Show that $a_n\to\sqrt{2}$ with $a_n$ defined by $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ 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.

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 $n> 1$.
