I want to show that the sequence defined by $x_{n+1}=\sqrt{2-x_n}$ with $x_0=\frac{2}{3}$ converges. To be honest, I have no idea how to prove that because this sequence is not even monotonic but it is bounded. Thanks in advance.
Show that the sequence $x_{n+1}=\sqrt{2-x_n}$ converges
cauchy-sequencesreal-analysissequences-and-series
Best Answer
It may help to know what this converges to (if it converges). The limit would have to fulfil $$ x=\sqrt{2-x},$$ i.e., be a solution of $x^2+x-2=0$. So $x=\frac{-1\pm \sqrt 9}2$, but only the positive solution (that is $1$) makes sense.
So we may try to show that the distance to $1$ decreases. Let $y_n=x_n^2-1$. Then
$$y_{n+1}=x_{n+1}^2-1=(2-x_n)-1=1-x_n=-\frac{y_n}{1+x_n} $$ and as certainly $x_n>0$, we conclude that $|y_{n+1}|<|y_n|$, i.e., $|y_n|$ is strictly decreasing. This also implies that $|y_n|\le |y_0|=\frac59$, so $x_n\ge\sqrt{1-\frac59}=\frac23$, and we find the fetter inequality $$|y_{n+1}|\le \frac{3}{5}|y_n|$$ so that $y_n\to 0$ at least geometrically.