Radius of convergence of $\sum \frac{z^n}{n \log n \log\log n}$ I'm trying to find the radius of convergence (along with consideration of boundary behavior) of $$\sum S_n=\sum_n^\infty \frac{z^n}{n \log n \log\log n}$$
I know that series of the form $\sum \frac{1}{n \log n \,(\log\log n)^p}$ converge when $p>1$ and diverge otherwise, but I don't know how to apply this knowledge. Does the fact that $\sum \frac {S_n}{z^n}$ diverges imply that $\sum S$ diverges (i.e. has radius of convergence 0)? If so, why?
Also, if so, how would I find the radius of convergence if $S_n$ were instead $\frac{z^n}{n \log n (\log\log n)^2}$?
 A: Radius of convergence for $\sum_n S_n=\sum_n\frac{z^n}{n \log n\log \log n}=1$.  Proof: when $z=1$ series diverges, when $|z|\lt 1$, use ratio test  $\lim_{n\to \infty}\frac{S_{n+1}}{S_n}=z \lt 1$, therefore convergence.
For $p=2$, the radius of convergence $=1$ also.  Use the ratio test for $|z|\gt 1$ to get divergence, while ratio test for $|z|\lt 1$ gives convergence, since in both cases the limit ratio $=z$.  Note that this applies for any $p\gt 0$.
A: Cauchy-Hadamard formula gives that the series converges in $(-1,1)$, i.e. the radius of convergence is $1$.   For $z=1$, the series diverges. For $z=-1$, the series converges as a Leibniz series. Thus $[-1,1)$ is the set of convergence.
UPDATE
Let the coefficient be $a_n$, then since
$$
\newcommand{\abs}[1]{\left| #1\right|}
n^{1/n} \leqslant \abs{a_n}^{-1/n}\leqslant (n \cdot n \cdot n)^{1/n} = n^{3/n}, 
$$
i.e.
$$
n^{-3/n} \leqslant \abs{a_n}^{1/n}\leqslant n^{-1/n},
$$
take $\limsup_n$ we get
$$
\limsup_n \abs{a_n}^{1/n} = 1,
$$
thus the radius is $R =1/1 =1$.
