Power series in the complex plane Let $c_{n}\in \mathbb{C}$, $n=0, 1, ...,$.
Question 1. There exists a power series, $\sum_{n=0}^{\infty}c_{n}x^{n}$ convergent for all real $x\in \mathbb{R}$ (eventually for all $x\ge 0$), but such that the complex series $\sum_{n=0}^{\infty}c_{n}z^{n}$ has a finite radius of convergence only (i.e. if $z\in \mathbb{C}$, then the previous complex series is convergent ONLY for $|z| < 3$, say) ?
Question 2. In words, if $\sum_{n=0}^{\infty}c_{n}x^{n}$ is convergent on the real axis, then replacing $x\in \mathbb{R}$ by $z\in \mathbb{C}$, the series $\sum_{n=0}^{\infty}c_{n}z^{n}$ is convergent in the whole complex plane ?
My guess is that the answer to Question 2 is NO, but I would need a concrete example mentioned by Question 1.
Thank you in advance.
All the best,
George
 A: Given any sequence $(c_n)$, you can prove, based on the Root Test, that there is a number $R$ (the radius of convergence) such that $\sum_n c_n z^n$ converges uniformly on compact subsets of the open disc of radius $R$, whereas it diverges for all $z$ with $|z|>R$. In particular, if $\sum _n c_n x^n $ converges for all $x\in [0,\infty)$, then $R=\infty$ so that $\sum_n c_n z^n$ converges for all $z\in \mathbb C$.
A: The answer to question $1$ is NO. If a power series converges for all real $x$, then it converges for all complex $x$.
Question $2$ is identical to question $1$, so... yes, by replacing real $x$ with a complex one, you still get a convergent power series.
A: Then answer to both questions is obtained by properly using the following observation:
If the series $\sum_{n=0}^\infty c_nz^n$ converges for a certain $z_0\in\mathbb C$, then the same series also converges for every $z\in\mathbb C$, with 
$\lvert z\rvert<\lvert z_0\rvert$.
Or equivalently
If the series $\sum_{n=0}^\infty c_nz^n$ DOES NOT converge for a certain $z_0\in\mathbb C$, then the same series DOES NOT converge for every $z\in\mathbb C$, with 
$\lvert z\rvert>\lvert z_0\rvert$.
To see this (the first formulation), assume that  $\sum_{n=0}^\infty c_nz^n_0$ converges, then $\{c_nz^n_0\}$ bounded, say by $M$, and if $\lvert z\rvert<\lvert z_0\rvert$, and set $\lvert z/z_0\rvert=a<1$. Then
$$
\left|\sum_{n=0}^\infty c_nz^n\right| = \left|\sum_{n=0}^\infty c_nz^n_0\left(\frac{z}{z_0}\right)^n\right|\le \sum_{n=0}^\infty Ma^n=\frac{M}{1-a}<\infty.
$$
