Limit of sequence given by $a_n = \frac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3})$?

Question

I have the sequence modeled by the equation $$a_n = \dfrac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3}), \:a_1 = 1,\: a_2 = 2,\: a_3 = 3.$$ Simply by manually calculating the sequence I know it converges around $2.3$, but I do not know how to express it in terms of a limit. Since it relies on the previous $3$ values of the sequence I do not know how to express this formulaically.

Answer 1

You will find that the limit is $\frac16 a_1 + \frac13 a_2 + \frac12 a_3$

which with $a_1 = 1,\: a_2 = 2,\: a_3 = 3$ gives a limit of $\frac73 \approx 2.3333$

More generally, if $a_n$ is the average of the previous $k$ terms then the limit is $$\frac{2}{k(k+1)}(a_1+2a_2+3a_3+\ldots + ka_k)$$

Answer 2

Let $$b_n=a_n+\dfrac23 a_{n-1}+\dfrac 13 a_{n-2},\tag{*}$$ we have $$b_n=b_{n-1},\qquad b_3=a_3+\dfrac23 a_{2}+\dfrac 13 a_{1}=\frac{14}3.$$ Obviously $b_n$ converges to $b_3=\frac{14}3$. If $a_n$ has a limit $l$, by taking limits on both sides of $(*)$, $$\frac{14}3=l+\frac 23 l+\frac 13 l\implies l=\frac 73.$$