Lemma Suppose that $\sum a_k z^k$ has radius of convergence $R>0$, then $$ \forall A>\tfrac{1}{R}\ \exists C >0\quad \big( \forall k\ |a_k|\le C A^k\big). $$
Proof By contradiction suppose that, given $A>\tfrac{1}{R}$, for every $C >0$ there is $k$ such that $|a_k|>C A^k$, then \begin{gather*} \tfrac{1}{R}=\limsup |a_k|^{1/k}\ge \limsup C ^{1/k}A, \end{gather*}
Is it possible to get a contradiction from this point? For instance $R=0$ by proving that $\limsup C ^{1/k}=\infty$. Or another approach, but should be simple.
Your first deduction implies, in particular, that there exist infinitely many $k$ such that $|a_k| > A^k$. Choose $B$ such that $\frac1A < B < R$. On the one hand, $B<R$, so that the series $\sum a_kB^k$ converges. On the other hand, $B>\frac1A$, so that $|a_kB^k|>1$ for infinitely many $k$; this implies that the sequence $a_kB^k$ does not tend to zero, which forces $\sum a_kB^k$ to diverge.