Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$ 
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$

It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.

the series expansion at $x=0$ of $ \ \cos(ax)=1-\frac{a^2 x^2}{2}+o(x^4) \ \ $ and
$\cos(a x) \cos(bx)=\frac{1}{2}\cos(x(a-b))+\frac{1}{2}\cos(x(a+b))$
the series of $ \ \frac{1}{2}\cos(x(a-b))+\frac{1}{2}\cos(x(a+b))=\frac{1}{2}(1-\frac{(a-b)^2 x^2}{2}+o(x^4) )+\frac{1}{2}(1-\frac{(a+b)^2 x^2}{2}+o(x^4) )=$
$\cos(a x) \cos(bx)=1-\frac{(a^2+b^2)x^2}{2}+o(x^4)$
so the series of $\cos(1 x) \cos(2x) \cos(3x)… \cos(n x) =1-\frac{(1^2+2^2+...+n^2) x^2}{2}+ o(x^4) = 1-\frac{n(n+1)(2n+1) x^2}{12}+ o(x^4)$
because $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
