Proving convergence of a sequence $a_{n+1} = \frac{1}{1+a_{n}}$ and finding the limit.

Question

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.

So, what is the limit? And if the sequence is not increasing, what kind of behaviour has it? — Angina Seng, Oct 15 '17 at 05:17
Hint: consider the auxillary sequence $b_n = \frac{\varphi a_n - 1}{\varphi + a_n}$ where $\varphi$ is the golden ratio. OR consider yet another auxillary sequence $F_n$ such that $a_n = \frac{F_n}{F_{n+1}}$.... — achille hui, Oct 15 '17 at 05:29
Show with $L=\frac{\sqrt{5}-1}{2}$ that $$a_{n+1}-L=\frac{1}{1+a_n}-\frac{1}{1+L}=\frac{L-a_n}{(1+L)(1+a_n)}$$, hence $|a_{n+1}-L|\leq q|a_n-L|$ with $q=1/(1+L)<1$. — Kelenner, Oct 15 '17 at 05:36
Answer here https://math.stackexchange.com/questions/1700421/finding-limit-of-a-sequence-a-n1-frac11a-n seem to address both convergence and limit. Then there is also https://math.stackexchange.com/questions/1544443/prove-that-sequence-a-n1-frac11a-n-is-bounded and https://math.stackexchange.com/questions/38739/convergence-of-a-n1-frac11a-n. — Sil, Oct 15 '17 at 06:21
And https://math.stackexchange.com/questions/435540/solution-of-recurrence-relation-x-n1-frac-1x-n-1, https://math.stackexchange.com/questions/2215002/let-a-00-if-a-n1-frac11a-n-show-that-the-sequence-a-n-converge, https://math.stackexchange.com/questions/332227/did-i-compute-the-limit-of-of-the-sequence-x-n1-fracx-nx-n1-x-o-1-pr, https://math.stackexchange.com/questions/962664/limit-of-a-given-sequence. Then there are couple of ones where $a_1=0$ but this should be enough. — Sil, Oct 15 '17 at 06:23

DeepSea · Answer 1 · 2017-10-15T06:08:08.403

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 $L = \dfrac{\sqrt{5} - 1}{2}$ .

Tsemo Aristide · Answer 2 · 2017-10-16T09:10:29.950

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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 $lim_na_n=l$.

edited Oct 16 '17 at 09:10

answered Oct 15 '17 at 05:25

Tsemo Aristide

89,587

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. – DeepSea Oct 16 '17 at 06:37

Michael Rozenberg · Accepted Answer · 2017-10-15T05:53:44.437

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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!

edited Oct 15 '17 at 05:53

answered Oct 15 '17 at 05:47

Michael Rozenberg

203,855

How do I determine the numerical value as you mentioned in your post? – Joseph Oct 15 '17 at 05:55
@Hugh What do you mean? Maybe do you want $\frac{\sqrt5-1}{2}=0.618...$? – Michael Rozenberg Oct 15 '17 at 05:58
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? – Joseph Oct 15 '17 at 06:00
@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. – Michael Rozenberg Oct 15 '17 at 06:08
@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. – Michael Rozenberg Oct 15 '17 at 06:38
Thank you. I appreciate your help. – Joseph Oct 15 '17 at 06:39
@Hugh You are welcome! – Michael Rozenberg Oct 15 '17 at 06:41

score 0 · Answer 4 · answered Oct 19 '17 at 18:43

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.

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

Proving convergence of a sequence $a_{n+1} = \frac{1}{1+a_{n}}$ and finding the limit.

4 Answers4