Suppose $\sum_{n=1}^{\infty}a_nx^n=f(x)$, then what can we say $\sum_{n=1}^{\infty}\frac{a_n}{n^3}x^n$ and the limit? Let $f(x)=\frac{1}{(1-x)(1-x^4)}$, let $a_n$ be the nth term of the maclaurin expansion of $f(x)$.
What can we say about the power series $a_0+\sum_{n=1}^{\infty}\frac{a_n}{n^3}x^n$? Can we express the power series in terms of $f(x)$, i.e. composition with some other function, or quotient with some other function? 
In particular, what can we say about the limit $\lim_{n\to \infty}\frac{a_n}{n^3}$?
 A: I observe that for $n \geq 1$, $\frac{1}{n^3} \leq 1$, so $|\frac{a_n}{n^3} x^n| \leq |a_n x^n|$ so your first series is a candidate for comparison with your second series, resolving your question of convergence.  (This uses the fact that all your $a_n$ have the same sign, in fact are positive integers, so $\sum a_n x^n$ is absolutely convergent on $(-1,1)$.)  This also addresses your question about $\frac{a_n}{n^3}$ as $n \rightarrow \infty$.
Also, $a_0 + \sum_{n=1}^\infty \frac{a_n}{n^3} x^n= \frac{1}{8} (2 \text{Li}_2(x)+\text{Li}_3(-x)+(1+i) \text{Li}_3(-i x)+(1-i)\text{Li}_3(i x)+5 \text{Li}_3(x)+8)  \text{,}  $
where the $\textrm{Li}_s(z)$s are polylogarithms.
A: If $$f(x) = \sum_{n=0}^{\infty} a_nx^n$$ then we have $$\frac{f(x) - f(0)}{x}=\sum_{n=1}^{\infty} a_nx^{n-1}$$ Integrating this we get $$g_1(x)=\int_{0}^{x}\frac{f(t)-f(0)}{t}\,dt=\sum_{n=1}^{\infty} \frac{a_n} {n} x^n$$ Repeating the same argument further we get $$g_2(x)=\int_{0}^{x}\frac{g_1(t)}{t}\,dt=\sum_{n=1}^{\infty} \frac{a_n} {n^2}x^n$$ and $$g_3(x)=\int_{0}^{x}\frac{g_2(t)}{t}\,dt=\sum_{n=1}^{\infty} \frac{a_n} {n^3}x^n$$ For your given function I don't think we can integrate beyond $g_1(x)$ (at least I can't).
For $g_1$ note that $$\frac{f(x) - f(0)}{x}=\frac{1+x^3-x^4}{(1-x)^2(1+x)(1+x^2)}$$ and you can use partial fractions to express it in the form $$\frac{A} {1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+x}+\frac{Dx+E}{1+x^2}$$ and integrate it. 
A: We have
$$\tag 1 \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n,\,\,\, \frac{1}{1-x^4}=\sum_{n=0}^{\infty}x^{4n}.$$
Let $b_n=1$ for all $n,$ $c_n=1$ if $n$ is a multiple of $4,$ and $c_n=0$ otherwise. Then the first power series in $(1)$ is $\sum_{n=0}^{\infty}b_nx^n,$ and the second power series is $\sum_{n=0}^{\infty}c_nx^n.$
We're set up to use the Cauchy product. We get
$$\tag 2\sum_{n=0}^{\infty}a_nx^n = \frac{1}{(1-x)(1-x^4)}=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}b_kc_{n-k}\right)x^n.$$
Because $b_k,c_k\in \{0,1\}$ for all $k,$ the inner sum on the right of $(2)$ is no more than $n+1.$ Thus $0\le a_n\le n+1$ for all $n.$ This gives $0\le a_n/n^3 \le (n+1)/n^3 \to 0.$ More than that, it shows $\sum_{n=1}^{\infty}(a_n/n^3)x^n$ is absolutely convergent and uniformly convergent on $[-1,1].$ (The last following from the Weierstrass M test.)
