Prove every point between $\liminf$ and $\limsup$ is accumulation point $x_n$ is bounded, $\lim_{n \rightarrow \infty}(x_{n+1}-x_n)=0$, $\liminf_{n \rightarrow \infty}=l$, $\limsup_{n \rightarrow \infty}=L$, show that every point in $[l,L]$ is an accumulation point.
I am trying to use Stolz Cesàro theorem to show that $\lim_{n \rightarrow \infty} \frac{x_n}{n}=0$, is it the right direction?
 A: Suppose $l<x<L$ but $x$ is not an accumulation point of $(x_n)$. Then there are $a,b$ such that $l<a<x<b<L$ and $(a,b)\setminus \{x\}\cap (x_n)=\emptyset.$ Now, there is an integer $N$ so large that $|x_{n+1}-x_n|<\frac{b-a}{2}$ whenever $n>N.$ But there is also an integer $M>N$ such that $x_M>b$ (why?). But then, if  $x_n\in \{x_M, x_{M+1},\cdots \},\ x_n>b$, from which it follows that $\liminf x_n\neq l$, which is a contradiction.
A: Suppose that $p\in[l,L]$. If $p=l$ or $p=L$ we are done, so assume $l<p<L$. We want to show that for every $\varepsilon>0$ and each positive integer $N$ there is $n_{\varepsilon,N}>N$ such that $|p-x_{n_{\varepsilon,N}}|<\varepsilon$. 
Take $m$ such that $|x_{n+1}-x_n|<\varepsilon$ for all $n>m$. Take $k>\max\{m,N\}$ such that 
$|x_k-L|<L-p$, so in particular $p<x_k$. Take $j>k$ such that 
$|x_j-l|<p-l$, so in particular $x_j<p$. Consider the finite sequence of points 
$x_k,x_{k+1},\dots,x_{j-1},x_j$. There is an index $b$ with $k\le b\le j-1$ such that 
$x_b\ge p>x_{b+1}$. Using that $|x_{b+1}-x_b|<\varepsilon$ we have that 
$|x_b-p|<\varepsilon$ (and also $|x_{b+1}-p|<\varepsilon$). So we could take 
$n_{\varepsilon,N}=b$ (or also we could take $n_{\varepsilon,N}=b+1$), which completes the proof. 
Regarding $\lim_{n \rightarrow \infty} \frac{x_n}{n}=0$ it holds regardless of the condition $x_{n+1}-x_n\to0$, as long as the given sequence is bounded, in particular if $\lim \inf_{n \rightarrow \infty}=l$ and $\lim \sup_{n \rightarrow \infty}=L$ exist and are finite numbers. But this condition $\lim_{n \rightarrow \infty} \frac{x_n}{n}=0$ does not necessarily imply that every $p\in(l,L)$ is an accumulation point. For example, take $x_n=0$ if $n$ is even, and take $x_n=1$ if $n$ is odd. Then $l=0$, $L=1$ and $l$ and $L$ are the only accumulation points, so no point $p\in(0,1)$ is an accumulation point. Thus Stolz Cesàro theorem that you refer to seems of no use for this problem. (As a matter of fact, after I reviewed the statement of Stolz Cesàro theorem, I do not quite see how it relates to the condition $\lim_{n \rightarrow \infty} \frac{x_n}{n}=0$, to begin with.) 
