This is from Ross's Elementary Analysis Textbook:
The series $(2^{-n})(x^n)$ from $n=1$ to $n= \infty$ represents a continuous function on $(-2,2)$, but the convergence isn't uniform.
He points out that for all $a$ with $0<a<2$, the series converges uniformly on $[-a,a]$ by the Weierstrass M Test. Thus $f$ is continuous on $(-2,2)$
But then he points out $f$ does not converge uniformly on $(-2,2)$ because the following condition doesn't apply: $\sup (2^{-n})(x^n)< \epsilon$ for all $n>N$ and for all $x \in (-2,2)$
Can someone explain these two seemingly contradictory results?
The rough intuition is that uniform convergence on $[-a,a]$ for all $0<a<2$ does not imply uniform convergence on the open interval $(-2,2)$ because the bound you need for uniform convergence on $[-a,a]$ blows up to infinity as $a$ approaches $2$. And this is actually what happens: in the interval $[-a,a]$ we can uniformly bound the partial sums of the series by $$ \sum_{k=0}^n 2^{-k}a^k = \frac{(a/2)^{n+1}-1}{a/2-1}. $$ What happens as $a\to2$?
What's happening here is that the series $$ \sum_{k=0}^\infty 2^{-k}x^k $$ diverges at $x=2$: we can make it arbitrarily large by choosing $x$ close enough to $2$. In other words, there is no way to uniformly bound the series as $x$ approaches $2$.
By comparison, the series $$ \sum_{k=1}^\infty \frac{x^k}{k^2} $$ converges uniformly in the interval $[-a,a]$ for every $0<a<1$, and also converges uniformly in the interval $[-1,1]$. Do you see why?