mean that $f$ is identically $0$.  [Just take this subsequence in your limit definitions for derivatives at $z_0$.]
Case 2: The sequence $(a_n)$ takes only finitely many values.  This means that the function is of the form $f(z) = p(z) + \sum_{n \geq N} z_0 (z-z_0)^n$ for some polynomial of degree at most $N$ (because eventually all the terms $a_n$ must equal $z_0$).  But this actually is perfectly reasonable.  Consider say $z_0 = 0$ and say $f(z) = 0 - z + 0 + z^3 + 0 +0+0+ \cdots = z^3 -z.$  this satisfies $f(0) = f(1) = f(-1) =0$ [so $f(a_n) = 0$ for all $n$], and $a_n \to 0 = z_0$.
So the thing seems to be false.  But I imagine the intent was not to allow case 2 as a possibility.
A: First, if $f$ is a power series with radius of convergence $R$, then $f$ is analytic on the ball $B(z_0,R)$. Next, the set $\{a_n: n\in\mathbb{N_0}\}$ is the set of zeroes of $f$, and has an accumulation point, $z_0\in B(z_0,R)$. By the Identity Theorem, $f\equiv 0$.
