Prove the following theorem: Look at the series $ \sum a_k x^k $ and denote its radius of convergence as $ R $. Suppose the series uniformly converges at $ [0,R) $, then it converges pointwise at $ R $.
One can prove the contrapositive, that is: If $ \sum a_k x^k $ does not converge pointwise at $ x_0 = R $. Then there is no uniform convergence of the series at $ [0,R) $
Proof from lecture notes: Suppose $ \sum a_k x^k $ does not converge pointwise at $ x_0 = R $. Suppose for the sake of contradiction that there is a uniform convergence at $ [0,R) $, then by Cauchy's criterion for uniform convergence we have that for all $ \epsilon>0 $ there exists $ n_0 $ s.t. for all $ m,n > n_0 $ it occurs that $ | \sup_{x\in [ 0,R) } \sum_{k=n}^{m} a_k x^k | < \epsilon $ and from continuity ( we talk about about a finite sum of continuous functions ) we get that $ | \sum_{k=n}^m a_k R^k | \leq \epsilon $.
Hence, at point $ R $ the series ( numerical series ) $ \sum a_k R^k $ satisfies Cauchy's criteria and specifically, converges - a contradiction.
I didn't understand the proof above from the line where it says " and from continuity ... ", and specifically, how they jumped to " $ | \sum_{k=n}^m a_k R^k | \leq \epsilon $ ", if you have any idea, can you please explain?
For each $x\in[0,R)$,$$\left|\sum_{k=n}^ma_kx^k\right|<\varepsilon.\tag1$$Now, suppose that$$\left|\sum_{k=n}^ma_kR^k\right|>\varepsilon.\tag2$$Then, since$$\begin{array}{ccc}[0,R]&\longrightarrow&\Bbb R\\x&\mapsto&\displaystyle\sum_{k=n}^ma_kx^k\end{array}$$is continuous at $R$, there is a $\delta>0$ such that$$|x-R|<\delta\implies\left|\sum_{k=n}^ma_kx^k-\sum_{k=n}^ma_kR^k\right|<\left|\sum_{k=n}^ma_kR^k\right|-\varepsilon.$$So, if $x\in[0,R]$ is such that $|x-R|<\delta$,\begin{align}\left|\sum_{k=n}^ma_kx^k\right|&=\left|\sum_{k=n}^ma_kx^k-\sum_{k=n}^ma_kR^k+\sum_{k=n}^ma_kR^k\right|\\&\geqslant\left|\sum_{k=n}^ma_kR^k\right|-\left|\sum_{k=n}^ma_kx^k-\sum_{k=n}^ma_kR^k\right|\\&>\left|\sum_{k=n}^ma_kR^k\right|-\left(\left|\sum_{k=n}^ma_kR^k\right|-\varepsilon\right)\\&=\varepsilon\end{align}and this is impossible, because of $(1)$. So, since $(2)$ doesn't hold, we have$$\left|\sum_{k=n}^ma_kR^k\right|\leqslant\varepsilon.$$