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I am trying to prove the sequence (for n = 1,2,3...) $$a_{1} = 1,\,a_{n+1} = \frac{1}{1+a_{n}}$$ converges and then find its limit. I can find its limit easily but I cannot find a way to prove that it converges. I know there exists a theorem stating that a monotone increasing and bounded sequence has a limit. I do not think this sequence is increasing. Also, how can I apply the theorem $$\lim_{x\to\infty} q^{n}=0 \,{if} |q| < 1$$ to this question? I think it is irrelevant but it is good to know different approaches to prove convergence of a sequence. Thank you.

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marked as duplicate by rtybase, Community Oct 19 '17 at 19:30

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$$a_{n+1}-\frac{\sqrt5-1}{2}=\frac{1}{1+a_n}-\frac{2}{\sqrt5+1}=\frac{2\left(\frac{\sqrt5-1}{2}-a_n\right)}{\sqrt5+1)(1+a_n)}.$$ Thus, for all $n\geq2$ we obtain $$\left|a_n-\frac{\sqrt5-1}{2}\right|=\frac{2\left|\frac{\sqrt5-1}{2}-a_{n-1}\right|}{\sqrt5+1)(1+a_{n-1})}<$$ $$<\frac{2}{\sqrt5+1}\cdot\left|\frac{\sqrt5-1}{2}-a_{n-1}\right|<...<\left(\frac{2}{\sqrt5+1}\right)^{n-1}\left|\frac{\sqrt5-1}{2}-a_1\right|\rightarrow0$$ because $\frac{2}{\sqrt5+1}<1$.

Id est, $$\lim_{n\rightarrow+\infty}a_n=\frac{\sqrt5-1}{2}$$ and we are done!

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  • $\begingroup$ How do I determine the numerical value as you mentioned in your post? $\endgroup$ – Joseph Oct 15 '17 at 5:55
  • $\begingroup$ @Hugh What do you mean? Maybe do you want $\frac{\sqrt5-1}{2}=0.618...$? $\endgroup$ – Michael Rozenberg Oct 15 '17 at 5:58
  • $\begingroup$ yes. Thank you, Michael. I am thinking I may assume the sequence converges to L, then I have $\lim_{x\to\infty} a_{n+1} = \lim_{x\to\infty} a_{n} = L$, then solve for L. Am I right? $\endgroup$ – Joseph Oct 15 '17 at 6:00
  • $\begingroup$ @Hugh I just calculated the limit by definition of the limit by reasoning, which you looked for. If we found the limit then the sequence converges automatically. $\endgroup$ – Michael Rozenberg Oct 15 '17 at 6:08
  • $\begingroup$ @Hugh Yes, of course if $a$ converges to $L$ then $\lim\limits_{n\rightarrow+\infty}a_{n}=\lim\limits_{n\rightarrow+\infty}a_{n+1}=L.$ But for this way you need to prove the convergence. $\endgroup$ – Michael Rozenberg Oct 15 '17 at 6:38
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One way to proceed is $a_{2n} - a_{2(n-1)} = \dfrac{1}{1+a_{2n-1}} - \dfrac{1}{1+a_{2n-3}}= - \dfrac{a_{2n-1} - a_{2n-3}}{(1+a_{2n-1})(1+a_{2n-3})}= \dfrac{a_{2n-2} - a_{2n-4}}{K}, K > 0$. Thus using induction you can prove $\{a_{2n}\}$ is a convergent sub-sequence ( either bounded above or below you can check initial values of $a_2, a_4$. Similarly you can find a similar expression for $a_{2n-1}- a_{2n-3}$ and by induction and Bolzanos theorem you can show $\{a_{2n-1}\}$ is a convergent sub-sequence. There is a theorem that you would have to prove that this lead to $\{a_n\}$ is also convergent, say to $L$,and you can solve for $L = \dfrac{1}{1+L}, 0 < L < 1$ and you can get the answer you are looking for....

Note: Both sub-sequences mentioned above converge to the same limit $L$ indeed as you can show that too. And the theorem I mentioned can be applied and $L = \dfrac{\sqrt{5} - 1}{2}$ .

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Write $f(x)={1\over{1+x}}$, compute it's derivative, it is negative, let $l>0$ such that $f(l)=l$, $a_0\geq l$ implies that $f(a_0)\geq l$ and recursively $f(a_n)\geq l$.

MVT implies that $|f(a_n)-f(a_{n-1})|=|a_{n+1}-a_n|\leq |f'(l)||a_n-a_{n-1}|$, since $f'(l)|<1$, we deduce that $(a_n)$ is a Cauchy sequence and $lim_na_n=l$.

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  • $\begingroup$ Your proof does not work. $a_{n+1} > a_{n+2} \implies a_{n+2} < a_{n+3}$,...etc.. you would have to use mine or Rosenzberg's trick to do it. $\endgroup$ – DeepSea Oct 16 '17 at 6:37
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As I saw the task

$a_{n+1}$=$\frac{1}{1+{a_n}}$=$\frac{1}{1+\frac{1}{a_{n-1}}}$=$\frac{1}{1+\frac{1}{1+\frac{1}{a_{n-2}}}}$ ...... making common denominators:

$a_{n+1}$=$\frac{1}{1+{a_n}}$=$\frac{1+a_{n-1}}{2+{a_{n-1}}}$=$\frac{2+a_{n-2}}{3+{a_{n-2}}}$=$\frac{3+a_{n-3}}{5+{a_{n-3}}}$=.....=$\frac{F_{n}+a_{n-n+1}}{F_{n+1}+{a_{n-n+1}}}$ (=$\frac{F_{n}+1}{F_{n+1}+1}$),

where $F_n$ the n. Fibonacci number.

$\lim_{n\rightarrow+\infty}{(a_{n+1})}$= $\lim_{n\rightarrow+\infty}$($\frac{1+{1\over{F_{n}}}}{1+{1\over{F_{n-1}}}}$)$\frac{F_{n}}{F_{n+1}}$=$\frac{\sqrt5-1}{2}$

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