Prove convergence of continuous fraction

Question

I am studying real analysis and I have to show that the following recursively defined sequence $$x_{1}=\frac{1}{2} ; x_{n+1}=\frac{1}{2+x_{n}}$$ converges to $L = \sqrt{2} - 1$. My intention was first to show that the sequence converges to later find the limit using $ L = \frac{1}{L+2} $, however I don't know how to prove that the sequence converges since I already know that it is not monotone so I can't use the Monotone Convergence Theorem. I would appreciate any hints so I can try to solve this problem, thanks.

Answer 1

$x_{n+1} =\frac{1}{2+x_{n}} $.

Let $c = \sqrt{2}-1$ and $y_n = x_n-c $.

Showing $x_n \to c$ is the same as $y_n \to 0$.

We have $y_{n+1}+c =\frac{1}{2+y_{n}+c} $ or, since $c^2 = 3-2\sqrt{2}$,

$\begin{array}\\ y_{n+1} &=\dfrac{1}{2+y_{n}+c}-c\\ &=\dfrac{1-c(2+y_{n}+c)}{2+y_{n}+c}\\ &=\dfrac{1-2c-c^2-2cy_{n}}{2+y_{n}+c}\\ &=\dfrac{1-2(\sqrt{2}-1)-(3-2\sqrt{2})-2cy_{n}}{2+y_{n}+c}\\ &=\dfrac{-2cy_{n}}{\sqrt{2}+1+y_{n}}\\ \end{array} $

Since $0 < 2c < 1$, if $|y_n| < 1$ then $|y_{n+1}| \lt \dfrac{|y_n|}{\sqrt{2}} $ so $y_n \to 0$.

Since $x_1 = \frac12$, $y_1 =\frac12-(\sqrt{2}-1) =\frac32-\sqrt{2} \approx 1.5-1.414 =0.086 $ so $|y_1| < \frac12 \lt 1$.

Therefore $y_n \to 0$.

Rigorously, $|\frac32-\sqrt{2}| < 1$ is the same as $-1 \lt \frac32-\sqrt{2} \lt 1 $ which is the same as $(\sqrt{2}\lt \frac52)$ and $(\sqrt{2} \gt \frac12)$ both of which follow from squaring.

Answer 2

Considering the form of the potential limit let set $u_n=x_n+1$

$u_1=x_1+1=\frac 32$ and $u_{n+1}=x_{n+1}+1=\frac{1}{2+u_n-1}+1=\frac{2+u_n}{1+u_n}$

Now the trick is to use $m^2=2$ so that $m=\sqrt{2}$ is the limit of $u_n$. (we are not interested in the negative one, since $u_n\ge 0$ is easy to prove).

Notice that $u_{n+1}-m=\frac{m^2+u_n}{1+u_n}-m=-\frac{(m-1)(u_n-m)}{1+u_n}$

In particular $\dfrac{u_{n+1}-m}{u_n-m}<0$ this means that the values of $u_n$ are one side of the limit then the other side (you have a spiral convergence).

So now prove $u_{2n}$ and $u_{2n+1}$ are monotone adjacent sequences (one $\searrow$, the other $\nearrow$) bounded (lower or upper) by $m$ and thus convergent to $m$.

Hint: study the sign of $f(f(u))-u=(m^2-u^2)\times(\text{positive stuff})$ still with $f(u)=\frac{m^2+u}{1+u}$

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