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**choovuck** (1) use ratio test:

$\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim\frac{(2n +2)(2n+1)}{(n+1)^2}=4$, so $\displaystyle R=4$ (I think you have the mistake in the theorem in part (a))

(2) denote $\displaystyle z^3=t$. Then we have the series $\displaystyle \sum_{j=1}^{\infty}t^j/2^j $, so use (b):

$\displaystyle \lim a_n^{1/n}=\lim 1/2=1/2$, so the series converge for any $\displaystyle |t|<2$. This means the original series converge for any $\displaystyle |z|=|t|^{1/3}<2^{1/3}$, i.e. $\displaystyle R=2^{1/3}$