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.
 A: Use $$1-\prod_{k=1}^n\cos{kx}=1-\prod_{k=1}^n\left(1-2\sin^2\frac{kx}{2}\right)\sim2\sum_{k=1}^n\left(\frac{kx}{2}\right)^2= \frac{x^2}{2}\sum_{k=1}^nk^2$$
A: 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 $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
A: Set $f_n(x)=\cos x\cos2x\cdots\cos nx$ for simplicity. We want to prove by induction that
$$
\lim_{x\to0}\frac{1-f_n(x)}{x^2}=\frac{1}{12}n(n+1)(2n+1)
$$
The statement is true for $n=1$. Assume it is true for $n$. Then
$$
f_{n+1}(x)=f_n(x)(\cos nx\cos x-\sin nx\sin x)
$$
Thus
\begin{align}
\frac{1-f_{n+1}(x)}{x^2}
&=\frac{1-f_n(x)\cos nx\cos x}{x^2}+\frac{\sin nx\sin x}{x^2}\\[6px]
&=\frac{1-f_n(x)}{x^2}\cos nx\cos x+\frac{1-\cos nx\cos x}{x^2}+\frac{\sin nx\sin x}{x^2}
\end{align}
The limit of the second summand is, by standard computations, $(n^2+1)/2$; the limit of the third summand is $n$.
Thus, by the induction hypothesis, the limit is
\begin{align}
\frac{1}{12}n(n+1)(2n+1)+\frac{n^2+1}{2}+n
&=\frac{1}{12}n(n+1)(2n+1)+\frac{(n+1)^2}{2}\\[6px]
&=\frac{n+1}{12}(2n^2+n+6n+6)\\[6px]
&=\frac{1}{12}(n+1)(n+2)(2n+3)
\end{align}
as required.
A: This is handled in the same manner as in this answer.
Split the numerator like $$1-\cos  x+\cos x(1-\cos 2x\cos 3x\dots\cos nx) $$ and the desired limit is equal to $$\lim_{x\to 0}\frac {1-\cos x} {x^2}+\lim_{x\to 0}\cos x\cdot \frac{1-\cos 2x\dots\cos nx} {x^2}$$ which is same as $$\frac {1} {2}+\lim_{x\to 0} \frac{1-\cos 2x\dots\cos nx} {x^2}$$ Applying same technique we see that the above is equal to $$\frac{1}{2}+\frac {2^2} {2}+\lim_{x\to 0} \frac{1-\cos 3x\dots\cos nx} {x^2}$$ and continuing in same fashion we see that the desired limit is equal to $$\frac{1^2+2^2+\dots+n^2}{2}=\frac{n(n+1)(2n+1)}{12}$$
A: You can use L'Hopital's rule: $$\lim_{x\rightarrow 0}\frac{1-\prod_{i=1}^n(\operatorname{cos}(ix))}{x^2}=\lim_{x\rightarrow 0}\frac{\frac{d(1-\prod_{i=1}^n(\operatorname{cos}(ix)))}{dx}}{\frac{d(x^2)}{dx}}=\lim_{x\rightarrow 0}\frac{\sum_{i=1}^ni\operatorname{sin}(ix)\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))}{2x}$$ 
You can derive the above expression by noting that, by applying the product rule to $$\prod_{i=1}^n(\operatorname{cos}(ix)))$$ 
the result will be a sum of terms of the form $$\operatorname{cos}(x)\operatorname{cos}(2x)...(\operatorname{cos}(ix))'...\operatorname{cos}(nx)=\operatorname{cos}(x)\operatorname{cos}(2x)...(-i\operatorname{sin}(ix))...\operatorname{cos}(ix)$$ 
Now, using the familiar limit $$\lim_{x\rightarrow 0}\frac{\operatorname{sin}(x)}{x}=1$$ 
We simplify as follows:
$$\lim_{x\rightarrow 0}\frac{\sum_{i=1}^ni\operatorname{sin}(ix)\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))}{2x}=\lim_{x\rightarrow 0}\sum_{i=1}^ni\Big(\frac{\operatorname{sin}(ix)}{2x}\Big)\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))=\\\lim_{x\rightarrow 0}\frac{1}{2}\sum_{i=1}^ni\Big(\frac{\operatorname{sin}(ix)}{x}\Big)\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))=\lim_{x\rightarrow 0}\frac{1}{2}\sum_{i=1}^ni^2\Big(\frac{\operatorname{sin}(ix)}{ix}\Big)\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))=\\\lim_{x\rightarrow 0}\frac{1}{2}\sum_{i=1}^ni^2\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))=\frac{1}{2}\sum_{i=1}^ni^2\bigg(\lim_{x\rightarrow 0}\prod_{j=1,j\neq i}^n(\operatorname{cos}(jx)))\bigg)=\frac{1}{2}\sum_{i=1}^ni^2\bigg(\prod_{j=1,j\neq i}^n(1)\bigg)=\\\frac{1}{2}\sum_{i=1}^ni^2*1=\frac{1}{2}\sum_{i=1}^ni^2$$ 
As Rhys Hughes notes in the comments, the formula for the sum of consecutive squares up to $n$, $\sum_{i=1}^ni^2$, is given by $\frac{n(n+1)(2n+1)}{6}$. You can find many proofs of this online. The easiest (which you can try for yourself) is by induction. Since, the limit is half of this sum, it equals $\frac{n(n+1)(2n+1)}{12}$ as desired.
A: Multiply by conjugate:
$$\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}=\\
\lim_{x \rightarrow 0} \frac{1-\cos^2 x\cos^2 2x\cdots \cos^2 nx}{x^2\cdot(1+\cos x\cos 2x\cdots \cos nx)}=\\
\frac12 \lim_{x \rightarrow 0} \frac{1-(1-\sin^2 x)(1- \sin^2 2x)\cdots (1-\sin^2 nx)}{x^2}=\\
\frac12 \lim_{x \rightarrow 0} \frac{(\sin^2 x+\sin^2 2x+\cdots +\sin^2 nx)-(\sin^2 x \sin^2 2x +\cdots)+\cdots}{x^2}=\\
\frac12 \left(1+2^2+\cdots +n^2\right)=\frac{n(n+1)(2n+1)}{12}.$$
Note: The expansion of the brackets is performed by the elementary symmetric polynomial rule (see also here) and the limits of the degrees $4$ and higher when divided by $x^2$ will vanish.
