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How to evaluate this problem:

If $a_n=1$ and $a_{n+1}= 1 + \frac{1}{1 + a_n}$ use part to show $\lim_{n\rightarrow\infty}a_n = \sqrt{2}$. This gives the development of a continuous fraction $$\sqrt{2}= 1 + \frac{1}{2 + \frac{1}{2+\frac{1}{2+ ...}}}$$

Andria
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    Welcome to MSE. Please read this text about [how to ask a good question](https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question). – José Carlos Santos Aug 14 '21 at 14:19
  • What have you tried? Hint: if $ \lim a_{n+1} = L$ then $ \lim a_n = L$ as well. – talbi Aug 14 '21 at 15:00
  • Replace $b_n=a_n+1$, leading to $b_{n+1}=2+\frac{1}{b_n}$ and ... [check this](https://math.stackexchange.com/questions/2566429/how-to-find-lim-limits-n-to-infty3-frac1a-n-where-a-n-3-frac/2566862#2566862) as an example. – rtybase Aug 15 '21 at 12:00

1 Answers1

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Suppose your sequence has a limit, say L.

Then $\lim_{n\to \infty}a_n=\lim_{n\to \infty}a_{n+1}=L$ and the recurrence relationship implies that L=1+$\frac{1}{1+L}$ which in term implies that (L+1)(L-1)=1 $\Rightarrow L^2-1=1 \Rightarrow L=\sqrt{2}$ or $L=-\sqrt{2}$ , assuming $1+L\neq-1$.

Now, notice that your sequence is bounded below by 1, bounded above by 2 and Cauchy, so it has a positive limit.

Indeed, let $n,m\in\mathbf{N} $ with $n\lt m$. Then $$|a_{m+1}-a_{n+1}|= |\frac{a_n-a_m}{(1+a_{m})(1+a_{n})}|=|\frac{a_{n-1}-a_{m-1}}{(1+a_{m})(1+a_{n})(1+a_{m-1})(1+a_{n-1})}|$$ and inductively you have that $|a_{m+1}-a_{n+1}|\leq|\frac{1}{2^n}|$ , with $lim_{m,n\to \infty}|\frac{1}{2^n}|=0$ , because $|a_{i-1}-a_{j-1}|\leq1$ and $(1+a_i)\geq2$ for every $i,j\in\mathbf{N}$.

User
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