Convergence of the $a_1 = a, a_{n + 1} = \frac{2}{a_n + 1}$ sequences by using subsequences I supposed to find a candidate limit and show the monotonicity of the $a_{2n}$ and $a_{2n + 1}$ subsequences of the given sequence:
$$a_1 = a$$
$$a_{n + 1} = \frac{2}{a_n + 1}$$
$$with \quad0 < a < 1$$
Finding a candidate limit is easy just supposing that $a_{n} \to L$ and with a uniqueness of the limit I can solve $a_{n+1} \to L$ and it gives me L = 1. 
But for proving that the subsequences $a_{2n}$ and $a_{2n + 1}$ are monotone decreasing and increasing respectively and converge to the same limit I need to find some type of recurrence relation between the members of that sequence. The best I found is:
$$a_{2(n+1)} = \frac{2a_{2n} + 2}{a_{2n} + 3}$$
$$a_{2(n + 1)+1} = \frac{2a_{2n + 1} + 2}{a_{2n + 1} + 3}$$
But this doesn't give me any good recurrence relation between the subsequence members. Any help?
Thank you
 A: .Since $a_1=a>0$ the recurrence relation implies that $a_n>0$ for all n. Note that $$a_{n+1}(a_n+1)=2$$ for all positive integers n.
Therefore $$  a_{n+1}(a_n+1)=a_{n+3}(a_{n+2}+1)=2$$ The above equality implies that  if $$a_{n+1}<a_{n+3}$$ then 
$$a_{n}>a_{n+2}.$$ That is, if one of your subsequences is increasing the other is decreasing.
Note that the function $$f(x)=\frac {2}{x+1}$$ has a fixed point at $ x=1 $ with  $f'(1)=-1/2.$ Thus $ x=1$ is an attractor. The interval$(0,1)$ is contained in the basin of attraction therefore starting at x=a, every subsequence of iterates converges to $ L=1.$    
A: It's easy to show that $a_n >0, \forall n$. Then let's look at the sequence $b_n=|a_n-1|$ then we have
$$0\leq b_{n+1}=|a_{n+1}-1|=\left|\frac{2}{a_n+1}-1\right|=\left|\frac{a_n-1}{a_n+1}\right|=\frac{b_n}{a_n+1}\leq b_n \tag{1}$$
$(b_n)_{n\geq1}$ is descending and lower bounded by $0$ so it's converging. If we assume that $\lim\limits_{n\rightarrow\infty}b_n=L>0$ then we have $\forall \varepsilon >0$
$$0<L<b_n<L+\varepsilon \tag{2}$$
for all $n$ from some $N(\varepsilon)$ onwards. But, some of the $a_n
\geq1$, it's easy to show that $0<a_n\leq 1 \iff 1<a_n+1\leq 2 \iff \frac{1}{2} < \frac{a_n+1}{2}\leq 1\iff 2 > a_{n+1} \geq 1$. By induction we can show that $0<a_{2k+1}\leq 1$ and $1\leq a_{2k} < 2$, since $0<a_1< 1$.
So, from the definition of the limit, we will have $n=2k > N(\varepsilon)$ and $n+1=2k+1 > N(\varepsilon)$ with $b_n$ and $b_{n+1}$ satisfying $(1)$ and $(2)$ while:
$$0<L<b_{2k+1}=\frac{b_{2k}}{a_{2k}+1}\leq \frac{b_{2k}}{2}<\frac{L+\varepsilon}{2}$$ 
which is clearly a contradiction for $0<\varepsilon<L$. So $(2)$ doesn't holf for $\forall \varepsilon >0$ and as a result
$$\lim\limits_{n\rightarrow\infty}b_n=0 \Rightarrow \lim\limits_{n\rightarrow\infty}a_n=1$$
