If $f(x)=\sum^{\infty}_{n=0}a_n x^n$ converges for $x=50$, can $f$ be uniformly continuous over $[0,10]$?

Question

I want to show that if \begin{align}f(x)=\sum^{\infty}_{n=0}a_n x^n\end{align} converges for $x=50,$ then $f(x)$ is uniformly continuous on $[0,10].$

MY TRIAL

Since the series \begin{align}f(x)=\sum^{\infty}_{n=0}a_n x^n\end{align} converges for $x=50$, then the radius of convergence, $R=50.$ Therefore, $f$ is continuous at $x=50.$

Hence, the series converges uniformly on $[0,10]$. Since $f$ is uniformly convergent over $[0,10]$, then it is uniformly continuous on $[0,10]$.

Please, I need someone to help me check this proof if it's correct. I'll also love to see alternative approaches. Thanks!

$[0,10]$, then

Answer 1

$\sum^{\infty}_{n=0}a_n x^n$ converges at $x=50$ means the convergent radius $R$ is at least $50$. And $f(x)$ is continuous at least in $(-50,50]$ and $\sum^{\infty}_{n=0}a_n x^n$ converges uniformly on any compact subset of $(-50,50]$ . From the condition you give, we only can get above conclusion. So $f(x)$ of course is continuous on $[0,10]$ and $\sum^{\infty}_{n=0}a_n x^n$ converges uniformly on $[0,10]$. BY Cantor theorem, $f(x)$ is continuous on $[0,10]$ implies $f(x)$ is uniformly continuous on $[0,10]$.