Convergence of sequence with $a_{n+2} < \frac{a_n + a_{n+1}}{2}$

Question

Let $(a_n)$ be a sequence with $a_n \in (0,1)$ and $a_{n+2} < \frac{a_n + a_{n+1}}{2}$. First question is following:

If $(a_n)$ is convergent than what is the set of possible limit points?

If $\alpha \in [0,1)$ and $\alpha +\varepsilon \in [0,1)$ than the sequence $\alpha, \alpha + \varepsilon, \alpha, \alpha + \frac{\varepsilon}{4}, \dots $ converges to $\alpha$.

But how to prove that $1$ can't be limit point (if it's true)?

And the second is:

Is $\lim a_n$ exists for any such $(a_n)$?

And i have no idea what to do with this..

Answer 1

Hint: For the second half, if $a_1 < a_2$ then show that $$a_1 < a_3 < a_5 < \dotsb < a_6 < a_4 < a_2.$$ Additionally show that $|a_{n+1} - a_n| \to 0$. Then conclude.

Answer 2

Since $a_n$ are in $(0,1)$, there is some $\epsilon > 0$ such that $a_1, a_2 < 1- \epsilon$. Now prove by induction that all $a_n < 1 - \epsilon$ and hence the limit is $\le 1- \epsilon$.

For the second part, define $b_n = \max(a_n, a_{n+1})$. Then $b_n$ is monotonic decreasing and bounded below, so converges to some $b$. Now, for each $n$, either $a_n = b_n$, or $a_{n+1} = b_n$ and hence $a_n > 2a_{n+1} - a_{n-1} \ge 2b_n - b_{n-1} \to b$. Of course $a_n \le b_n$ by construction, so squeeze theorem says $a_n$ converges to $b$ as well.