Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k$$ Now I try to calculate it the following way:
\begin{align}
& {}\qquad \sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k \\[8pt]
& =(-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\cdots)\cdot(-x^2+x^4-x^6+x^8-x^{10}+\cdots) \\[8pt]
& =x^3-x^4+0 \cdot x^5+0 \cdot x^6 +x^7-x^8+0 \cdot x^9 +0 \cdot x^{10} +x^{11}+\cdots
\end{align}
And now I conclude that it is equal to $\sum_{k=0}^{\infty}(x^{3+4 \cdot k}-x^{4+4 \cdot k})$ ($|x|<1$)
Is it correct? Are there any faster ways to solve that types of tasks? Any hints will be appreciated, thanks in advance.

Let $x\ne 1$.  By the usual formula for the sum of a finite geometric series, we have $1+x+x^2+x^3=\frac{1-x^4}{1-x}$. So your expression is equal to $\frac{1-x}{1-x^4}$.
Expand $\frac{1}{1-x^4}$ in a power series, multiply by $1-x$. Unless we are operating purely formaly, we will need to assume that $|x|\lt 1$.
For details, note that $\frac{1}{1-t}=1+t+t^2+t^3+\cdots$.
Substituting $x^4$ for $t$, we get that our expression is equal to 
$$(1-x)(1+x^4+x^8+x^{12}+\cdots).$$ 
Now multiply through by $1-x$. 
