Can we find radius of convergence for a complex power series without referring to the Cauchy-Hadamard formula?I found d'Alembert's ratio test: $$R=\lim_{n\to \infty}|\frac{a_n}{a_{n+1}}|$$ Provided that the limit exists.
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Yes you can, e.g $$\sum_{n=1}^{\infty}\frac{(i z)^n}{1+2^n}$$
$$C_n=\frac{i^n}{1+2^n}$$
$$\lim\limits_{n\to \infty}\bigg|\frac{C_n}{C_{n+1}}\bigg|=\lim\limits_{n\to \infty}\bigg|\frac{i^n\big/[1+2^n]}{i^{n+1}\big/[1+2^{n+1}]}\bigg|=2=R$$