How to show $x_n=2^na_n$ converges, where $a_{n+1}=\sqrt{\frac{1-\sqrt{1-a_n^2}}{2}}$
The question originated from Professor David McKinnon.
Attempt:
I did prove it is increasing but failed to show it is bounded above. The sequence should converge to $\frac{\pi}{2}$, if I'm not mistaking.
In addition, ${a_n}$ is a decreasing sequence, and it is bounded below by 0, so I tried to use the fact that ${a_n}$ is a Cauchy to prove that ${x_n}$ is also a Cauchy.
Best Answer
Since $$\frac{a_{n+1}^2}{a_n^2} = \frac{1 - \sqrt{1-a_n^2}}{2a_n^2} = \frac{1}{2(1+\sqrt{1-a_n^2})} \le \frac{1}{2}$$
We have
$$a_n^2 = a_0^2 \prod_{k=1}^n \frac{a_k^2}{a_{k-1}^2} \le 2^{-n} \quad\implies\quad \sum_{k=0}^\infty a_k^2 \le \sum_{k=0}^\infty 2^{-k} = 2$$
Since
$$\frac{x_{n+1}^2}{x_n^2} = \frac{4a_{n+1}^2}{a_n^2} = \frac{2}{1+\sqrt{1-a_n^2}} = 1 + \frac{1-\sqrt{1-a_n^2}}{1+\sqrt{1-a_n^2}} = 1 + \frac{a_n^2}{(1+\sqrt{1-a_n^2})^2}$$
We find
$$1 \le \frac{x_{n+1}^2}{x_n^2} \le 1 + a_n^2 \le e^{a_n^2}$$
LHS tells us the sequence $(x_n)$ is increasing while RHS tells us it is bounded from above. $$x_n^2 = x_0^2 \prod_{k=1}^n \frac{x_k^2}{x_{k-1}^2} \le \prod_{k=1}^n e^{a_{k-1}^2} = e^{\sum_{k=0}^{n-1} a_k^2} < e^{\sum_{k=0}^\infty a_k^2} = e^2$$
As a result, sequence $(x_n)$ converges to some number $\le e$.