Radius of Convergence of power series of Complex Analysis

(i) Let $u_n=\dfrac{z^{3n}}{2n}$ so $$|\frac{u_{n+1}}{u_n}|=\frac{2n|z|^3}{2n+2}\to |z|^3<1\iff|z|<1$$ so $$R=1$$ (ii) Let $v_n=(2^n+i)(z-i)^n$ so $$|\frac{v_{n+1}}{v_n}|=|\frac{(z-i)(2^{n+1}+i)}{2^n+i}|\to2|z-i|<1\iff z\in D(i,\frac{1}{2})$$ so $$R=\frac{1}{2}$$ (iii) Let $w_n=z^{n!}$ so $$|\frac{w_{n+1}}{w_n}|\to \ell=0\iff |z|<1$$ so $$R=1$$

In the following write $\,z^{3n}=(z^3)^n\,$ if you want, which perhaps makes things a little clearer, so by the quotient rule:

$$\left|\;\frac{z^{3n+3}}{2n+2}\cdot\frac{2n}{z^{3n}}\;\right|=|z|^3\frac{2n}{2n+2}\xrightarrow[n\to\infty]{}|z|^3<1\iff |z|<1\;\;\ldots$$

Now, perhaps putting $\,w:=z-i\,$ can do things clearer:

$$\left|\;\frac{(2^{n+1}+i)w^{n+1}}{(2^n+1)w^n}\;\right|=|w|\left|\;\frac{2+\frac i{2^n}}{1+\frac i{2^n}}\;\right|\xrightarrow[n\to\infty]{}2|w|<1\iff |w|=|z-i|<\frac12\;\ldots$$

You have to find the formula of $a_n$. For the first one: $$\displaystyle\sum_{n=1}^\infty \bigg(\dfrac{1}{2n}\bigg)z^{3n}=0+0\cdot z+0\cdot z^2+\frac{1}{2}z^3+0\cdot z^4+0\cdot z^5+\frac{1}{4}z^6+\dots $$ Therefore$$a_n=\begin{cases}0 &,n=0\text{ or }n\ne 3k,k\geq1\\\frac{1}{2k} &,n=3k,k\geq 1\end{cases}$$ Thus, we now find the radius of convergence: $$\limsup\limits_{n\rightarrow \infty} a_n^{1/n}=\lim\limits_{k\rightarrow \infty}(a_{3k})^{1/3k}=\lim\limits_{k\rightarrow \infty}(\frac{1}{2k})^{1/3k}=1$$