Can a power series converge uniformly or even normally in all $\mathbb{R}$ (or $\mathbb{C}$ if we consider complex power series)?
My consideration: given a power series $\sum_{n \geq 0} a_n z^n$ with radius of convergence $\rho$ I think that, indipendently from $a_n$,
$$\sup_{z \in \mathbb{C}} |a_n| |z|^n=|a_n| |\rho|^n$$
Therefore if $\rho=+\infty$ a power series cannot converge normally in all $\mathbb{C}$.
Is the previous consideration wrong?
And is uniform convergence in all $\mathbb{C}$ of a power series possible? If it is, then can you suggest an example?
Yes, that's a fine short argument. As Robert notes, you need infinitely many $n$ such that $a_n\neq 0$.
The rigorous proof would be something like the follow.
Let $s_n(x)=\sum_{k=0}^{n} a_kx^k$, and assume $s_n(x)\to s(x)$ for all $x$.
Let $\epsilon>0$ and $N$ be any integer.
Find $n>N$ such that $a_{n+1}\neq 0.$ Then $|s_n(x)-s_{n+1}(x)|=|a_{n+1}||x|^{n+1}$. This is we can find $x_0$ so that $|s_n(x_0)-s_{n+1}(x_0)|>2\epsilon$. But then $$2\epsilon<|s_n(x_0)-s_{n+1}(x_0)|<|s_n(x_0)-s(x_0)|+|s_{n+1}(x_0)-s(x_0)|$$
So it is not possible for both $|s_n(x_0)-s(x_0)|<\epsilon$ and $|s_{n+1}(x_0)-s(x_0)|<\epsilon$ and hence $s_n(x)\to s(x)$ can't be uniform.
More generally a necessary but not even close to sufficient condition for $f_n\to f$ uniformly is that $|f_{n+1}-f_n|$ is bounded for all but finitely many $n$.