For $\sum_{0}^{\infty}a_nx^n$ with an infinite radius of convergence, $a_n \neq0$ ,  the series does not converge uniformly  in $(-\infty , \infty)$. I'd like your help with the following question:

Let $\sum_{0}^{\infty}a_nx^n$ be a power series with an infinite
  radius of convergence, such that $a_n \neq 0 $ , for infinitely many values of
  $n$. I need to show that the series does not converge uniformly   in
  $(-\infty , \infty)$.

I quite don't understand the question, since I know that within the radius of convergence for a power series, there's   uniform convergence, and since I know that the radius is "infinite", it says that the uniform convergence is infinite, no?
Thanks!
 A: Hint: put $s_N(x):=\sum_{n=0}^Na_nx^n$. If the sequence $\{s_N\}$ is uniformly convergent on $\mathbb R$ then the sequence $\{s_{N+1}-s_N\}$ converges uniformly on $\mathbb R$ to $0$ so $\{a_{N+1}x^{N+1}\}$ converges uniformly to $0$. Do you see the contradiction?
A: First, your argument is flawed. I assume the theorem you are referring to is that you have uniform convergence on $[−a,a]$ for any real number $a$ with $a<R$, where $R$ is the radius of convergence of the power series. 
But you can not take $a=\infty$ in your problem...

What you could use is the following, very useful, fact:

A series $\sum\limits_{k=1}^\infty f_k(x)$ converges uniformly on $I$
  if and only if the sequence of partial sums $S_n=\sum\limits_{m=1}^n
 f_m$ is uniformly Cauchy. That is, for every $\epsilon>0$,
  there is an integer $N$ such that 
       $$|S_m(x)-S_n(x)|=\Bigl|\sum_{k=n+1}^m f_k(x)\Bigr|<\epsilon,\quad\text{ for all }x\in I\ \ \text{and all }m\ge n \ge N.$$


An immediate result of this (and  a result that, by itself, is easily proven, as in Davide Giraudo's answer) is:

If $\sum\limits_{k=1}^\infty f_k(x)$ converges uniformly on $I$, then
  for every $\epsilon>0$, there is an integer $N$ such that   $$|f_n(x)|
  =| S_{n }(x)-S_{n-1}(x)|<\epsilon,\quad\text{ for all  }x\in I\ \ \text{  and all  }n \ge N;$$ that is, the sequence $(f_n)$ converges uniformly
  to the zero function on $I$.

 


One can use the first result I mentioned directly to show that your series is not uniformly Cauchy over $\Bbb R$:
Since $a_i\ne 0$ for infinitely many $n$,  given any positive integer $N$, you can choose $m>n\ge N$   such that$\sum\limits_{k=n}^m a_k x^k$ is not a constant polynomial. Taking the limit as $x$ tends to infinity will verify that this sum is not uniformly close to 0. 
A: Let $X$ be a set. If $(f_n)$ is a sequence of functions from $X$ to $\mathbb C$, and if the series $\sum_{n\ge0}f_n$ converges uniformly on $X$, then the sequence $(f_n)$ converges uniformly to $0$.
Proof: Write
$$
f_n=\sum_{k=0}^n\ f_k-\sum_{k=0}^{n-1}\ f_k
$$
and let $n$ tend to $\infty$.
