Let $\sum_0^\infty$ $a_k$ be a real convergent series. Show that the power series $f(x) := \sum_0^\infty a_k x^k$ is continuous on $[0, 1]$.
I know that within the radius of convergence it is true by uniform continuity of power series. But on the boundary (circle of convergence), this may not be true. How should this be proved?
Also, it should be noted that the series $ \sum_0^\infty|a_k|$ may be divergent.
If $\sum_0^\infty a_k$ converges, $\sum_0^\infty a_k x^k$ converges uniformly over $[0,1]$.
Let $\epsilon>0$. Let $r_n=\sum_{k=n}^\infty a_k$.
$r_n\to 0$ so there's some $N$ such that $n\geq N \implies |r_n|\leq \epsilon$
Note that
$$\sum_{k=n}^\infty a_k x^k=\sum_{k=n}^\infty (r_k-r_{k+1})x^k=r_nx^n + \sum_{k=n+1}^\infty r_k(x^k-x^{k-1})$$
If $n\geq N$ and $x\in [0,1]$ is arbitrary,
$|r_nx^n|\leq |r_n|\leq \epsilon$ and $$\big|\sum_{k=n+1}^\infty r_k(x^k-x^{k-1}) \big| \leq \sum_{k=n+1}^\infty \big|r_k(x^k-x^{k-1})\big| \leq \epsilon \sum_{k=n+1}^\infty |x^k-x^{k-1}| = \epsilon \sum_{k=n+1}^\infty (x^{k-1} - x^k) = \epsilon x^n \leq \epsilon.$$ Therefore, $$n\geq N \implies \forall x\in [0,1], \left|\sum_{k=n}^\infty a_k x^k \right|\leq 2\epsilon $$
Uniform convergence implies continuity.