<s>[INST] A recursive sequence is defined by... A sequence is defined recursively by $a_1=1$ and $a_{n+1} = 1 + \frac{1}{1+a_{n}}$. 
Find the first eight terms of the sequence $a_n$. What do you notice about the odd terms and the even terms? By considering the odd and even terms separately, show that $a_n$ is convergent and deduce that its limit is $\sqrt{2}$.

EDIT: Ok, so I was being an idiot and forgot how to basic math for a little. The first 8 terms are as follows:
EVENS:
a2 = 1.5
a4 = 1.46666...
a6 = 1.415731...
a8 = 1.41426...

ODDS:
a1 = 1
a3 = 1.4
a5 = 1.4054...
a7 = 1.41395...

From these I see that the even terms are decreasing while the odd terms are increasing. How can I use these to prove that $\{a_n\}$ converges? Does it have something to do with alternating series or something similar?
 [/INST] We have
$a_{n+1} 
= 1 + \frac{1}{1+a_{n}}
= \frac{2+a_n}{1+a_{n}}
$.
Therefore,
$a_n > 1$
for all $n$.
Also,
$\begin{array}\\
a_{n+1}-\sqrt{2}
&= \dfrac{2+a_n}{1+a_{n}}-\sqrt{2}\\
&= \dfrac{2+a_n-\sqrt{2}(1+a_n)}{1+a_{n}}\\
&= \dfrac{2-\sqrt{2}-a_n(\sqrt{2}-1)}{1+a_{n}}\\
&= \dfrac{(\sqrt{2}-1)(\sqrt{2}-a_n)}{1+a_{n}}\\
\end{array}
$
so
$\dfrac{a_{n+1}-\sqrt{2}}{\sqrt{2}-a_n}
=\dfrac{\sqrt{2}-1}{1+a_{n}}
$.
Therefore
(1)$a_n-\sqrt{2}$
alternates in sign
and
(2)$\big|\dfrac{a_{n+1}-\sqrt{2}}{\sqrt{2}-a_n}\big|
<\sqrt{2}-1
$.
This implies that
$a_n-\sqrt{2}
\to 0$
so that
$\lim_{n \to \infty} a_n
=\sqrt{2}
$.
</s>