TITLE: How find the value $\lim_{x\to 0}\dfrac{1}{x^2}\left(1-\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\right)$
QUESTION [8 upvotes]: find the value
$$I_{n}=\lim_{x\to 0}\dfrac{1}{x^2}\left(1-\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\right)$$
This is my methods:
\begin{align*}I_{n+1}-I_{n}&=\lim_{x\to 0}\dfrac{1}{x^2}\left(\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\left(1-\sqrt[n+1]{\cos{(n+1)x}}\right)\right)\\
&=\lim_{x\to 0}\dfrac{1-\sqrt[n+1]{\cos{(n+1)x}}}{x^2}\\
&=\dfrac{n+1}{2}
\end{align*}
so
$$I_{n}=\dfrac{n(n+1)}{4}$$
Have you other nice methods? Thank you

REPLY [10 votes]: $$
\cos(nx) = 1 - \frac{n^2 x^2}{2} + O(x^4)
$$
as $x \to 0$ so
$$
\sqrt[n]{\cos(nx)} = 1 - \frac{n x^2}{2} + O(x^4)
$$
by the binomial theorem, thus
$$
\begin{align}
\cos(x)\sqrt{\cos(2x)}\cdots\sqrt[n]{\cos(nx)} &= 1 - \frac{x^2}{2} \sum_{k=1}^{n} k + O(x^4) \\
&= 1 - \frac{n(n+1)}{4} x^2 + O(x^4).
\end{align}
$$