Sub-sequences of odd and even terms convergent implies the sequence converges? Given the sequence $$ a_{n+1} = 1 + \frac1{1+a_n} $$ where $a_1 = 1$
I've determined that the odd terms of the sequence are increasing, and the even terms are all decreasing.  I've also stated that the sequence is bounded below by 0, and bounded above by 2.  Thus both odd and even subsequences can be said to be convergent by the Monotonic Convergence Theorem.  
I realize that if both even and odd subsequences converge to the same limit L, I can find an $ N_0 $ that accounts for both even and odd terms being sufficiently close to the limit L.  
What I'm stuck on however is finding a way to show that the even and odd terms of the sequence converge to the same L.
Thanks for your time!
 A: Let $L$ be the positive solution to $L = 1+\frac{1}{1+L}$. It's really easy to solve for $L$, but even without explicitly solving for $L$ we can still show that the sequence converges to $L$ if we know that $a_n\ge 0$ for all $n$.
To do so, notice that
$$a_{n+1}-L = \left(1+\frac{1}{1+a_n}\right) - \left(1 + \frac{1}{1+L}\right) = \frac{1}{1+a_n} - \frac{1}{1+L} = \frac{L-a_n}{(1+a_n)(1+L)}$$
and hence
$$|a_{n+1}-L|\le\frac{1}{1+L}\frac{1}{1+a_n}|a_n-L|\le\frac{1}{1+L}|a_n-L|$$
since $a_n\ge 0\implies 1+a_n\ge 1$. Since $L$ is positive, $\frac{1}{1+L}<1$, so $$|a_n-L|\le\left(\frac{1}{1+L}\right)^{n-1}|a_1-L|\xrightarrow{n\rightarrow\infty}0.$$
A: $$
\begin{align}
& \because \space 1 \lt a_{2n-1} \lt  a_{2n+1} \lt 2 \Rightarrow \lim_{n\rightarrow\infty} \left( a_{2n+1} - a_{2n-1} \right) = 0 \Rightarrow L_{o} = \lim_{n\rightarrow\infty} a_{2n+1} = \lim_{n\rightarrow\infty} a_{2n-1} \\
& \& \space\space\space 1 \lt a_{2n+2} \lt  a_{2n \space\space\space\space\space} \lt 2 \Rightarrow \lim_{n\rightarrow\infty} \left( a_{2n \space\space\space\space\space} - a_{2n+2} \right) = 0 \Rightarrow L_{e} = \lim_{n\rightarrow\infty} a_{2n \space\space\space\space\space} = \lim_{n\rightarrow\infty} a_{2n+2} \\
& \\
& a_{n+1} = 1 + \frac1{1 + a_{n}} \Rightarrow (a_{n+1} - 1) (a_{n} + 1) = 1 \text{.} \quad \text{For} \space\space \color{red}{n\rightarrow\infty} \Rightarrow \\
& (a_{\small2 n+1 \normalsize} - 1) (a_{\small 2n \normalsize} + 1) = (L_{o} - 1) (L_{e} + 1) = 1 \small\space\&\space\normalsize (a_{\small 2n \normalsize} - 1) (a_{\small 2n-1 \normalsize} + 1) = (L_{e} - 1) (L_{o} + 1) = 1 \\
& \Rightarrow (L_{o} - 1) (L_{e} + 1) = (L_{e} - 1) (L_{o} + 1) \Rightarrow L_{o} - L_{e} = L_{e} - L_{o} \Rightarrow 2 L_{o} = 2 L_{e} \Rightarrow \color{red}{L_{o} = L_{e}} \\
\end{align}
$$
