Prove that $\lim_{n\to \infty} a_n=\frac{\sqrt{5}-1}{2}$ if $a_{n+1}=\sqrt{1-a_n}$ and $0<a_0<1$.

Question

If $0<a_0<1$ and $a_{n+1}=\sqrt{1-a_n}$, prove that: $$\lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$$

Here what I do is that when $n\to \infty$, $a_{n+1}=a_n$

$\therefore a_n=\sqrt{1-a_n}\Rightarrow a_{n}^2+a_n-1=0$

which implies that when $n\to \infty$, $a_n\to \dfrac{\sqrt{5}-1}{2}$

$\therefore \lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$

My doubt here is, was I correct in simply assuming that $a_{n+1}=a_n$ when $n\to \infty$? (I'm sorry for I may be weak in basics.)

If not, then please suggest alternative method.

Answer 1

You have the right idea but you are building your house on a foundation of mud.

There is not any $n$ where $a_{n+1} = a_n$. So it is not true that $n\to \infty; a_{n+1} = a_n$. It isn't even true that $n\to \infty$ that $a_{n+1} \to a_n$. (In the latter the $n$ sleeps away to infinity so there is no fixed $a_n$ for $a_{n+1}$ to approach.)

What is really going on is that !!!IF!!! (put a huge pin in that "if") the $\lim_{n\to \infty} a_n$ exist and $\lim_{n\to \infty} a_n = L$. Then for any continuous function $f$ then $\lim_{n\to \infty} f(n)=f(L)$.

And so as $a_{n+1}=\sqrt{1-a_n}$ then $\lim_{n\to \infty} a_{n+1} = \sqrt{1-L}$.

But note: If $\lim_{n\to \infty} a_n$ exists then $\lim_{n\to \infty} a_{n}=\lim_{n\to \infty}a_{n+1}$

[Slight diversion: If $\lim_{n\to \infty} a_n = L$ that means, by definition that for any $\epsilon > 0$ there exists and $N_1$ so that for every $n > N_1$ we have $|a_n - L|< \epsilon$ That means if $N_2 = N_1 + 1$ then for every $n > N_2$ then $n+1 > N_1$ and $|a_{n+1} - L| < \epsilon$. So that means $\lim_{n\to\infty} a_{n+1} = L$ as well. So $\lim_{n\to\infty} a_{n+1}=\lim_{n\to \infty} a_n$]

[Another (casual) way of thinking of it is: $a_n \to L$ means the sequence gets infinitely close to $L$ so all the terms are close. So the the terms with the indexing off by just a few will also get close to $L$.]

So $L = \lim_{n\to \infty} a_{n} = \lim_{n\to \infty}a_{n+1} = \sqrt{1-L}$

So $L = \sqrt{1-L}$ and if we solve for $L$ we geta $L =\frac {\sqrt 5-1}2$

And so we are done. $\lim_{n\to \infty} a_n = L = \frac {\sqrt 5 - 1}2$

.....

Except for that pin.

We don't know that $\lim_{n\to \infty} a_n$ exists.

Consider $b_n = 2^n$. Then if $\lim_{n\to \infty} b_n = L$ we would have that $b_{n+1} = 2^{n+1} = 2\cdot 2^n = 2b_n$ so $L = 2L$ so $L = 0$. SO $\lim_{n\to \infty} 2^n = 0$. That's obviously absurd.

But we must prove $\lim_{n\to \infty} a_n$ does exists first.

But if it exists the limit, $L$, must be $\frac {\sqrt 5 -1}2$

Proving that $\lim a_n$ exists really is the heart of the matter.

Which isn't all that easy..... See Michael Rozenbergs answer.

Answer 2

hint It is easy to prove by induction that for all $ n\ge 0$, $0\le a_n\le 1$.

Let $$f(x)=\sqrt{1-x}$$ from $ [0,1)] $ to $[0,1)$.

$$f'(x)=\frac{-1}{2\sqrt{1-x}}<0$$ $ f$ is decreasing at $ [0,1) $, so the subsequences $ (a_{2n})$ and $( a_{2n+1})$ are monotonic and convergent.

I let you to prove they have the same limit.

Answer 3

$$a_{n+1}-\frac{\sqrt5-1}{2}=\frac{\frac{\sqrt5-1}{2}-a_n}{\sqrt{1-a_n}+\frac{\sqrt5-1}{2}}$$ and $$a_{n+2}-a_n=\sqrt{1-\sqrt{1-a_n}}-a_n=\frac{a_n\sqrt{1-a_n}\left(\frac{\sqrt5-1}{2}-a_n\right)\left(\frac{\sqrt5+1}{2}+a_n\right)}{(\sqrt{1-\sqrt{1-a_n}}+a_n)((1+a_n)\sqrt{1-a_n}+1)}.$$ Now, let $a_1<\frac{\sqrt5-1}{2}.$

Thus, for any odd $n$ we have $a_n<\frac{\sqrt5-1}{2}$ and $a_{n+2}>a_n$,

which says that for odd $n$ there is $\lim\limits_{n\rightarrow+\infty}a_n$ and by your work

(we just need to solve the following equation: $a=\sqrt{1-\sqrt{1-a}}.$) it's equal to $\frac{\sqrt5-1}{2}.$

Now, $$a_2=\sqrt{1-a_1}>\sqrt{1-\frac{\sqrt5-1}{2}}=\frac{\sqrt5-1}{2},$$ which says that for even $n$ we have $a_n>\frac{\sqrt5-1}{2}$ and $a_{n+2}<a_n$,

which gives that for even $n$ there is $\lim\limits_{n\rightarrow+\infty}a_n.$

Let this limit is equal $b$.

Thus, since $$a_{n+2}=\sqrt{1-\sqrt{1-a_n}}$$ and $f(x)=\sqrt{x}$ is a continuous function, we obtain: $$b=\sqrt{1-\sqrt{1-b}},$$ which gives $$b\in\left\{0,1,\frac{\sqrt5-1}{2}\right\}.$$ Since, $a_{n+2}<a_n$, we see that $b\neq1$ and since, $a_n>\frac{\sqrt5-1}{2},$ we see that $b\neq0$.

Thus, $b=\frac{\sqrt5-1}{2}.$

Can you end it now?