Find the limit

$\begin{aligned} a_n & = \frac 13 \left(a_{n-1} + a_{n-2} + a_{n-3}\right) \\ 3 a_n & = a_{n-1} + a_{n-2} + a_{n-3} \\ 3 \sum_{k=4}^n a_k & = \sum_{k=4}^n a_{k-1} + \sum_{k=4}^n a_{k-2} + \sum_{k=4}^n a_{k-3} \\ 3 \sum_{k=4}^n a_k & = \sum_{k=3}^{n-1} a_k + \sum_{k=2}^{n-2} a_k + \sum_{k=1}^{n-3} a_k \\ 3a_n + 2a_{n-1} + a_{n-2} & = 3a_3 + 2a_2 + a_1 \\ \lim_{n \to \infty} \left(3a_n + 2a_{n-1} + a_{n-2}\right) & = \frac {16}9 + \frac 89 + \frac 13 = 3 \\ 6 \lim_{n \to \infty} a_n & = 3 \\ \implies \lim_{n \to \infty} a_n & = \frac 12 = \boxed{0.5} \end{aligned}$

$3a_{n}+2a_{n-1}+a_{n-2}=\frac{3}{3}(a_{n-1}+a_{n-2}+a_{n-3})+2a_{n-1}+a_{n-2}=3a_{n-1}+2a_{n-2}+a_{n-3}=…=3a_{3}+2a_{2}+a_{1}$ .Hence $\large \lim_{n\to\infty}(3a_{n-1}+2a_{n-2}+a_{n-3})=3a_{3}+2a_{2}+a_{1}=3$ or $3l+2l+l=3$ . Therefore $l=\frac{1}{2}$ where $\large\lim_{n\to\infty}a_{n}=l.$