Let $f(x) = \sum a_n x^n$. Let's assume that $f(x)$ has a radius $R=\infty$ and $f(x)$ converges uniformly.
Now, obviously $f(0) = 0$. Meaning, $f(x)$ pointwise converging at $x=0$. Since we assumed that $f(x)$ converges uniformly on $\mathbb{R}$, then it must be that $f(x)$ converges uniformly to $0$.
In short, if a power series converges uniformly, it must be to $0$.
Is that all right?
EDIT:
Here where the question is coming from:
Let $\sum_n a_nx^n$, a power series with $R=\infty$ and $a_n \ne 0$ infinitely many times . Prove the series doesn't converge uniformly for $(-\infty, \infty)$.
Now, the solution starts with using Cauchy's criteria for uniform convergence. It claims that if we would like to show the series doesn't converge uniformly we want to show that for every $N\in\mathbb{N}$, there are $m,n > N$ such that:
$$\left| \sum_{n=0}^\infty a_nx^n \right| > \varepsilon$$
Could you explain it please? My understanding is that $\sum_{n=0}^\infty a_n 0^n = 0$. Meaning, the series pointwise converges to $0$ at $x=0$. So if the series converged uniformly, it would have been to $0$.
Isn't it right?
Almost. It must be a polynomial:
Given $\epsilon=1$, there exists $N$ such that $f(x)$ differs from $\sum_{k=0}^n{a_kx^k}$ by less than $1$ for all $x$ and $n>N$. Especially, $x\mapsto a_kx^k$ is bounded if $k>N$. This is only possible with $a_k=0$.