Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}$ How to prove that 
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$

Often when $\sin(x)$ and $x$ appear in a formula such as this, it is based on the fact that
$$
\lim_{a\to0}\frac{\sin(ax)}{a}=x\tag{1}
$$
Since the summands are $\dfrac{4^{-k}}{\cos^2(2^{-k}x)}$ , it appears that an identity involving a double angle formula is being exploited. Therefore, with an eye toward a telescoping sum, I started with
$$
\begin{align}
\frac4{\sin^2(2x)}-\frac1{\sin^2(x)}
&=\frac{4-4\cos^2(x)}{4\sin^2(x)\cos^2(x)}\\
&=\frac{4\sin^2(x)}{4\sin^2(x)\cos^2(x)}\\
&=\frac1{\cos^2(x)}\tag{2}
\end{align}
$$
Substitute $x\mapsto 2^{-k}x$ and multiply by $4^{-k}$ and sum (and telescope)
$$
\begin{align}
\sum_{k=1}^n\frac{4^{-k}}{\cos^2(2^{-k}x)}
&=\sum_{k=1}^n\left(\frac{4^{-k+1}}{\sin^2(2^{-k+1}x)}-\frac{4^{-k}}{\sin^2(2^{-k}x)}\right)\\
&=\frac1{\sin^2(x)}-\frac{4^{-n}}{\sin^2(2^{-n}x)}\\
&\to\frac1{\sin^2(x)}-\frac1{x^2}\tag{3}
\end{align}
$$
since, by $(1)$,
$$
\lim_{n\to\infty}\frac{4^{-n}}{\sin^2(2^{-n}x)}=\frac1{x^2}\tag{4}
$$
