If $\sum a_n r^n$ converges, then $\sum a_n x^n$ converges uniformly in $[0,r]$ Hello this problem I have no idea what can I do.

Let $\sum a_n x^n$ be a power series with finite convergence radius $r$. Prove that if $\sum a_n r^n$ converges, then $\sum a_n x^n$ converges uniformly in $[0,r]$.

 A: From the convergence of $\sum a_n x^n$ in $x = r$ it follows that $a_n r^n \to 0$. So in particular, the sequence is bounded, say by $M > 0$, that is $|a_n r^n| \leqslant M$ or $|a_n| \leqslant M r^{-n}$. Now let $l \in [0, r)$, then for each $k$ we have
$$|a_k l^k| \leqslant |a_k| |l|^k \leqslant \left (\frac{|x|}{r} \right )^k$$
By assumption we have $|z|/r < 1$. Therefore by the comparison test the series converges absolutely, hence converges.
Now, if a series converges uniformly Cauchy, then it converges uniformly. Now the above series converges uniformly if for any $\varepsilon > 0$, there is a positive integer $N$ such that if $N_0 \leqslant j \leqslant m$ then
$$\left |\sum_{k = j}^m a_n x^n \right | < \varepsilon$$
for all $x \in [0, r)$. Now
$$\left |\sum_{k = j}^m a_n x^n \right | \leq \sum_{k = j}^m |a_n| |x|^n.$$
This can be made easily smaller than any $\varepsilon > 0$. Do you see how?
So, now we should extend the result to $x = r$. As said in the comments, Abel's theorem allows to extend the convergence to the boundary if $r = 1$.
