prove that $\sum_{n=0}^{\infty} {\frac{n^2 2^n}{n^2 + 1}x^n} $ does not converge uniformly in its' convergence radius In calculus:

Given $\displaystyle \sum_{n=0}^{\infty} {\frac{n^2 2^n}{n^2 + 1}x^n} $, prove that it converges for $-\frac{1}{2} < x < \frac{1}{2} $, and that it does not converge uniformly in the area of convergence.

So I said:

$R$ = The radius of convergence $\displaystyle= \lim_{n \to \infty} \left|\frac{C_n}{C_{n+1}} \right| = \frac{1}{2}$ so the series converges $\forall x$ : $-\frac{1}{2} < x < \frac{1}{2}$.

But how do I exactly prove that it does not converge uniformly there? I read in a paper of University of Kansas that states: "Theorem: A power series converges uniformly to its limit in the interval
of convergence." and they proved it.
I'll be happy to get a direction.
 A: Let's prove a more general statement: if the coefficients of power series $\sum a_n x^n$ are such that the limit
$$\lim_{n\to\infty} \frac{a_n}{R^n}  $$
is finite and nonzero, then 


*

*the interval of convergence is $(-R,R)$

*the series does not converge uniformly on $(-R,R)$


The first part is immediate from ratio test and the fact that $a_nx^n\not\to 0$ when $|x|\ge R$. For the second part, use the fact that for every $n$
$$\sup_{|x|< R} |a_n x^n| = \sup_{|x|\le  R} |a_n x^n| = |a_n| R^n$$ 
which by our assumption does not approach zero. Therefore, the uniform convergence fails: even individual terms of the series are not uniformly small, let alone the remainder.  
A: According to Hagen von Eitzen theorem, you know that it will uniformly (and even absolutly) converge for intervals included in ]-1/2;1/2[. So the problem must be there.
The definition of uniform convergence is : 
$ \forall \epsilon>0,\exists N_ \varepsilon \in N,\forall n \in N,\quad [ n \ge N_ \varepsilon \Rightarrow \forall x \in A,d(f_n(x),f(x)) \le \varepsilon] $ 
So you just have to prove that there is an $\varepsilon$ (1/3 for instance or any other) where for any $N_ \varepsilon$ there will always be an $x$ (look one close from 1/2) where $d(f_n(x),f(x)) \ge \varepsilon $ for an $n \ge N_ \varepsilon$ 
