Prove: if $\sum^\infty_{n=0}a_nx^n$ converges for every $x$, then $\sum^\infty_{n=0}a_n$ converges absolutely

Question

Prove:

if $\sum^\infty_{n=0}a_nx^n$ converges for every $x$, then $\sum^\infty_{n=0}a_n$ converges absolutely.

I get why the statement is correct (because it means that the convergence of the series doesn't dependent on $x$), but can't find a formal way to prove it.

Answer 1

If suffices to require that $\sum^\infty_{n=0}a_nx^n$ converges for one $x$ of absolute value larger than one. Then $$ |a_n x^n| \to 0 $$ implies that $$ |a_n| \le \frac{C}{|x|^n} $$ for some $C > 0$. The conclusion follows because $\sum_{n=1}^\infty \frac{1}{r^n}$ is convergent for $r > 1$.

In the context of power series: If $\sum^\infty_{n=0}a_nx^n$ converges for all $x$ then its radius of convergence is infinity, and that implies absolute convergence for all $x$, in particular for $x=1$.