# proving convergence of $a_{n+1}=1+\frac{1}{1+a_{n}}$ [duplicate]

$$a_1=1.$$ $$a_{n+1}=1+\frac{1}{1+a_{n}}$$

Prove that the sequence is convergent.

I'm trying to prove the convergence of this sequence but having trouble. At first I thought this might be a monotone sequence since then I can try monotone convergence theorem to prove its convergence.

But after checking some terms, I realized it seemed the sequence is oscillating. So I'm not sure how to prove the convergence of this sequence.

Thanks.

• The sequence is oscillating, but you should be able to show monotonicity of $a_{2n}$ and (separately) $a_{2n+1}$. Nov 26 '20 at 6:48
• This sequence is Cauchy sequence so it converges Nov 26 '20 at 6:53
• Oh this is cauchy sequence? I will have to check it out...
– kim
Nov 26 '20 at 6:57
• Nov 26 '20 at 8:14

This sequence is a Cauchy sequence so it converges.

First you see $$a_n>0, \forall n \in \mathbb{N}$$ from recursive relation. [$$a_1=1$$ and $$a_{n+1}$$ is defined to added positive terms]

Second since $$a_n>0$$ thus $$a_{n+1} = 1 + \frac{1}{1+a_n} \leq 2$$

Now consider \begin{align} |a_{n+1} - a_n| = \left| \frac{1}{1+a_n} - \frac{1}{1+a_{n-1}} \right| = \frac{|a_n - a_{n-1}|}{(1+a_n)(1+a_{n-1})} \leq \frac{1}{4} | a_n - a_{n -1}| \end{align} and this is cauchy sequence. [Series with this form is called contractive and after repeatively apply the same procedure continuing to $$|a_2-a_1|$$, and by Squeeze theorem you can easily guess $$a_n$$ is a Cauchy sequence]

In $$\mathbb{R}$$ cauchy sequence implies convergences so it converges. Then by taking limits $$\lim_{n\rightarrow \infty} a_n = \alpha$$ we have $$\alpha^2 = 2$$ and from $$a_n>0$$, $$\alpha = \sqrt{2}$$.

• thanks a lot! Cauchy sequence is quite useful as always...
– kim
Nov 26 '20 at 7:09
• You need $a_n \ge 1$ for the last estimate, not $a_n > 0$. (The question is a multiple duplicate, though). Nov 26 '20 at 8:16

A method often useful with oscillating sequences: Let $$b_n=|(a_n)^2-2|.$$ Then $$0\le b_{n+1}=\frac {b_n}{(1+a_n)^2}\le \frac {b_n}{4}$$ because $$1+a_n\ge 2$$ by induction on $$n$$.

So $$b_n$$ decreases to $$0$$. So $$(a_n)^2\to 2$$ with each $$a_n>0.$$

The motivation for the "$$2$$" in the definition of $$b_n$$ is that IF $$a_n$$ converges to a limit $$L$$ then $$L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty} 1+\frac {1}{1+a_n}=1+\frac {1}{1+L},$$ implying $$L^2=2.$$