If $\abs{a_{k+1}/a_k}\to L$, then $\sum a_k$ converges absolutely when $L<1$ and diverges when $L>1$.Suggest a slogan
Let \((a_k)_{k\ge0}\) be nonzero complex numbers such that \(L\coloneqq\lim_{k\to\infty}\abs{\frac{a_{k+1}}{a_k}}\) exists in \([0,\infty]\).
If \(L<1\), then \(\sum_k a_k\) converges absolutely.
If \(L>1\), then \(\sum_k a_k\) diverges.
If \(L=1\), the test gives no information: \(\sum_k 1/k\) diverges and \(\sum_k 1/k^2\) converges, and both have \(L=1\).
In particular, if \(c_k\neq0\) and \(\lim_k\abs{c_{k+1}/c_k}=\ell\in[0,\infty]\), applying the test to \(a_k=c_kz^k\) shows that \(\sum_kc_kz^k\) has radius of convergence \(1/\ell\), with \(1/0=\infty\) and \(1/\infty=0\).