Limit of $a_{n}^{2}$ when the sequence is recursive

Question

The problem states:

Given the sequence:

$$a_0 \in \left(0,1\right)$$ $$a_{n+1}= 1 - \sqrt {1-a_{n}^{2}}$$

First part.

Check whether it converges, and if so find its limit.

Second part.

Find: $$\lim_{n \to \infty} \frac {a_{n}^{2}}{a_{n+1}}$$

The first part is clear to me, and I've done it. I've proven that the sequence is bounded (0 from below and 1 from above), and also that it is monotone decreasing, so the sequence converges to 0.

The second part is what confuses me. If the sequence is given recursively, how do I know what $a_n$ even is? Furthermore, how do I put it in a limit that I'm supposed to evaluate?

Any ideas?

Thanks.

Answer 1

I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $L\geqslant 0$. In fact, in this case $L$ must be $0$, but this must be deduced, for instance via

\begin{align} L = 1 - \sqrt{1-L^2} &\implies (L-1) = -\sqrt{1-L^2} \\ &\implies (L-1)^2 = 1-L^2 \\ &\implies 2L^2-2L = 0 \\&\implies L(L-1) = 0 \implies L=0 \text{ or } L=1. \end{align}

---

As for the second part, consider that your recurrence relation implies

\begin{align} &(a_{n+1}-1) = -\sqrt{1-a_n^2} \\ \implies &(a_{n+1}-1)^2 = 1-a_n^2 \\ \implies &a_n^2 = 1 - (a_{n+1}^2-2a_{n+1}+1) = 2a_{n+1} - a_{n+1}^2. \end{align}

It follows that

\begin{align} \lim_{n\to\infty}\frac{a_n^2}{a_{n+1}} &= \lim_{n\to\infty}\frac{2a_{n+1} - a_{n+1}^2}{a_{n+1}} \\&= \lim_{n\to\infty} 2 - a_{n+1} = 2, \end{align}

where the last step follows from your previous calculation.

Answer 2

Since $a_n \to 0$ we have

$$\frac{a_n^2}{a_{n+1}}=\frac{a_n^2}{1-\sqrt{1-a_n^2}}=\frac{a_n^2}{1-\sqrt{1-a_n^2}}\,\frac{1+\sqrt{1-a_n^2}}{1+\sqrt{1-a_n^2}}=$$$$=\frac{a_n^2\left(1+\sqrt{1-a_n^2}\right)}{1-1+a_n^2}=1+\sqrt{1-a_n^2}\to 2$$