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

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.

• Hint: If $\displaystyle\lim_{n\rightarrow\infty}a_n=l$ then what happens when you take limit on both sides? – Yadati Kiran Nov 17 at 2:22
• There is a very standard way of solving linear recurrence relation. In this case, if $\alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = \sum_{k=1}^{3}c_k \alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values. – Sangchul Lee Nov 17 at 2:33
• @SangchulLee And one of $\alpha_k=1$ in this problem. – i707107 Nov 17 at 2:35
• @YadatiKiran I'm not sure it will help here because all you will get is $l=\frac{1}{3}[l+l+l]=l$. – Anurag A Nov 17 at 3:01
• @AnuragA: Initially the problem was $a_n = \dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post. – Yadati Kiran Nov 17 at 3:36

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 $$\frac{2}{k(k+1)}(a_1+2a_2+3a_3+\ldots + ka_k)$$
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 $$(*)$$, $$\frac{14}3=l+\frac 23 l+\frac 13 l\implies l=\frac 73.$$
• If I may ask, how did you come up with general term $b_n$? – Yadati Kiran Nov 17 at 3:57