Convert $f(x)=(\cos(x))^3$ to powers of x and find if converges. I started out by writing the Taylor series for $x_0=0$ (Maclaurin series) of $f(x)=(\cos(x))^3$.
If my calculations are correct $$f(x)=1-\frac{3x^2}{2!}+\frac{21x^4}{4!}-\frac{183x^2}{6!}+...$$
and after some simplifications $$f(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}a_n$$ where $a_n$ is a sequence of $a_0=1, a_1=3, a_2=21,a_3=183, ..$ 
Is there any way to find the $a_n$ sequence or a better way to solve the whole problem?

By using the identity $\cos(3x)=4\cos^3 x-3\cos x$ it follows 
\begin{align}
\cos^3x&=\frac{1}{4}\cos (3x)+\frac{3}{4}\cos x\\
&=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(3x)^{2k}}{(2k)!}+\frac{3}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(x)^{2k}}{(2k)!}\\
&=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{3(3^{2k-1}+1)}{(2k)!}x^{2k}
\end{align}
