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.
 A: 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 $1-x$. 
A: Hint: Another method to approach the problem (more by brute force than Andre's clever technique) is $$f(x)=\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)(x^2+1)}.$$ Now use partial fraction decomposition and geometric series.
A: If I recall, if you have two power series, based at the same point, then the radius of convergence of their product is at least the smaller of the two radii of convergence. Generally it will be the smaller of the two radii of convergence. The at least part comes from the possibility of numerators in one cancelling with denominators in another. This isn't the case in this problem.
Note that $1+x+x^2+x^3 \equiv (1+x)(1+x^2)$, and so:
$$\frac{1}{1+x+x^2+x^3} \equiv \frac{1}{1+x} \times \frac{1}{1+x^2}$$
Both factors of the right are got from geometric series with initial terms $1$, and then common ratios of $-x$ and $-x^2$ respectively. The converge for $|x|<1$ and $|x^2|<1$ respectively. These both have radius of convergence $\rho=1$, so the radius of convergence of their product is $\rho=1$.
A: Use the Cauchy product: 
$$\sum_{k=0}^\infty a_kx^k\cdot \sum_{k=1}^\infty b_kx^k=\sum_{k=0}^\infty c_kx^k$$
where $$c_k=\sum_{n=0}^k a_n\cdot b_{k-n}$$
In your case: $a_k=(-1)^k$ and $$b_k=\begin{cases}0 & ,k =2l+1 \\(-1)^k&,k=2l\end{cases}$$
