What should I do when the factor $a_n$ in a power series is a series itself? We were given the following exercise:

calculate the convergence radius of the following series and examine their behaviour on the edge of their respective convergence circles:
$$\sum_{n=0}^{\infty}(1+\frac 1{8}+\frac 1{27}+\ldots+\frac 1{n^{3}})x^n$$
$$\sum \frac {x^{n^2}}{2^n}$$

now correct me if I'm wrong, but you can calculate a power series' convergence radius by converting it to $\sum_{n=0}^{\infty}a_n (x-x_0)^n$, where $x_0$ is the centre of the convergence circle. The convergence radius $r$ can then be calculated with $\frac1{r}=\lim \limits_{n \to \infty}|\frac {a_{n+1}}{a_n}|$. But In the first series, $a_n=\sum_{k=0}^{n}\frac 1{k^3}$ is a series itself. How should I go about solving this exercise? 
Also, does "examining their behaviour on the edge of their respective convergence circles" mean I should see if the series has a limit for $n \to (x_0-r)$ and $n \to (x_0+r)$?  Thank you in advance for your help!
 A: The simplest is probably to use Hadamard's formula:
$$\frac 1 R=\limsup\sqrt[n]{a_n\vphantom{b}}=\limsup\sqrt[n]{a_n\vphantom{b}}.$$
Observe that
$$1<a_n <\zeta(3),\quad\text{ so }\quad 1 < \sqrt[n]{a_n} <\sqrt[n]{\zeta(3)}.$$
Can you take it from there?
A: I think that using formulae to calculate the radius of convergence is misguided, and that the comparison test is in general a better way to proceed.
Let $a_n:=\sum_{k=1}^{n}\frac{1}{k^3}$, and consider the series $\sum_n a_n x^n$ whose radius of convergence is $R$.
Case (i) $x>1$. We have that $a_n x^n >x^n$, and $\sum x^n$ is divergent; so by comparison  $\sum a_n  x^n$ is also divergent. Hence $R\leqslant 1$.
Case (ii) $x<1$. Note that $\sum_{k=1}^{\infty}\frac{1}{k^3}$ is convergent, to $S$ say. Then the sequence $a_n$ is monotone increasing and bounded by $S$. Hence for each $x$ we have that $a_n x^n < Sx^n$. Now for such $x$ the series $\sum Sx^n$ is convergent, and so by comparison $\sum a_n  x^n$ is also convergent. Hence $R\geqslant 1$.
That is, the radius of convergence is $1$.
It remains to discuss convergence on the circle of convergence. As $|a_n x^n|=a_n$, and $a_n\not\to 0$, the series can't converge at these points.
Your second series can also be done by comparison.
