Uniform convergence on the interval of convergence

Question

I am with a lot of doubts about the uniform convergence of a power series. For example, consider the series

$$\sum \frac{1}{n} x^n$$

1. It is easy to find that the radius of convergence of this series is $R=1.$ I.e, the series converges absolutely on the interval $(-1,1)$

2. There is a theorem that states that this series converges uniformly on any interval of the form $[-r,r], 0<r<1.$ Does that implies that the series converges uniformly at $(-1,1)?$(I mean, for all $x \in (-1,1)$, we can find a $0<r<1$ such that $-r\leq x\leq r$

3. In another topic here, someone used that $\lim_{x \to 1} \frac{1}{n}x^n =\frac{1}{n}$ to justify that the series does not converges uniformly on $[-1,1)$ (since the harmonic series does not converges). But I don't understand this approach. Should we use that "$\lim \sum = \sum \lim$" to see this implication? If it is true, the series cannot converges uniformly in $(-1,1)$, right? If one could give a more detailed explanation, I would be glad!

Thanks in advance!

Answer 1

It doesn't converge uniformly on $(-1,1)$, essentially because by continuity it would have to converge uniformly on $[-1,1]$ too, but as you say it takes some work as you can't just swap sum and limit.

More precisely, let $S_n(x) = \sum_{k=1}^n \frac{1}{n} x^n$ for $x\in (-1,1)$. For $n>m$ and every $x\in (0,1)$, $$| S_n(x) - S_m(x)| = \sum_{k=m+1}^n \frac{1}{k}x^k $$ Hence, for every $x \in (0,1)$, $$\| S_n - S_m \|_\infty \geq \sum_{k=m+1}^n\frac{1}{k}x^k,$$ hence by continuity of $x \mapsto \sum_{k=m+1}^n \frac 1n x^k$, (i.e. a finite sum of terms from the harmonic series) we have $$\|S_n - S_m \|_\infty \geq \sum_{k=m+1}^n \frac{1}{k},$$ so because $\sum_{k=1}^n \frac{1}{k}$ is not a Cauchy sequence, the above inequality tells us $S_n$ is not uniformly Cauchy, hence you have no uniform convergence on $(-1,1)$.

Answer 2

Let $$s_n (x) =\sum_{k=1}^n \frac{1}{k} x^k$$ and $$s(x) =\sum_{k=1}^{\infty} \frac{1}{k} x^k .$$ Take any $1>\varepsilon >0$ then we have $$\sup_{x\in (-1,1)} |s_n (x) -s(x) |=\sup_{x\in (-1,1)}\left|\sum_{k=n+1}^{\infty} \frac{1}{k} x^k\right|\geq \sum_{k=n+1}^{2n} \frac{1}{k} \left((1-\varepsilon)^{\frac{1}{2n}}\right)^k\geq (1-\varepsilon) \cdot n\cdot \frac{1}{2n} =\frac{1-\varepsilon}{2} .$$ Hence the series does not converges uniformly on $(-1,1).$