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

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.

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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 $a_n>0$, $\alpha = \sqrt{2}$.

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  • $\begingroup$ thanks a lot! Cauchy sequence is quite useful as always... $\endgroup$
    – kim
    Nov 26 '20 at 7:09
  • $\begingroup$ You need $a_n \ge 1$ for the last estimate, not $a_n > 0$. (The question is a multiple duplicate, though). $\endgroup$
    – Martin R
    Nov 26 '20 at 8:16
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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 $L^2=2.$

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