If $0<u_1<1$ and $u_{n+1}=1-\sqrt{1-u_n}$ for $n\geq 1$ prove that sequence is convergent

Question

If $0<u_1<1$ and $u_{n+1}=1-\sqrt{1-u_n}$ for $n\geq 1$ prove that

1. sequence $\{u_n\}$ converges to $0$
2. $\lim_{n\rightarrow \infty} \frac{u_{n+1}}{u_n}=\frac{1}{2}$

I know that Monotonic Increasing + Bdd above implies convergent

But don't know how to prove those for this particular problem. Unable to show $u_{n+1}>~ or ~< u_n$ Also unable to solve the 2nd one. Please help.

Answer 1

By induction we can show: $0 < u_n < 1$ for all $n \ge 1$. Thus: $u_{n+1} = \dfrac{u_n}{1+ \sqrt{1- u_n}} \implies \dfrac{u_{n+1}}{u_n} = \dfrac{1}{1+\sqrt{1- u_n}} \to \dfrac{1}{1+1} = \dfrac{1}{2}$ as $u_n \to 0$. To show $u_n \to 0$ observe that: $ 0 < u_{n+1} < u_n$ means $\{u_n\}$. is monotonically decreasing and bounded below by $0$ thus converges to $x$.So $x = 1 - \sqrt{1-x} \implies x = 0,1$. But $u_n < u_1 \implies x \le u_1 < 1 \implies x < 1 \implies x = 0.$ To show that $0 < u_n < 1$ by induction, we have $0 < u_1 < 1$ as given. So assume $0 < u_n < 1 \implies 0 < 1-u_n < 1 \implies 0 <\sqrt{1-u_n} < 1\implies 0 < 1 - \sqrt{1-u_n} < 1 \implies 0 < u_{n+1} < 1 \implies 0 < u_n < 1, \forall n \ge 1$.

Answer 2

prove that $$u_n>0$$ for all $n$ and show that $$u_{n+1}<u_n$$ this is equivalent to $$1-\sqrt{1-u_n}<u_n$$ can you proceed?

Answer 3

Let $f(x)=1-\sqrt{1-x}$ for $0\leq x\leq 1$. We have

• $\forall x\in[0,1]\;\; f(x)\in [0,1]$ $\implies \forall n\in \mathbb N \;\;0\leq u_n \leq 1$

• $f'(x)=\frac{1}{2\sqrt{1-x}}>0$ $\implies (u_n) $ monotonic.

• $f(x)-x=-x\sqrt{1-x}\leq0 $ $\implies (u_n)$ is decreasing

thus, $(u_n)$ converges to the fixed point $0$.

for the second

$$1-u_{n+1}=(1-u_n)^{\frac{1}{2}}=(1-u_0)^{\frac{1}{2^{n+1}}}$$

$$\implies \frac{u_{n+1}}{u_n}=\frac{1-(1-u_0)^{\frac{1}{2^{n+1}}}}{1-(1-u_0)^{\frac{1}{2^n}}}$$

$$\frac{1}{1+(1-u_0)^{ \frac{1}{2^{n+1}} } }$$

and the limit is $$\frac{1}{2}.$$

For the last simplification, we used

$$\frac{1-A}{1-A^2}=\frac{1}{1+A}$$

Answer 4

Alternative method for the first part.

If $u_{n+1} = 1 - \sqrt{1-u_n}$ then $1-u_{n+1} = \sqrt{1-u_n}$. Defining $v_n := 1-u_n$, we see that $v_n = \sqrt{v_{n-1}} = v_{n-2}^{\frac12· \frac12}= … = v_0^{\frac{1}{2} … \frac12} = v_0^{1/(2^{n})}$. This is a subsequence of the well-known convergent sequence $v_0^{1/k} → 1$; hence $v_n → 1$, and hence $u_n → 0$.