In this problem we use mathematical induction and trigonometric manipulations. Finally we also compute the limit of the resulting sequence.
Let $a_0=\frac{1}{2}$ and $a_n$ be defined inductively by $a_n=\sqrt{\frac{1+a_{n-1}}{2}}$, $n \geq 1.$
1. Show that for $n=0,1,2, \dots a_n=cos\theta_n$ for some $0<\theta_n<\pi/2$ and determine $\theta_n$.
2. Using (a) or otherwise, calculate $lim_{n \to \infty} 4^n\left(1-a_n\right)$
Using Inductive hypothesis and basic trigonometry, try to deduce the $n$-th term of the sequence.
Deduce the limit after substituting the $n$-th term that is found previously.
Part a
Here $a_o =\frac{1}{2}=cos\frac{\pi}{3}=cos\theta_{0}$(say).
Now, Suppose $a_{n-1}=cos \theta_{n-1}$ where $0<\theta_n<\frac{\pi}{2}$.
Now, $a_n=\sqrt{\frac{1+a_{n-1}}{2}}$
$=\sqrt{\frac{1+cos \theta_{n-1}}{2}}$
$=\sqrt{\frac{2cos^{2} \theta_{n-1}}{2}}$
$=cos \frac{\theta_{n-1}}{2}$
So, $\theta_n=\frac{\theta_{n-1}}{2}$
Again, $\theta_n=\frac{\theta_{n-1}}{2}=\frac{\theta_{n-2}}{2^{2}}$=...=$\frac{\theta_{0}}{2^{n}}$.
So, $\theta_n=\frac{\pi}{3.2^{n}}$.
Part b
$\lim_{n\to\infty} 4^n\left(1-a_n\right)$
$ =\lim_{n\to\infty}2^{2n}.\{1-cos\frac{\pi}{3.2^{n}}\} $
$ =\lim_{n\to\infty}2^{2n}.2sin^{2}\frac{\pi}{3.2^{n+1}}$
$ = \lim_{\frac{1}{2^{n+1}} \to 0} \left(\frac{sin\frac{\pi}{3.2^{n+1}} }{\frac{\pi}{3.2^{n+1}}}\right)^{2} . \left(\frac{\pi}{3}\right)^{2} $
$=\frac{1}{2}.\frac{{\pi}^2}{9}$
$=\frac{{\pi}^2}{18}.$
