Does this bounded sequence converge? The sequence $(a_n)$ is bounded for $n=1, 2, \dots$, such that
$$a_n \leq \frac{1}{2} \left(a_{n-1} + a_{n+1}\right)$$
for $n \geq 2$. I want to prove the sequence $(a_n)$ converges.
Since I am told the sequence is bounded, I was trying to prove it is monotonic so that I could use a known theorem to claim it converges.
However, I can prove it is bounded, but I am unable to prove it is monotonic. 
I tried to use series and use the fact that it is telescopic but nothing useful came out.
Any ideas? Am I going in the wrong track?
 A: Let's define the sequence $b_n = a_{n + 1} - a_n$.
The condition $a_n \le \frac{1}{2}(a_{n - 1} + a_{n + 1})$ can be rearranged to $a_n - a_{n - 1} \le a_{n + 1} - a_n$, or put another way $b_{n - 1} \le b_n$. So the sequence $b_n$ is monotonically increasing. This implies that $sign(b_n)$ is eventually constant (either - or $0$ or +).
This in turn implies that the sequence $a_{n + 1} - a_1 = b_1 + ... + b_n$ is eventually monotonic. More precisely, it's eventually decreasing if $sign(b_n)$ is eventually -, it's eventually constant if $sign(b_n)$ is eventually $0$, it's eventually increasing if $sign(b_n)$ is eventually +. Since the sequence $a_{n + 1} - a_1$ is also bounded, we get that it converges. This immediately implies that the sequence $a_n$ converges.
A: Let us fix $k\ge 1$.
Define $$
b_1 = a_k\\
b_2 = a_{k+1}\\
b_{n+1} = 2b_n - b_{n-1}
$$
Then you can prove that $a_{n+k-1} \ge b_n$ and that $b_n = (a_{k+1}-a_k) n + 2a_k - a_{k+1}$.
As $a$ is bounded, $b_n\nrightarrow+\infty$ and $a_{k}\ge a_{k+1}$.
Hence the sequence is monotonic.
