If $\sum a_n$ converges then radius of convergence $\sum a_n z^n$ is at least $1$. The question reads as follows:
Suppose $\sum_{n=0}^{\infty} a_n$ converges. Show then that $\sum a_n z^n$ has radius of convergence $R\geq 1$.
I wanted to use the comparison test here:
If $|z|<1$ then $|a_n z^n| \leq |a_n|$ so $\sum |a_n z^n|$ converges if $\sum |a_n|$ does. But all we know is that $\sum a_n$ converges, so that doesn't actually work. 
How would I advance? Should the question say that $\sum a_n$ converges absolutely?
 A: If $\sum_{n=0}^{\infty} a_n$ converges then $a_n \to 0$ and in particular, it is bounded. Write $|a_n| < M$ for some $M > 0$. Then
$$ \sum_{n=0}^{\infty} \left| a_n z^n \right| \leq M\sum_{n=0}^{\infty} |z|^n $$
and so if $|z| < 1$, the series $\sum_{n=0}^{\infty} a_n z^n$ converges (absolutely and uniformly on compact subsets) and in particular, $R \geq 1$.
A: This results from Abel's lemma:

Let $r_0>0$ a real number. If the sequence  $(\lvert a_n\rvert r_0^n)$ is bounded, the series $\displaystyle \sum_{n\ge 0} a_n  z^n$ is absolutely convergent for $\lvert z\rvert <r_0$. 

Hence the radius of convergence,, which is the supremum of such $r_0$s,  is $\;\ge r_0$.
Here, since the series  $\displaystyle \sum_{n\ge 0} a_n $ is convergent, its general term $a_n$ tends to $0$, hence there exists  $N$ such that, say, $\lvert a_n\rvert<1$ for all $n>N$. There results that every $\lvert a_n\rvert$ is bounded by $M=\max\bigl(\lvert a_n\rvert,\lvert a_n\rvert,\dots,\lvert a_n\rvert,1\bigr)$, and $r_0=1$ satisfies the hypothesis of Abel's lemma.
