The sequence $x_{n+m}\le \frac{x_n+x_{n+1}+\cdots+x_{n+m-1}}{m}$. I have problem

Let $m \ge 2 -$ fixed positive integer. The sequence of non-negative real numbers $\{x_n\}_{n=1}^{\infty}$ is that for all $n\in \mathbb N$
  $$x_{n+m}\le \frac{x_n+x_{n+1}+\cdots+x_{n+m-1}}{m}$$
  Prove that this sequence has a limit.

Here is the solution of this problem.

Q.:Why the sequence is not monotone?

 A: Let $m=2$ then $x_{n+2} \le \frac { x_n + x_{n+1}} 2$. Just take $x_1 = 1, x_2=3, x_3=2$
A: Note that, given the unilateral z-transform
$$
A(z) = \sum\limits_{0\, \leqslant \,n} {a_{\,n} \,z^{\,n} } \quad \left| {\;a_{\,n < 0}  = 0} \right.
$$
then the Moving Window Sum has a z-transform:
$$
\begin{gathered}
  W(A(z),h) = \sum\limits_{0\, \leqslant \,n} {\left( {a_{\,n - h + 1}  +  \cdots  + a_{\,n - 1}  + a_{\,n} } \right)\,z^{\,n} }  =  \hfill \\
   = \left( {1 + z +  \cdots  + z^{\,h - 1} } \right)A(z) = \frac{{1 - z^{\,h} }}
{{1 - z}}A(z) \hfill \\ 
\end{gathered} 
$$
Now, what you have defined is a sequence which is
moving-window average of  itself. So, if we put

$$
\left\{ \begin{gathered}
  x_{\,n}  = 0\quad \left| {\,n < 0} \right. \hfill \\
  x_{\,0}  = 1 \hfill \\
  x_{\,n}  = \frac{1}
{h}\left( {x_{\,n - h}  + x_{\,n - h + 1}  +  \cdots  + x_{\,n - 1} } \right)\quad \left| {\;1 \leqslant h} \right. \hfill \\ 
\end{gathered}  \right. \tag{1}
$$  

we get

$$
\begin{gathered}
  X_{\,h} (z) = \sum\limits_{0\, \leqslant \,n} {x_{\,n} \,z^{\,n} }  = 1 + \frac{1}
{h}\,\sum\limits_{1\, \leqslant \,n} {\left( {x_{\,n - h + 1}  +  \cdots  + x_{\,n - 1} } \right)\,z^{\,n} } \quad  \Rightarrow  \hfill \\
   \Rightarrow \quad X_{\,h} (z) = 1 + \frac{z}
{h}\,\frac{{1 - z^{\,h} }}
{{1 - z}}X_{\,h} (z)\quad  \Rightarrow \quad \left( {1 - \frac{{z\left( {1 - z^{\,h} } \right)}}
{{h\left( {1 - z} \right)}}} \right)X_{\,h} (z) = 1\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad X_{\,h} (z) = \frac{{h\left( {1 - z} \right)}}
{{\left( {h - \left( {h + 1} \right)z + z^{\,h + 1} } \right)}} = \frac{h}
{{\left( {h + \left( {h - 1} \right)z +  \cdots  + 2\,z^{\,h - 2}  + z^{\,h - 1} } \right)\left( {1 - z} \right)}} \hfill \\ 
\end{gathered}  \tag{2}
$$  

$X_h(z)$ is a meromorphic function with a single real pole in $z=1$,
which has a well defined Taylor expansion.
The Final Value theorem tells us that the $x$ sequence is limited
$$
\mathop {\lim }\limits_{z\; \to \,1} \left( {1 - z} \right)X_{\,h} (z) = x_\infty   = \frac{h}
{{\left( {h + \left( {h - 1} \right) +  \cdots  + 2\, + 1} \right)}} = \frac{2}
{{\left( {h + 1} \right)}}
$$
Example with $h=3$
$$
x = \left\{ {1,\;\frac{1}
{3},\;\frac{4}
{9},\,\frac{{16}}
{{27}},\,\frac{{37}}
{{81}},\,\frac{{121}}
{{243}},\, \cdots } \right\}
$$
where $3^n x_n$ is the OEIS sequence A103770.
and the Taylor series of $X_3(z)$ effectively corresponds to the $x$ sequence.
As also demonstrated in the OEIS article, the $x$ sequence converges to $1/2$.
