If $\sum A_n$ converges, does $\sum A_n x^n$ converge uniformly on $[0,1]$? 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?
 A: 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 $[0,1]$.
A: We may use Abel's theorem itself here.
$ \sum A_n $ is uniformly convergent (as it is free from 'x').
And sequence {$ {x^n} $} is monotonically decreasing and bounded in (0,1).
Therefore, by Abel's theorem, $\sum A_nx^n$ is uniformly convergent.
