There is an exercise in Lay's Real Analysis (8.3.7) which runs:
Suppose that a sequence $(a_n)$ is bounded but that the series $\sum a_n$ diverges. Prove that the radius of convergence of the power series $\sum a_n x^n$ is equal to $1$.
Certainly, if we plug $x=1$ into the power series, we have a divergent series, so it's trivial that the radius of convergence $R\leq 1$. I don't see how (in fact I don't even believe that) $R$ is forced to equal one. I'm really hoping someone else has run into this exercise and can testify to it being misstated. But if you can prove $R=1,$ fine.
The sequence is bounded by a constant $M$. Then you have:
$\sum_{k=0}^{\infty} |a_kx^k|$ $\leq \sum_{k=0}^{\infty} |Mx^k|$=$M\sum_{k=0}^{\infty} |x^k|$. What is the radius of convergence of the power series $\sum_{k=0}^{\infty} x^k$?