Suppose $ \sum a_k x^k $ uniformly converges at $ [0,R) $, then it converges pointwise at $ R $.

Solution 1:

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$$\left|\sum_{k=n}^ma_kR^k\right|\leqslant\varepsilon.$$