My question is following.
Let $(\xi_n)_{n\in\mathbb{N}}$ be a positive sequence which satisfies the following inequalities, for every $n\in\mathbb{N}$
$\xi_{2n+1}\leq \frac{1}{2}\xi_{2n}+\frac{1}{2}\xi_{2n-1}$
$\xi_{2n+2}\leq \frac{1}{2}\xi_{2n+1}+\frac{1}{2}\xi_{2n-1}$
and I want to prove the convergence of this sequence. Ideally, I want a generalisable proof with more terms, or with different comportements. For example,
$\xi_{3n+1}\leq \frac{1}{2}\xi_{3n}+\frac{1}{2}\xi_{3n-1}$
$\xi_{3n+2}\leq \frac{1}{2}\xi_{3n+1}+\frac{1}{2}\xi_{3n-1}$
$\xi_{3n+3}\leq \frac{1}{2}\xi_{3n+2}+\frac{1}{2}\xi_{3n-1}$
Thank you.
First, let us derive a weaker, but simpler set of inequalities by plugging in the first inequality into the second to arrive at \begin{align*} \xi_{2n+1} &\leq \frac{1}{2}\xi_{2n} + \frac{1}{2}\xi_{2n-1} \\ \xi_{2n+2} &\leq \frac{1}{2}\xi_{2n+1} + \frac{1}{2}\xi_{2n-1} \leq \frac{1}{2}\left( \frac{1}{2}\xi_{2n} + \frac{1}{2}\xi_{2n-1} \right) + \frac{1}{2}\xi_{2n-1} = \frac{1}{4}\xi_{2n} + \frac{3}{4}\xi_{2n-1}. \end{align*} Now, let $M_n = \max\{\xi_{2n+1},\xi_{2n+2}\}$ and $m_n = \min\{\xi_{2n+1},\xi_{2n+2}\}$. Then, since \begin{align*} \xi_{2n+1} &\leq \frac{1}{2}\xi_{2n} + \frac{1}{2}\xi_{2n-1} \leq \frac{1}{2}M_{n-1} + \frac{1}{2}M_{n-1} = M_{n-1} \ \text{and} \\ \xi_{2n+2} &\leq \frac{1}{4}\xi_{2n} + \frac{3}{4}\xi_{2n-1} \leq \frac{1}{4}M_{n-1} + \frac{3}{4}M_{n-1} = M_{n-1}, \end{align*} the sequence $M_n$ is monotonously decreasing and, since it is bounded below by $0$, converges to some $M \in \mathbb{R}$. On the other hand, at least one of $\xi_{2n}$, $\xi_{2n-1}$ equals $m_{n-1}$, therefore, \begin{align*} \xi_{2n+1} &\leq \frac{1}{2} m_{n-1} + \frac{1}{2}M_{n-1} \\ \xi_{2n+2} &\leq \frac{1}{4} m_{n-1} + \frac{3}{4}M_{n-1}, \end{align*} i.e. $M_n \leq \frac{1}{4} m_{n-1} + \frac{3}{4}M_{n-1}$, which implies $M_{n} - m_{n-1} \leq 3 (M_{n-1}-M_n)$, which converges to $0$. Thus, $m_n \rightarrow M$ as well.
This method will always work as long as the right hand sides of the inequalities are recurrent convex combinations of the type presented in your two examples.
It might be worth noting that, if the sequence is not bounded below, it will still either converge or diverge to $-\infty$, since if $M_n$ is monotonously decreasing but not bounded below, $\xi_n \rightarrow -\infty$ trivially.