Radius of convergence of two power series I am trying to find the radius of convergence and trying to figure out the behaviour on the frontier of the disk of convergence of the following power series:
a) $\sum_{n=1}^{\infty} \dfrac{n!}{(2-i)n^2}z^n$
b) $\sum_{n=1}^{\infty} \dfrac{1}{1+(1+i)^n}z^n$
I know that the radius of convergence of a power series is $R$ where $\dfrac{1}{R}=\overline{\sup \lim} \sqrt[n]{|a_n|}$
So, in a), we have $\overline{\sup \lim} \sqrt[n]{|a_n|}=\dfrac{\sqrt[n]{n!}}{\sqrt[n]{\sqrt{5}n^2}}$, I have no idea how to calculate this limit. And in b) I have the same problem, how do I calculate $\lim_{n \to \infty} \dfrac{1}{\sqrt[n]{|1+(1+i)^n|}}$
I would really appreciate help calculating these limits. Thanks in advance.

One would rather use the ratio test.
For a), one obtains, as $n \to \infty$,
$$
\left|\dfrac{(n+1)!}{(2-i)(n+1)^2}\times \dfrac{(2-i)n^2}{n!}\right|=\frac{n^2}{(n+1)} \to \infty
$$ thus $R=0$.
For b), one obtains, as $n \to \infty$,
$$
\left|\dfrac{1}{1+(1+i)^{(n+1)}}\times \dfrac{1+(1+i)^n}{1}\right|=\left|\dfrac{1}{1+i}\right| \times \left|\dfrac{1+(1+i)^{-n}}{1+(1+i)^{-(n+1)}}\right|\to \frac1{\sqrt{2}}
$$ thus $R=\sqrt{2}$. On the frontier of the disk of convergence, one may write $z=\sqrt{2}e^{i\theta}$, $\theta \in [0,2\pi]$ then one may observe the behaviour of the general term,
$$
\dfrac{\left(\sqrt{2}\right)^n}{1+(1+i)^n}\,e^{in\theta}
$$ giving, as $n \to \infty$:
$$
\left|\dfrac{\left(\sqrt{2}\right)^n}{1+(1+i)^n}\,e^{in\theta}\right|=\left|\dfrac{\sqrt{2}}{1+i}\right|^n \times \left|\dfrac1{1+(1+i)^{-n}}\right|\to 1 \neq0,
$$ the series is divergent everywhere on the frontier of the disk.
