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I have a sequence defined as such

$$a_1 = 1 \text{ and } a_{n+1} = \sqrt{1+a_n}$$

And I'm asked to prove that the sequence converges to the golden ratio. My first guess was to try $|a_n - L|<\epsilon$, as per usual when trying to determine convergence, but I don't actually have a definition for any $a_n$ (or do I?). If I rearrange for $a_n$, I get $$a_n = a_{n+1}^2-1$$

But does it make sense to define $a_n$ in terms of its later terms? It doesn't to me.

humdiddlediddle
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2 Answers2

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Here are the steps you need to prove.

  • First prove that $1 \leq a_n < \dfrac{\sqrt5 + 1}2$ using induction.
  • Nest prove that $a_n$ is a monotone increasing sequence using induction.
  • Now recall that due to completeness of the real numbers a monotone increasing sequence bounded above converges. Call the limit as $L$.
  • Make use of limit laws/rules to conclude that $L = \sqrt{1+L}$
  • Prove that $L = \dfrac{\sqrt{5}+1}2$
  • Could you please elaborate a little bit on how to make use of limit laws? I am solving the same question as well, and I have already showed that the sequence is increasing and bounded above. So by the MCT the sequence converges to some limit L. –  Sep 25 '13 at 03:07
  • @mespebjidom By limit laws, I mean the following: If $a_n \to L$, then the following are true: (i) $1+a_n \to 1+L$, (ii) $\sqrt{1+a_n} \to \sqrt{1+L}$ and (iii) $a_{n+1} \to L$. –  Sep 25 '13 at 03:08
  • I am thinking now that if {$a_n+1$} and {$a_n$} converge to the same limit, then by the sequence of arithmetic theorem we have L = sqrt(1 + L), and then we just solve for L for the golden ratio. –  Sep 25 '13 at 03:09
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Notice the golden ratio $\varphi$ is a fix point of the map $x \mapsto \sqrt{x+1}$. If $a_m = \varphi$ for some $m$, then

$$a_n = \varphi\;\;\text{ for all } n \ge m\quad\implies\quad\lim_{n\to\infty} a_n = \varphi$$

We will consider the case what if $a_n \ne \varphi$ for all $n$ below.

One way to check the sequence converges to $\varphi$ is show near $\varphi$, the map $x \mapsto \sqrt{x+1}$ is a contraction mapping and $a_n$ does get close to $\varphi$ for some $n$.

We can do this using mean value theorem

$$|a_{n+1} - \varphi | = | \sqrt{a_n+1} - \sqrt{\varphi+1} | = \frac{1}{2\sqrt{\xi+1}} | a_n - \varphi | \le \frac12 |a_n - \varphi |$$

where $\xi$ is some number between $a_n$ and $\varphi$. We can also do this pure algebraically

$$\frac{a_{n+1}-\varphi}{a_n - \varphi} = \frac{\sqrt{1+a_n} - \sqrt{1+\varphi}}{a_n - \varphi} = \frac{1}{\sqrt{1+a_n}+\sqrt{1+\varphi}} < \frac12$$

Both approaches lead to $$|a_n - \varphi| \le 2^{-(n-1)} |a_1 - \varphi|\;\;\text{ for all }n \quad\implies\quad \lim_{n\to\infty} a_n = \varphi\;\;\text{ again! }$$

achille hui
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