Well, the elements of the ring $\mathbb{C}[x]$ are easy to understand for me. They can be thought of as polynomial functions from $\mathbb{C} \rightarrow \mathbb{C}$ and as a result they are infinitely differentiable and defined over the whole complex plane.
Coming to the field $\mathbb{C}(x)$, what can we say about the elements in the field $\mathbb{C}(x)$ ? They also are functions from $\mathbb{C} \rightarrow \mathbb{C}$ but with some singularities. What else ?
What about the elements that are algebraic over $\mathbb{C}(x)$? Are these elements precisely the fractional power series with coefficients in $\mathbb{C}$ since via Puiseux theorem,the field of Puiseux series is algebraically closed.
So,my question is :
If I have a formal power series with coefficients in $\mathbb{C}$ which converges within a radius of convergence $R$. Does this element belong to $\mathbb{C}(x)$ or is algebraic over $\mathbb{C}(x)$
Your question is much deeper than you seem to realize. A function $y$ that’s algebraic over $\Bbb C(x)$ comes from a polynomial of two variables $F(X,Y)\in\Bbb C[X,Y]$. That is, $(x,y)$ will be a zero of such a two-variable polynomial $F$. Thus this question takes in the whole subject of the complex algebraic geometry of curves.
As to formal power series, let’s leave aside for a moment all questions of convergence. Then you have the ring $\Bbb C[[x]]\supset\Bbb C[x]$, as you know. If you look at the fraction fields, conventionally written $\Bbb C((x))\supset\Bbb C(x)$, the small field has transcendence degree one over $\Bbb C$, while the big one has uncountable transcendence degree over $\Bbb C$. In other words, unimaginably bigger!
Now, even restricting to convergent series, as Patrick has pointed out, our good friend the exponential series is way beyond the field $\Bbb C(x)$: not in it, not algebraic over it.