Proving the continued fraction representation of $\sqrt{2}$
Solution 1:
You have shown that the sequence of $a_{2n}$ converges. Call $e$ (for even) its limit, and call $o$ the limit of $a_{2n+1}$. Then the recursive relation giving $a_{n+1}$ in terms of $a_n$ implies that $$e=1+1/(1+o)$$ and $$o=1+1/(1+e),$$ from which it follows that $e=o=\sqrt2$.
[For example, take the limit as $n\to\infty$ of $a_{2n}=1+1/(1+a_{2n-1})$ to get the first relation, and similarly you get the second exchanging odd and even. From the first equation, you get $2+o=e+eo$ and from the second $2+e=o+oe$. Subtracting, you get $o-e=e-o$, so $o=e$. From the first equation, you now get $2=o^2$.]
Solution 2:
As I explained in a previous similar question, one can use the Banach fixed point theorem to solve these kinds of problems. For the special case of iterating a fractional linear transformation you can also (as I think is also explained in answers to that question) write down an explicit closed form for the $n^{th}$ term by diagonalizing a $2 \times 2$ matrix.
Solution 3:
I don't think you need to break it up into two cases. In stead we can do this:
- Establish that the sequence $\langle a_i \rangle$ is cauchy and therefore convergent to some $L$. For this establish the sequence $\langle f^i(0)\rangle$ converges to zero, where $f^i$ is the 'i'th iterate of $\frac{1}{2+x}$.
- Notice that the relation: $a_{n+1}=1+\frac{1}{1+a_n}$ is the same as $a_{n+1}a_n + a_{n+1} - a_n -2 = 0$.
- Take limits at both ends to get: $L^2+L-L-2=0$.