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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 $$\sum_{n=1}^{\infty}\frac{(i z)^n}{1+2^n}$$

$$C_n=\frac{i^n}{1+2^n}$$

$$\lim\limits_{n\to \infty}\bigg|\frac{C_n}{C_{n+1}}\bigg|=\lim\limits_{n\to \infty}\bigg|\frac{i^n\big/[1+2^n]}{i^{n+1}\big/[1+2^{n+1}]}\bigg|=2=R$$

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  • $\begingroup$ @Nehorai...I got it now. Thanks! $\endgroup$
    – UserAb
    Feb 21, 2016 at 22:53
  • $\begingroup$ You are welcome $\endgroup$
    – 3SAT
    Mar 7, 2016 at 7:34

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