# Is a power series uniformly convergent in its interval of convergence?

Let $R>0$ be the radius of convergence of a power series $Σa_nx^n$. Is it not uniformly convergent in $(-R,R)$? My book goes out of its way to say that if $[a,b]⊂(-R,R)$, then the power series converges uniformly in $[a,b]$. Can't we just say that it is uniformly convergent in $(-R,R)$?

• This mode of convergence is sometimes called "uniform convergence on compact subsets", "local uniform convergence", or "normal convergence". As shown below it's not the same as uniform convergence. Commented Nov 30, 2017 at 22:16

No. Think about the geometric series, which converges (but not uniformly) to $$1 + x + x^2 + \cdots = \frac{1}{1-x}$$ on $(-1,1)$.

• This seems so obvious now that you've pointed it out. Thank you. Commented Nov 30, 2017 at 13:29
• It's an occupational hazard among mathematicians: things often look obvious after you understand them. That doesn't mean they were obvious beforehand. Commented Nov 30, 2017 at 13:31
• +1 for the consolation. The confusion came from the fact that we proved certain properties of a power series in $(-R,R)$, from the uniform convergence in a closed subinterval $[-R+δ,R-δ]$. For instance, the sum function of the power series is continuous on the subinterval due to uniform convergence in said subinterval. We then proceed to conclude the continuity in $(-R,R)$ since $δ$ was arbitrary. I thought if we can do this, why not use the same reasoning for uniform convergence? Commented Nov 30, 2017 at 13:53
• @HritRoy Yes, pretty much. To be a little more technical, continuity is a property regarding values near a single point, not at a single point. Commented Nov 30, 2017 at 14:00
• Also note that if the series converges uniformly in $(-R,R)$, then it in fact converges uniformly in $[-R,R]$, and in particular, it converges to a continuous function on $[-R,R]$. Thus, any series that does not have finite limits at the endpoints of the interval must be a counterexample. Commented Nov 30, 2017 at 17:31

No. Example: $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1$.

Suppose that $s_n(x):=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$ converges uniformly on $(-1,1)$ to $\frac{1}{1-x}$.

Then , to $\epsilon=1$ we get $N \in \mathbb N$ such that

$\frac{|x|^{N+1}}{1-x}=|s_N(x)-\frac{1}{1-x}|<1$ for all $x \in (-1,1)$.

Hence $\lim_{x \to 1-}\frac{|x|^{N+1}}{1-x} \le 1$. But $\lim_{x \to 1-}\frac{|x|^{N+1}}{1-x}= \infty$, a contradiction.

• Ah, of course. Silly of me. Thanks alot. Commented Nov 30, 2017 at 13:28