\({\color{Yellow} .} \displaystyle\lim_{n \to +\infty }\; \left [4^n\cdot \left ( 1-cos\, \frac{1}{2^n} \right ) \right ] \quad \begin{matrix} {\color{Yellow} .} \\ _{(1-\cos 2\alpha =2\sin^2\alpha )} \end{matrix} \\ =\displaystyle\lim_{n \to +\infty }\;\left [ 4^n\cdot 2\cdot \sin ^2\frac{1}{2^{n+1}} \right ] \\ =2\cdot \displaystyle\lim_{n \to +\infty }\;\left [ 4^n\left ( \frac{1}{2^{n+1}}\cdot \frac{\sin \frac{1}{2^{n+1}}}{\frac{1}{2^{n+1}}} \right ) ^2\right ] \\e =2\cdot \displaystyle\lim_{n \to +\infty }\;\left [ 4^n\cdot \frac{1}{2^{2n+2}}\cdot 1^2 \right ] \\ =2\cdot \displaystyle\lim_{n \to +\infty }\;\frac{4^n}{4^n\cdot 4}=\frac{1}{2}\)