TITLE: Convergent Sequence, mean of previous three numbers
QUESTION [5 upvotes]: I am given three real numbers $x_0, x_1, x_2$ and the next number in the sequence is defined as the mean of the three real numbers.
So: $$x_a=\tfrac{1}{3} \cdot (x_{a-3} + x_{a-2} + x_{a-1})$$
I am just getting started with learning about converging sequences and wanted to find out how to proof that this sequence is converging and if possible to what value given $x_0, x_1$ and $x_2$
Thanks in advance 
Prankster

REPLY [2 votes]: Call the weights of the first three terms $a,b,c$ in $x_n$ as $a_n,b_n,c_n$
We can see that $(a_1,a_2,a_3)=(1,0,0),(b_1,b_2,b_3)=(0,1,0),(c_1,c_2,c_3)=(0,0,1)$:
$$a_{n+3}=\frac{a_{n+2}+a_{n+1}+a_n}{3}$$
Let's call the roots of $3x^3=x^2+x+1$ as $1,\alpha,\beta$. Now, solving the linear system in $k_{a_1},k_{a_2},k_{a_3}$:
$$k_{a_1}+k_{a_2}\alpha^1,k_{a_3}\beta^1=a_1=1$$
$$k_{a_1}+k_{a_2}\alpha^2,k_{a_3}\beta^2=a_2=0$$
$$k_{a_1}+k_{a_2}\alpha^3,k_{a_3}\beta^3=a_3=0$$
This allows us to write $a_n=k_{a_1}+k_{a_2}\alpha^n,k_{a_3}\beta^n$. Taking $\lim_{n\to +\infty} a_n$ is then easy by noting that $|\alpha|,|\beta|<1$, and we get:
$$\lim_{n\to +\infty}a_n=\lim_{n\to +\infty}k_{a_1}+k_{a_2}\alpha^n,k_{a_3}\beta^n=k_{a_1}=\frac16$$
Simlarly
$$\lim_{n\to +\infty}b_n=\frac13$$
$$\lim_{n\to +\infty}c_n=\frac12$$
So in fact, our limit converges to:
$$\lim_{n\to \infty} x_n=\frac{a+2b+3c}{6}=\frac{x_1+2x_2+3x_3}{6}$$
Which is such a nice result that it begs for a nicer proof.