I am almost successful in this proof but my argument needs some touches. It suffices to show that the sequence of partial sums, being continuous, converges uniformly, Then, the limit function must be continuous. Here is the question
If $\sum^{\infty}_{k=0}a_k$ converges, then $\sum^{\infty}_{k=0}a_k x^k$ converges to a continuous function on $(-1,1)$
Let $\epsilon>0$ be given. Convergence of $\sum^{\infty}_{k=0}a_k$, implies that there exists $N$ such that
$$\left| \sum^{m}_{k=n+1}a_k \right|=\left| \sum^{m}_{k=0}a_k-\sum^{n}_{k=0}a_k \right|<\epsilon,\;\;\forall\;m\geq n\geq N.$$ Define $$A_n=\sum^{n}_{k=0}a_k $$ Since $\{A_n\}_n$ converges then it is bounded. So, there exists $M>0$ such that $$\left|A_n\right|\leq M,\;\;\forall\;n\geq 1. $$ Convergence of $\{\left|x\right|^m\}_m$ to zero implies that there exists $N_0$ such that $$\left|x\right|^m\leq \frac{\epsilon}{M},\;\;\forall\;n\geq N_0. $$ Let $m,n\in \Bbb{N}$ such that $m\geq n$ and $\bar{N}=\max\{N_0,N \}$. Then,
\begin{align}\left|\sum^{n}_{k=n+1}a_k x^k \right|&=\left|\sum^{m-1}_{k=n+1}A_k (x^k-x^{k+1}) -A_m x^m\right|\\\leq &\left|\sum^{m-1}_{k=n+1}A_k (x^k-x^{k+1}) -A_m x^m\right|\\\leq &\sum^{m-1}_{k=n+1}\left(\left|A_k\right| \left|x^k-x^{k+1}\right|\right) +\left|A_m\right| \left|x\right|^m\\<&\epsilon\sum^{m-1}_{k=n+1}\left|x^k-x^{k+1}\right|+M\left(\frac{\epsilon}{M}\right)\end{align} My problem now, is how to bound $$\sum^{m-1}_{k=n+1}\left|x^k-x^{k+1}\right|.$$ The sequence $\{ x^k\}_k$ is decreasing only in $[0,1)$ and not in $(-1,0)$. So, I cannot have $$\sum^{m-1}_{k=n+1}\left|x^k-x^{k+1}\right|=\sum^{m-1}_{k=n+1}\left(x^k-x^{k+1}\right).$$ If this were possible, the proof is done. Is there any way I can bound this? Any help will be appreciated.