For the series $$ \sum_{n=1}^{\infty} 6^n \tanh(n^2) \, z^n, $$ the radius of convergence is $$ \lim_{n \to \infty} \left| \frac{6^n \tanh(n^2)}{6^{n+1} \tanh\bigl((n+1)^2\bigr)} \right| = \frac{1}{6} \lim_{n \to \infty} \left| \frac{\tanh(n^2)}{\tanh\bigl((n+1)^2\bigr)} \right| $$
But this is where I get stuck: $\tanh$ asymptotes to $1$, but I'm not sure how to prove what the ratio is. L'Hôpital's doesn't seem to accomplish anything, and I've been mucking with $\tanh$ identities and haven't come up with anything useful.
as stated above, the simplest answer was the limit of the ratio is equal to the ratio of the limits (barring certain cases). Thanks!