I am working with the generating function $f(x) = \frac{1}{1-5x^4}$. In terms of factoring, I would imagine it would be most reasonable to factor $1-5x^4$, and the use partial fractions to get each individual term. We see that, $1-5x^4 = -(5x^4 - 1) = -(\sqrt{5}x^2 - 1)(\sqrt{5}x^2 + 1),$ and thus $$\frac{1}{1-5x^4} = \frac{A}{-(\sqrt{5}x^2-1)} + \frac{B}{\sqrt{5}x^2 + 1}.$$ $$\implies 1 = A(\sqrt{5}x^2+1) + B(-(\sqrt{5}x^2-1))$$ We see that when $x = \frac{1}{5^{1/4}},$ $$\implies 1= A(2) \implies \frac{1}{2} = A$$, and thus $$\implies 1 = \frac{1}{2}(\sqrt{5}x^2+1) + B(-(\sqrt{5}x^2-1))$$ $$\implies \frac{1-\frac{1}{2}(\sqrt{5}x^2+1)}{-(\sqrt{5}x^2-1)} = B.$$
However, this gives me a rather difficult pair of terms to try to algebraically derive a closed form for the $n$th coefficient. Is there a simpler way to solve for this closed form? Any recommendations would be appreciated.
The easiest approach is start with the geometric series $$ \frac{1}{1-z}=\sum_{n=0}^{\infty}z^n $$ and replace $z$ with $5x^4$. This leads to $$ \frac{1}{1-5x^4}=\sum_{n=0}^{\infty}5^nx^{4n}$$ Since the original series is valid for $|z|<1$, this series is valid for $5|x|^4<1$, i.e. $|x|<5^{-\frac{1}{4}}$.