I am having trouble determining the radius of convergence for the following sums. I have tried ratio test, but havnt gotten anywhere. I don't know what other methods i can apply:
$\sum_{n=0}^{\infty}n!z^{n!}$
$\sum_{n=0}^{\infty}z^{3n}2^n/n!$
What other methods could be used to find the radius of convergence?
On
Let's try the ratio test for the first series. One has
$${a_ {n+1}\over a_n}={(n+1)!\over n!}=n+1\to +\infty$$
And the radius of convergence is $0$. For the second one, we can rewrite it
$$\sum_{n=0}^{+\infty} {\left(\sqrt[3]{2}z^3\right)^n\over n!}=e^{\sqrt[3]{2}z^3}$$
And the radius of convergence of the exponential is $+\infty$. If we definitely want to use the ratio test the series is the composition with $z\to z^3$ with the series
$$\sum_{n=0}^{+\infty}{2^n\over n!}Z^n\,\,\text{with}\,\,Z=z^3$$
Now apply the ratio test to the series in $Z$
$${b_{n+1}\over b_n}={2\over n+1}\to 0$$
And the radius of convergence is $+\infty$ for $Z$ and the same with $z$
I agree the $n! z^{n!}$ term can be scary, but note that $f(z) = \sum_{n=0}^\infty n! z^{n!-1}$ is just the derivative of $g(z) = \sum_{n=0}^\infty z^{n!}$ whose radius of convergence is obvious.
For the second one, think to a well-known (entire) function.