Find limit of $(a_n)$ defined by $a_1 = 1$ and $a_{n+1} = \frac{1}{3+a_n}$

Question

Let $(a_n)$ be the sequence defined by $a_1 = 1$ and $a_{n+1} = \frac{1}{3+a_n}$ for all $n \in \mathbb N.$ Show that $(a_n)$ converges and find its limit.

I have no idea on how to handle this though so I would definitely appreciate some help for a rigorous proof.

For part b are you able to show that the sequence is bounded above by $\frac 13$ and below by $0$? — abiessu, Sep 02 '18 at 20:45
Part $b)$ of your post is a question found in Mathematics Magazine about 2 years ago. Look that up. To do that sequence, you have to consider subsequence... — DeepSea, Sep 02 '18 at 20:49
@Shaun Its an exercise consisted of 2 parts and I only asked one question. Please review the posts better before commenting (and potentially downvoting). — Rebellos, Sep 02 '18 at 20:51
@DeepSea A subsequence approach seems rather interesting ! I'll check it up, thanks ! — Rebellos, Sep 02 '18 at 20:52

Eclipse Sun · Accepted Answer · 2018-09-02T21:03:14.787

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First we note that $a_n\in (0,1/3)$ for $n>1$.

Since $$a_{n+1}-a_{n}=\frac{1}{3+a_n}-\frac{1}{3+a_{n-1}}=\frac{a_{n-1}-a_n}{(3+a_n)(3+a_{n-1})},$$ and $(3+a_n)(3+a_{n-1})>9$, we see that $$|a_{n+1}-a_n|<\frac19 |a_n-a_{n-1}|.$$ This shows that $\{a_n\}$ is Cauchy, so it converges to some number $a\in[0,1/3]$. To find $a$, take limits on both sides of $$a_{n+1}=\frac1{3+a_n}.$$

edited Sep 02 '18 at 21:03

answered Sep 02 '18 at 20:57

Eclipse Sun

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@Rumpelstiltskin I missed that. It is true when $n>1$. – Eclipse Sun Sep 02 '18 at 21:03
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Yes I saw that, I deleted my comment. You could also use Banach's theorem about contractions here, maybe you should mention that. – Jakobian Sep 02 '18 at 21:08

score 1 · Answer 2 · answered Sep 02 '18 at 21:11

Using contraction mapping technique. We will look at the function $f(x)=\frac{1}{3+x}$ s.t. $a_{n+1}=f(a_n)$.

$\forall x \in \left[0,1\right] => f(x) \in \left[0,1\right]$. Indeed $$0\leq x\leq 1 \Rightarrow 3\leq 3+x\leq 4 \Rightarrow \frac{1}{3}\geq \frac{1}{3+x}\geq \frac{1}{4}>0$$ or $$1\geq f(x)\geq 0$$
$f(x)$ is a contraction mapping on $\left[0,1\right]$, from MVT $\forall x,y \in \left[0,1\right], \exists c$ in between them s.t. $$|f(x)-f(y)|=|f'(c)||x-y|=\left|-\frac{1}{(3+c)^2}\right||x-y|\leq\frac{1}{9}|x-y|$$

Since $a_1 \in \left[0,1\right]$, from Banach fixed-point theorem, the sequence has a limit on $\left[0,1\right]$ which you can find from $L=\frac{1}{3+L}$.

Find limit of $(a_n)$ defined by $a_1 = 1$ and $a_{n+1} = \frac{1}{3+a_n}$

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