Question: If $\sum_{n=0}^\infty A_n$ converges, does the series $\sum A_n x^n$ converge uniformly on $[0,1]$?
Pointwise convergence is easy enough to see, and intuitively I think the series should converge uniformly as well, but I'm having trouble showing that in proof. Indeed, if we fix $\varepsilon > 0$ and procure a $N$ by the convergence of $\sum A_n$, it is not universally true that $|\sum_{n=q}^p A_n x^n| \leq |\sum_{n=q}^p A_n| < \varepsilon$ whenever $p \geq q \geq N$. I also tried to use the statement $|\sum_N^\infty A_n x^n| < |\sum_N^\infty x^n|$ past some point where $|A_n| < 1$, but this only gives pointwise convergence.
Perhaps we can break up the interval in $[0, \xi]$ and $[\xi, 1]$, the first interval converging uniformly using the geometric property and the second somehow else?
The answer is yes. We use partial summation (used to prove Abel theorem) to get an estimate of Cauchy sum.
Let $B_n=\sum_{k=m}^n A_k$. So by Cauchy Criterion, for any $\epsilon>0$, there is a $N$ such that $$ |B_n|<\epsilon \quad\text{whenever }\quad n,m>N\tag1 $$ We have \begin{align} \sum_{k=m}^n A_kx^k&=\sum_{k=m}^n (B_k-B_{k-1})x^k \\ &=\sum_{k=m}^n B_kx^k -\sum_{k=m}^n B_{k-1}x^k \\ &=\sum_{k=m}^{n-1} B_k(x^k-x^{k+1})+B_nx^n\tag{2} \end{align} Note: $B_{m−1}=0$.
Since $\lim_{n\to\infty}x^n$ exists and $\:x^n \downarrow$ on $[0,1]$, for all $k>0$ and $x\in[0,1]$ we have $$ x^k-x^{k+1}\geqslant0\: $$ Since for all $k>m$, $-\epsilon<B_k<\epsilon$ $$ |B_k(x^k-x^{k+1})|<\epsilon(x^k-x^{k+1})\tag3 $$ So for all $n,m>N-1$ and $x\in[0,1]$, by $(1)$, $(2)$ and $(3)$ there is \begin{align} \left|\sum_{k=m}^n A_kx^k\right|&\leqslant\sum_{k=m}^{n-1} |B_k(x^k-x^{k+1})|+|B_nx^n| \\ &\leqslant\sum_{k=m}^{n-1} \epsilon\:(x^k-x^{k+1})+\epsilon \:x^n \\ &=\epsilon \:(x^m-x^n+x^n) \\ &=\epsilon \:x^m \\ &\leqslant \epsilon \end{align} So by Cauchy Criterion, $\sum_{k=1}^{\infty} A_kx^k$ converges uniformly on $[0,1]$.