Does a sequence in $\mathbb R$ with this condition $2a_n \leq a_{n-1}+ a_{n+1}$ satisfy this convergence? Suppose I have a sequence {$a_n$} $\subseteq \mathbb{R}$ which is bounded and satisfies the following condition: $2a_n \leq a_{n-1}+ a_{n+1}$. Why is it that $lim_{n\rightarrow\infty}(a_{n+1}-a_{n})$= 0?
 A: Inequality $$2a_n \leq a_{n-1}+a_{n+1}$$ can be rearranged to $$a_n-a_{n-1} \leq a_{n+1}-a_n,$$ so we have that $b_n=a_{n+1}-a_n$ is bounded monotone sequence, and thus convergent (boundedness is due to triangle inequality).
Let $L$ be the limit of $(b_n)$. If $L>0$, then there exists $n_0$ such that $b_n > 0$, for all $n\geq n_0$, and thus $(a_n)$ is monotone for $n\geq n_0$. But, $(a_n)$ must be convergent as well in that case, so $$\lim_n b_n = \lim_n (a_{n+1}-a_n) = 0$$ Contradiction. Similarly for $L<0$. We conclude that $\lim_n (a_{n+1}-a_n) = 0$, as wanted.
A: Let $b_n=a_n-a_{n-1}$. From the condition $2a_n \leq a_{n-1}+ a_{n+1}$ we get $b_n \leq b_{n+1}$. {$b_n$} is bounded because there exists $M$ such that $b_n=a_n-a_{n-1} \leq M+M$.
To conclude, {$b_n$} is increasing and bounded above, thus it is convergent.
A: Case 1. $a_{n-1}<a_{n+1}$ Then
$$
\eqalign{
& 2a_n<a_{n-1}+a_{n+1} < 2a_{n+1} \cr
& \rightarrow a_n < a_{n+1} \cr
}
$$
Case 2. $a_{n-1} > a_{n+1}$ Then
$$
\eqalign{
& 2a_n < a_{n-1}+a_{n+1} < 2a_{n-1} \cr
& \rightarrow a_n < a_{n-1} \cr
}
$$
In both cases monotonicity is implied, therefore convergence.
