TITLE: Radius of Convergence of power series of Complex Analysis
QUESTION [10 upvotes]: I have come across the following few questions on past exams papers.. I know how to solve these type when it is of the form $a_nz^n$ but don't have a clue what to do with these. Any help would be greatly appreciated!
(i) $\displaystyle\sum_{n=0}^\infty \bigg(\dfrac{1}{2n}\bigg)z^{3n}$.
(ii) $\displaystyle\sum_{n=0}^\infty(2^n+i)(z-i)^n$.
(iii) Show that the radius of convergence of $\displaystyle\sum_{n=0}^\infty z^{n!}$ is $1$ and that there are infinitely many $z \in \mathbb{C}$ with $|z|=1$ for which the series diverges.

REPLY [2 votes]: 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:
$$\limsup\limits_{n\rightarrow \infty} a_n^{1/n}=\lim\limits_{k\rightarrow \infty}(a_{3k})^{1/3k}=\lim\limits_{k\rightarrow \infty}(\frac{1}{2k})^{1/3k}=1$$