For arbitrary real number $x_0, x_1, ....,x_{p-1}$, define $x_n=\frac{x_{n-1}+x_{n-2}+....+x_{n-p}}p $ ($n\geq p $)
If $x_n$ converges, then the limit is equal to $$\frac {2X}{p(p+1)}\quad \text{while}\quad X=px_{p-1}+(p-1)x_{p-2}+...+x_0$$ For example, In the case of $p=3$, one can show that $3x_n+2x_{n-1}+x_{n-2}$ is constant. Adding $\frac 23x_{n-1}+\frac 13x_{n-2}$ on both sides of
$$x_n=\frac13x_{n-1}+\frac13x_{n-2}+\frac13x_{n-3}$$, we get
$$x_n+\frac 23x_{n-1}+\frac 13x_{n-2}=x_{n-1}+\frac23 x_{n-2}+\frac 13x_{n-3}$$
So all I have to do is show that $x_n$ is convergent. But how?? I'm completely stuck at this point.
One method: Clearly $$ \{x_n\mid n\geq N\}\subseteq[\min\{x_N,\dots,x_{N+p-1}\},\max\{x_N,\dots,x_{N+p-1}\}] $$ for all $N$. Let $$ d_n=\max\{x_n,\dots,x_{n+p-1}\}-\min\{x_n,\dots,x_{n+p-1}\} $$ and show that there is an $r<1$ such that $d_{n+p}\leq r d_n$