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?
 A: 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}
$.
A: Hint: $\sqrt{2}$ is a piece of cheese, $a_{2n}$ is a slice of bread, $a_{2n+1}$ is another slice of bread. What would you do next?
A: Hint: Applying the recurrence relation twice, we find that
$$
a_{n+2} = 1 + \frac{1}{1+a_{n+1}} =
1 + \frac{1}{1+\left(1 + \frac{1}{1+a_{n}}\right)} = \\
1 + \frac{1}{2 + \frac{1}{1+a_{n}}}=\\
1 + \frac{1+a_n}{2 + 2a_n + 1} =\\
1 + \frac{1+a_n}{3 + 2a_n} = \\
\frac{3 + 2a_n + 1+a_n}{3 + 2a_n} = \\
\frac{4 + 3a_n}{3 + 2a_n}
$$
This recurrence gives us both the even and odd sequences.  We need to show one of the following:


*

*The evens decrease to $\sqrt 2$ and the odds increase to $\sqrt 2$

*The evens stay above $\sqrt 2$, the odds stay below $\sqrt 2$, and the difference between the evens and odds converges to $0$.

