Finding the radius of convergence of following power series I was doing practice problems and wanted to check my work.
Find the radius of convergence of the following power series:
(1) $\Sigma_{n=0}^\infty n^p z^{n}$
(2) $\Sigma_{n=0}^\infty z^{n!}$
(3) $\Sigma_{n=0}^\infty q^{n^2} z^{n}$, $q \in \Bbb C, |q|<1$
My answers: 
I used Ratio Test for all of them.
(1) 
$$\lim_{n\to \infty} \frac {|(n+1)^pz^{n+1}|} {|n^pz^n|}= \lim_{n\to \infty}|z|^n \left|1+\frac 1 n \right|^p= |z|^n=L$$
Thus $R=1$.
(2)
$$\lim_{n\to \infty} \frac {|z^{(n+1)!}|} {|z^{n!}|}= \lim_{n\to \infty}|z|^{n+1}=L$$
Thus $R=1$.
(3)$$\lim_{n\to \infty} \frac {|q^{(n+1)^2}z^{n+1}|} {|q^{n^2}z^n|}= \lim_{n\to \infty}|z|^n \left|\frac {q^{n^2 +2n+2}} {q^{n^2}} \right|= \lim_{n\to \infty}|z|^n \left| {q^{2n+2}}\right|=L$$
$L < 1 $ when $|z|< 1$ because $ |q|<1$. So $R=1$.
These did not take me much time, so I am worried I am missing something.

(1) looks right, in (2) you can't really divide powers like that, but the answer is right anyway, and in (3) you should consider other possible combinations of parameters which lead to $L<1$. That's right, $|z|<1$ is sufficient, but is it necessary?
