Suppose that a sequence $(x_n )$ in $R$ satisfies
$x_{n+1} = 1 −\sqrt{1 − x_n}$
for all n ∈ N. Show that $(x_n )$ converges. To what does it converge? Does
$(x_{
n+1}/
x_{n})
$
converge?
I have solved the first and found that except for $x_1=1$, all other initial values makes the sequence converge to $0$. For the second part, (if the ratio converges) : if the initial value is 1, the ratio for sure converges to $1$ as all terms are $1$. If the initial value is $0$, it doesn't as none of the terms of the ratios are defined. For the intermediate values I am not sure how to prove that the ratio converges to some number( which my intuition tells me to be 1)? As I can't use ratios of limit result as the limits are $0$ $$$$ Constraint: Can't use taylor series as the book didn't reach there yet.
Best Answer
If the sequence converges to some limit $z$, it must satisfy $$ z = 1-\sqrt{1-z} \Rightarrow (1-z)^2 = 1-z \Leftrightarrow z=0 \vee z=1. $$ So, if the sequence converges, if must converge to either $0$ or $1$. In fact, the sequence will converge to $0$, unless $x_0=1$, in which case it will converge to 1.
Now, if $x_0<1$,
$$ \lim \frac{x_{n+1}}{x_n} = \lim \frac{1-\sqrt{1-x_n}}{x_n}=\lim\frac{x_n}{x_n (1+\sqrt{1-x_n})}=\frac 12. $$
If $x_0=1$, $\lim \frac{x_{n+1}}{x_n} = \frac{1}{1}=1$.
If $x_0 = 0$, the sequence $\frac{x_{n+1}}{x_n}$ is not even defined.