Ta có:
$u_{n+2}=\dfrac{5u_{n+1}-2u_n}{3} \Leftrightarrow u_{n+2}-u_{n+1}=\dfrac{2u_{n+1}-2u_n}{3} \Leftrightarrow v_{n+1}=\dfrac{2}{3}v_n$
Suy ra $(v_n)$ là một cấp số nhân.
Khi đó:
$u_n=u_n-u_{n-1}+u_{n-1}-u_{n-2}+\ldots+u_2-u_1+u_1$
$=v_{n-1}+v_{n-2}+\ldots+v_1+u_1$
$=\left(\dfrac{2}{3}\right)^{n-2}v_1+\left(\dfrac{2}{3}\right)^{n-3}v_1+\ldots+v_1+u_1$
$=v_1.\dfrac{1-\left(\dfrac{2}{3}\right)^{n-1}}{1-\dfrac{2}{3}}+u_1$
$=9-9\left(\dfrac{2}{3}\right)^{n-1}+2006$
Từ đó suy ra: $\lim\mathop{} u_n=2015$