Proof about power series without using the continuity property

Question

I took a picture of what I've tried to do and the proof I'm asking to provide. I feel like the hint just made me doubt what I would have done intuitively. What do we use abs(a(n)).

Answer 1

$$ \left | \sum_n a_nx^n\right|\\ \stackrel{\text{triangle ineq.}}{\leq} \sum_n |a_n||x|^n\\ \stackrel{|x|\leq K}{\leq}\sum_n |a_n|K^n\\ \stackrel{K<R}{<}+\infty $$ Where the final inequality is true because power series converge absolutely (and uniformly) on compact subsets of the interval of convergence (see @David C. Ullrich's proof above). So $|f(x)|$ for $x\in [-K,K]$ is indeed bounded, specifically by $\sum_n |a_n|K^n$.

Answer 2

The point is to show that $\sum|a_n|K^n<\infty$; once you know that you're done because...

Hint: Say $K<x<R$. Then the series $\sum a_nx^n$ converges, so the terms are bounded; say $$|a_nx^n|\le c.$$

But $$|a_n|K^n=\left(\frac Kx\right)^n|a_nx^n|,$$so...