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{$a_n$} is recursively defined as $a_1=1$ and $a_{n+1}=1+1/a_n$.

(a) Verify that ${a_n}$ is not monotone,
(b) Use the contraction principle to prove that ${a_n}$ converges,
(c) Show that the sequence converges to $\frac{1+\sqrt5}{2}$.

Now I'm having trouble with (a) and (b). I think I know how to solve (c). Since $a_n$ converges, the limits as n goes to infinity of $a_n$ and $a_{n+1}$ should be equal. That gives the equation $A=1+\frac{1}{A}$. This has the solution $\frac{1+\sqrt5}{2}$, which we know from an earlier problem. (And which one can check by substitution).

Now on (a). I don't know how to solve this. It's clear to me that the values go up and down, converging to $\frac{1+\sqrt5}{2}$. However how to prove this?

QuestionMaker
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2 Answers2

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$a_1=1;a_2=2;a_3=\frac{3}{2}$

Here $a_1<a_2$ but $a_2>a_3$..

So it can't be monotone.

EDITS:

Consider the function $f(x)=1+\frac{1}{x}$. Hence $a_{n+1}=f(a_n)$

Check that $|f^{'}(x)|=|\dfrac{-1}{x^2}|\le 1$ as $x\ge 1$ as every term of the sequence $\ge 1$.

Now $|f(x)-f(y)|=|f^{'}(c)||x-y|\le |x-y|$ .So $f$ is a contraction mapping on the complete metric space $\Bbb R$. Hence by Banach Fixed Theorem $f$ has a fixed point .

So there exists $a$ such that $a=1+\dfrac{1}{a}$ which is a point of convergence of $a_n$.

Learnmore
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  • So really just this example is enough? I thought I would have to prove it would never be eventually monotone. Thanks. – QuestionMaker Nov 20 '16 at 17:02
  • Yes. To prove a sequence is monotone, show that for every value of $n$, $a_n < a_{n+1}$ or show that for every value of $n$, $a_n > a_{n+1}$. Negate the logic: to show a sequence is not monotone, show that there exist values $n$ and $m$ such that $a_n < a_{n+1}$ and $a_m > a_{m+1}$. – Browning Nov 20 '16 at 17:10
  • All right seems quite clear! Any idea on how to solve b? I want to proof $|a_{n+2}-a_{n+1}|<k|a_{n+1}-a_n|$ for any $n$, and a certain $0<k<1$. Now $|a_{n+2}-a_{n+1}|=|1+\frac{1}{a_{n+1}}-a_{n+1}|$. How do I continue? – QuestionMaker Nov 20 '16 at 17:28
  • @QuestionMaker;I have added new things .Check it – Learnmore Nov 20 '16 at 17:43
  • Hey, I can follow your reasoning but I don't understand where it's leading. You've proven that the increasement $a_{n+1}-a_n$ is smaller than $1$, but how does that relate to earlier increasements? – QuestionMaker Nov 20 '16 at 19:02
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    Please check back my edits @QuestionMaker – Learnmore Nov 21 '16 at 02:47
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Assume $a_n>\phi$, the golden ratio. What does that tell you about $a_{n+1}?$ what about $a_{n+2}?$ working with your definition of $a_n$, you should get

$$a_{n+1}<\phi<a_{n+2}$$

Doing a bit more effort, you can find that

$$a_{n+1}<\phi<a_{n+2}<a_n$$

So the sequence is clearly not monotone.

By inductive reasoning, you will find

$$a_1<a_3<a_5<\dots<\phi<\dots<a_6<a_4<a_2$$

Simply Beautiful Art
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  • Hey thanks for your response. But I don't think I am supposed to use the Golden Ratio in (a) when you're supposed to prove it in (c). I think it's really as easy as learnmore's answer. But I appreciate your insights! – QuestionMaker Nov 20 '16 at 17:04
  • Well, assume the limit exists. And I think it's ok, you have to start somewhere. – Simply Beautiful Art Nov 20 '16 at 17:06