If a power series has positive radius of convergence and is non-constant within radius of convergence , then is all the zeroes of the series isolated? Let $f(x)=\sum_{n=0}^\infty a_n x^n$ be a real power series with positive radius of convergence $R$
(including $R=+\infty$) , then we know that $f$ is continuous in $(-R,R)$ , so the zero set of $f$ i.e. 
$Z(f):=\{x\in(-R,R):f(x)=0\}$ is closed . My question is ; is it true that if $f$ is non-constant in
$(-R,R)$ , then $Z(f)$ has no cluster point i.e. all the points of $Z(f)$ are isolated ?
 A: This is true (at least if you mean $Z(f)$ has no cluster point in $(-R,R)$; zeroes might accumulate at $\pm R$) and is a standard theorem in complex analysis.  Here is a sketch of a proof using only real methods.  Suppose $a\in (-R,R)$ is a cluster point of $Z(f)$.  It can be shown that $f(x)$ also has a power series expansion centered at $a$ which converges in a neighborhood of $a$ (basically, solve for a formal power series $g(x)$ such that $f(x)=g(x-a)$, and then check that $g$'s coefficients don't grow too fast).  Now note that $a$ is also a cluster point of $Z(f')$, since otherwise $f$ would be strictly monotonic on each side of $a$ so $a$ would be an isolated zero.  By induction, we then see that $a$ is a cluster point of $Z(f^{(n)})$ for all the higher derivatives of $f$.  In particular, this means $f^{(n)}(a)=0$ for all $n$.  But these derivatives are (up to factors of $n!$) the coefficients of the power series expansion of $f$ near $a$.  So we conclude that actually, $f$ is identically zero in a neighborhood of $a$.
We have now shown that the set of cluster points of $Z(f)$ is an open set.  It is also a closed set, so by connectedness, if it is nonempty, then it is all of $(-R,R)$.  That is, if $Z(f)$ has a cluster point, then $f$ is identically $0$.
A: It is true. Actually, this follows from a stronger result; from Knopp's Theory of functions Part 1, chapter 7, Theorem 2:

If both power series $$\sum a_n(z-z_0)^n\quad \text{and}\quad \sum
 b_n(z-z_0)^n$$ have a positive radius of convergence, and if their
  sums coincide for all points of a neighborhood of $z_0$, or only for
  an infinite number of such points (distinct from one another and from
  $z_0$) with the limit point $z_0$, then they are identical.

(emphasis mine.)
The proof goes like this: First, for $z=z_0$ it follows that $a_0=b_0$. Assume the first $m$ coefficients have been proven to be respectively equal. Then we have $$a_{m+1}+a_{m+2}(z-z_0)+\dots=b_{m+1}+b_{m+2}(z-z_0)+\dots$$ for all those infinitely many points. Let $z\to z_0$ along those points. Then, by continuity of the functions defined by the power series, $a_{m+1}=b_{m+1}$. Hence, both expansions are identical.
In your case, the second power series is just the null series, and this would mean $a_k=0$ for all $k$. 
This result is referred to as the identity theorem for power series, and is used to prove the identity theorem for analytic functions in complex analysis. This means roughly that if two complex differentiable functions are locally equal, then they are equal everywhere(!).
