Define a sequence of real numbers $(x_n)$ by $x_1 = 1, x_{n+1}= \frac{1}{2+x_n}$. Show that $(x_n)$ converges, and evaluate its limit.
Attempt: clearly,the sequence is bounded above by $2$ and $x_{n+1} - x_n = \frac{1}{2+x_n} - \frac{1}{2+x_{n-1}} = \frac{x_{n-1}-x_n}{(2+x_n)(2+x_{n-1})}$. Since $x_n >0$ $\forall n\in \mathbb{N}$, the sign of $x_{n+1} - x_{n} $ depends on the sign of $x_{n-1} - x_n$ and the sign alternates, after plugging in certain values of odd terms and even terms I feel like the odd terms decreases while the even terms increases. i.e We can see that $x_1 =1, x_2 = \frac{1}{3}, x_3 = \frac{1}{2+ \frac{1}{3}}, x_4 = \frac{1}{2+\frac{3}{7}}$ from here we see that $x_1 - x_3>0$ and $x_4 - x_2>0$.
If this is true then how do I prove that $x_{2n}$ increases and $x_{2n-1}$ decreases? I'm stuck
