Finding radius of convergence for series with $z^{n^2}$? Usually I see power series such as $$\sum_{n=0}^\infty \frac{n!}{(n)^n}z^n$$ Now, I am asked to find the radius of convergence for $$\sum_{n=0}^\infty 2^{-n}z^{n^2} \quad \text{ and } \quad \sum_{n=1}^\infty (3+4i)^n(z-4i)^n$$ 
How would I find the radius of convergence for those two power series?
 A: The second series is a geometric series with common ratio $(3+4i)(z-4i)$.  It converges if the absolute value of the common ratio is less than $1$ and diverges if the common ratio is more than $1$.
$$
|3+4i||z-4i| = \sqrt{3^2+4^2} |z-4i| = 5|z-4i|.
$$
That is $<1$ iff $|z-4i|<1/5$, i.e. iff the distance between $z$ and $4i$ is less than $1/5$.  The circle of convergence is therefore centered at $4i$ and has radius $1/5$.
If you apply the ratio test to the first series (recalling that $(n+1)^2-n^2=2n+1$), you get
$$
\frac{|z|^{(n+1)^2}/2^{n+1}}{|z|^{n^2}/2^n} = \frac{|z|^{2n+1}}{2} \to 0\text{ as }n\to\infty\text{ if } |z|<1 \text{ and }\to\infty\text{ if } |z|>1.
$$
So the center is $0$ and the radius is $1$.
A: HINTS:
Look at $\displaystyle \sum \frac{1}{2^n} z^{n^2}$. Firstly, note that $z^{n^2}$ is a lot like $z^n$, but more-so. When $|z|=1$, though, it's a constant. And the resulting series is something we understand. I assume for ease of writing that $z$ is real. Then if $z > 1$, what do we know about $\lim_{n \to \infty} z^n$? Using that, consider what happens to $(z^n)^n$. I love geometric series.
Look at the second one. Argue by absolute values. $|zz'| = |z||z'|$.
Isn't it funny that after learning all those cute series techniques in calculus, everything reduces to geometric series?
