uniform convergence of a series of function This is an example from the book Elementary Analysis by Kenneth Ross that I don't understand.
Example 
Show that $\sum_{n=1}^\infty 2^{-n}x^{n}$ represents a continuous function $f$ on (-2,2) but that the convergence is not uniform.
Solution
This is a power series with radius of convergence 2. Clearly the series does not converge at $x=2$ or at $x=-2$, so its interval of convergence is (-2,2).
Consdier $0<a<2$ and note that $\sum_{n=1}^\infty 2^{-n}a^{n}$ converges. Since $|2^{-n}x^{n}|\le2^{-n}a^{n}$ for $x\in[-a,a]$, the Weierstrass M-test shows that the series $\sum_{n=1}^\infty 2^{-n}x^{n}$ converges uniformly to a function on $[-a,a]$. Since each $2^{-n}x^{n}$ is continuous on $[-a,a]$, and the series converges uniformly on $[-a,a]$, the limit function $f$ is continuous at each point of the set$[-a,a]$. Since $a$ can be any number less than 2, we conclude that $f$ represents a continuous function on $(-2,2)$. 
Since we have sup{|$2^{-n}x^{n}$|:$x\in(-2,2)$}$=1$ for all n, the convergence of the series cannot be uniform on (-2,2). (This is proven by the lemma that if the series $\sum g_n$ converges uniformly on a set S, then $\lim_n$[sup${|g_n(x)|:x \in S}$]=$0$.)
The problem I have with this example is that it is shown in the second paragraph that the series converges uniformly on $[-a,a]$, for any $a\in(0,2)$. So we can say that it converges uniformly on $(-2,2)$. Indeed isn't this how we get the conclusion in the bolded part that $f$ is continuous on $(-2,2)$. Yet the last paragraph shows an argument that the convergence of the series cannot be uniform on $(-2,2)$. I don't understand how the argument in the 2nd paragraph does not prove that convergence of the series is uniform on $(-2,2)$. Can anyone clear me up please?
 A: Uniform convergence on all intervals of the form $[-a,a]$ does not imply uniform convergence on $(-2,2)$; this is not a property that can pass through limits in such a manner. Perhaps one has to go $100$ steps into the sequence to satisfy a certain error bound on the interval $[-1,1]$, then $1000$ steps for the interval $[-1.9, 1.9]$, then $10000$ on the interval $[-1.99, 1.99]$, and so on - the closer we get to the endpoints, the more steps might be required to get a "good enough" approximation from partial sums.
For a different example, consider the sequence $x^n$ on $[0,1]$. It's pretty easy to see that this converges to the function which is $1$ at $x = 1$ and $0$ on $[0,1)$; less obvious is that the convergence is, in fact, uniform on $[0,a]$ for all $a < 1$. But the convergence can't be uniform on the full interval, because the limit function isn't continuous - but a uniform limit of continuous functions is continuous. Again, the issue is that for points close to $1$, we have to go a huge number of steps into the sequence before we get a "good" approximation of the limit function on the appropriate interval.
