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I want to show that the real sequence $\{a_n\}_{n=1}^\infty$, defined recursively by $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$, converges to $\sqrt{2}$. The crux of matter is to prove that this sequence is convergent. By considering two subsequences consisting of the terms of even and odd indices, respectively, I was able to use the monotone sequence theorem to conclude that the original sequence truly converges. And a little algebra gives us the limit of $\sqrt{2}$. Now I'm wondering if it is possible to arrive at convergence by showing that $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence. Then I consider the difference $a_{n+k+1}-a_{n+1}$ because our sequence is recursively defined. And it's like: $$a_{n+k+1}-a_{n+1}=\frac{a_n-a_{n+k}}{(1+a_{n+k})(1+a_n)}.$$ This seems to add no further information. Did I go the wrong way? Are we doomed to failure in consideration of Cauchy sequences? Thank you.

Boar
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    Establish first that $1 \le a_n \le \frac{3}{2}$, then $$ \left| {a_{n + k + 1} - a_{n + 1} } \right| = \frac{{\left| {a_{n + k} - a_n } \right|}}{{(1 + a_{n + k} )(1 + a_n )}} \le \frac{{\left| {a_{n + k} - a_n } \right|}}{2} \le \cdots \le \frac{{\left| {a_{k + 1} - a_1 } \right|}}{{2^n }} \le \frac{3}{{2^n }}. $$ – Gary Oct 15 '21 at 06:49
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    @Gary Thank you, Gary. I finally made out your argument. It's beautiful and effective. – Boar Oct 16 '21 at 03:55

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Clearly all of the terms $a_n > 0$. Now look at the difference $a_n- \sqrt{2}$ and get $$a_{n+1} - \sqrt{2} = - \frac{\sqrt{2}-1}{1+ a_n} \cdot (a_n - \sqrt{2})$$ and from here we see that $$|a_{n} - \sqrt{2}| < (\sqrt{2}-1)^{n-1} |a_1 - \sqrt{2}| = (\sqrt{2}-1)^n$$ for all $n> 1$.

orangeskid
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