How to Prove that if $\sum_{k=0}^{\infty}a_k$ converges, $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges uniformly on $[0, 1]$. 
Prove that if $\sum_{k=0}^{\infty}a_k$ converges, $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges uniformly on $[0, 1]$.

I already know that $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges pointwise on $[0, 1]$ 
by either Abel's or Dirichlet's Test for pointwise convergence.
Now, can I do this proof without Dirichlet's Test for uniform convergence? 
I don't want to use it because, in the text, 
this question came before Dirichlet's Test for uniform convergence.

Since we don't know that $\sum_{k=0}^{\infty}a_k$ converges absolutely, 
we can't say that for all $x \in [0, 1]$, $|{a_k}{x^k}| \le |a_k|$ and apply the Weiertrass M-test.

Also, since we don't know if $\sum_{k=0}^{\infty}a_k$ is an alternating series, 
we can't say that that since $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges pointwise, 
let $\sum_{k=0}^{\infty}{a_k}{x^k} = A(x)$ and observe that for all $x \in [0, 1]$,
$$|A(x) - \sum_{k=0}^{n}{a_k}{x^k}| \le |a_{n+1}|x^{n+1} \le |a_{n+1}| \rightarrow 0$$
as $n$ goes to infinity.

A clue is sufficient.  I really want to think this problem through.  TY!
 A: For $n \in \mathbb{N}$, define
$$R_n = \sum_{k = n}^{\infty} a_k$$
and $M_n = \sup \{ \lvert R_k\rvert : k \geqslant n\}$. Then $M_n$ decreases to $0$, and using $a_k = R_k - R_{k+1}$ a summation by parts yields
\begin{align}
\sum_{k = m}^n a_k x^k &= \sum_{k = m}^n (R_k - R_{k+1})x^k \\
&= R_m x^{m-1} + \sum_{k = m}^n R_k(x^k - x^{k-1}) - R_{n+1}x^n \tag{$\ast$}
\end{align}
for $1 \leqslant m \leqslant n$. For $0 \leqslant x \leqslant 1$, we obtain
\begin{align}
\Biggl\lvert \sum_{k = m}^n a_k x^k\Biggr\rvert &\leqslant \lvert R_m\rvert + (1-x)\sum_{k = m}^n \lvert R_k\rvert x^{k-1} + \lvert R_{n+1}\rvert \\
&\leqslant \Biggl( 2 + (1-x)\sum_{k = m}^n x^{k-1}\Biggr) M_m \\
&= (2 + x^{m-1} - x^n)M_m \\
&\leqslant 3 M_m
\end{align}
uniformly in $x\in [0,1]$.
If we don't restrict to $x \in [0,1]$ and allow arbitrary complex $z$ with $\lvert z\rvert < 1$, taking the absolute value of each term in $(\ast)$ yields the estimate
\begin{align}
\Biggl\lvert \sum_{k = m}^n a_k z^k\Biggr\rvert
&\leqslant \Biggl( 2  + \lvert 1-z\rvert \sum_{k = m}^n \lvert z\rvert^{k-1}\Biggr) M_m \\
&\leqslant \biggl( 2 + \frac{\lvert 1-z\rvert}{1-\lvert z\rvert}\biggr) M_m.
\end{align}
Thus, for every $K \geqslant 1$ (and $1 \leqslant m \leqslant n$) we have
$$\Biggl\lvert \sum_{k = m}^n a_k z^k\Biggr\rvert \leqslant (2 + K)M_m$$
and hence uniform convergence on the (closed) Stolz region $S_K = \{ z : \lvert 1-z\rvert \leqslant K(1 - \lvert z\rvert)\}$. The interval $[0,1]$ is the special case $S_1$.
