In general, a power series is a sum $\sum_{n=1}^\infty a_n z^n$ for $z\in\mathbb{C}$.
In this question, I am intereseted in power series with $a_n=0$ for even or odd $n$ (these are typically written as: $\sum_{n=1}^\infty b_{n} z^{2n}$ (only even terms) or $\sum_{n=1}^\infty b_{n} z^{2n+1}$ (only odd terms) ).
Can we find the radius of convergence of this using the ratio test?
In general, we know that the radius of convergence can be found from the Cauchy-Hadamard relation:
$$R = \begin{cases}0 & C=\infty\\ \infty &C= 0\\ C^{-1} & otherwise\end{cases}; \quad\mathrm{for}\quad C=\limsup_{n\rightarrow\infty} (|a_n|^{1/n})$$
But if the following limit exists, we also know that it is easier to find the radius of convergence with the ratio test:
$$R=\lim_{n\rightarrow \infty} \frac{|a_n|}{|a_{n+1}|}$$
Obviously, with every odd or even term being 0, the limit doesn't exist but my question is, do we really have to resort to the Cauchy-Hadamard relation? or is there any way of modifying the limit test to get the radius of convergence anyway?
For "even" functions ($\sum_{n=1}^\infty b_{n} z^{2n}$). I do think that $R^2 =\lim_{n\rightarrow \infty} \frac{|b_n|}{|b_{n+1}|}$, because I think that this is just a normal power series, inside which we insert $z^n$. I can not find an example where this definitely did not work, but I am still not convinced that this is guaranteed to work.