I tried to prove:
If $g(x) = \sum_{n=0}^\infty a_n x^n$ is a power series that converges at $x= R > 0$ then it converges uniformly on $[0,R]$.
Please can you check my proof?
Let $\varepsilon > 0$. Goal is to find $N$ with $n,m > N$ implies
$$ |a_{m+1}R^{m+1}{x^{m+1}\over R^{m+1}} + \dots + a_n R^n {x^n \over R^n}|<\varepsilon$$
Because by assumption $g(R) = \sum_{n=0}^\infty a_n R^n < \infty$ there is $N$ such that $n,m>N$ implies
$$ |a_{m+1}R^{m+1}+ \dots + a_n R^n |<\varepsilon$$
Also, for $x \in [0,R]$, ${x \over R }\le 1$. Therefore for $n,m > N$
$$ |a_{m+1}R^{m+1}{x^{m+1}\over R^{m+1}} + \dots + a_n R^n {x^n \over R^n}| \le |a_{m+1}R^{m+1}{x^{m+1}\over R^{m+1}} + \dots + a_n R^n {x^{m+1} \over R^{m+1}}|\le |a_{m+1}R^{m+1}\cdot 1 + \dots + a_n R^n \cdot 1| <\varepsilon$$