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.

Answer 1

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 $$a_{n+1}=\frac1{3+a_n}.$$

Answer 2

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

1. $\forall x \in \left[0,1\right] => f(x) \in \left[0,1\right]$. 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 $$1\geq f(x)\geq 0$$
2. $f(x)$ 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 $L=\frac{1}{3+L}$.