How do we show the limit? We have that $(a_n)$ is a bounded sequence of real numbers that satisfy $2a_{n+1}\leq a_n+a_{n+2}, \forall n\in \mathbb{N}$. 
I want to show that the sequence $b_n=a_{n+1}-a_n$ converges and that the limit is $0$. 
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I have shown that $b_n$ is increasing and bounded. 
So, $b_n$ converges. 
How do we find the limit? 
 A: Note that from $$2a_{n+1}\leq a_{n+2}+a_n$$
we obtain $$a_{n+1}-a_{n+2}\leq a_n-a_{n+1}$$
So $$-b_{n+1}\leq -b_n$$
thus $b_{n+1}\geq b_n$, so $b_n$ is an increasing sequence.
Also, since $b_n=a_{n+1}-a_n$, and $a_n$ is bounded, $b_n$ is bounded.
A little more rigurous: $a_n$ being bounded means that there exists a $B\in\mathbb{R}$ with $|a_n|\leq B$, thus $|a_{n+1}-a_n|\leq |a_{n+1}|+|a_n|=2B$.
We cannot prove that $\lim b_n=0$, since this is not true. $a_n$ is not necessarily converging, and so for example, $a_n=n$, and now $b_n=(n+1)-n=1$. We know $b_n=1$ is converging and, in some sense, increasing, but the limit is $1$. We can choose $a_n=Ln$ to get $\lim b_n=L$, and so the limit of $b_n$ is not fixed by the properties given. Given that $a_n$ converges, though, proves $$\lim b_n=\lim (a_{n+1}-a_n)=\lim a_{n+1} - \lim a_n=0$$.
A: Note that $$a_{n+2}-a_{n+1}\ge a_{n+1}-a_n \,\,\,\,\forall n\in\mathbb{N}.$$ First
assume that $a_0\lt a_1,$ then the sequence $(a_n)$ is increasing and bounded. Therefore it is convergent.  
Next
if $a_0\gt a_1,$ Consider the sequence $(-a_n).$
In both cases $\lim b_n=0.$
