Đáp án: $\lim u_n=0$
Giải thích các bước giải:
Ta có :$u_n=\dfrac12\left(u_{n-1}+u_{n-2}\right)$
$\to u_n-u_{n-1}=\dfrac12\left(u_{n-1}+u_{n-2}\right)-u_{n-1}$
$\to u_n-u_{n-1}=-\dfrac12\left(u_{n-1}-u_{n-2}\right)$
$\to v_{n-1}=-\dfrac12v_{n-2}$
$\to \dfrac{v_{n-1}}{v_{n-2}}=-\dfrac12$
$\to v_n$ là cấp số nhân
Mà $v_1=u_2-u_1=1$
$\to v_n=\left(-\dfrac12\right)^{n-1}v_1=\left(-\dfrac12\right)^{n-1}$
$\to \begin{cases}u_{n}-u_{n-1}=\left(-\dfrac12\right)^{n-2}\\u_{n-1}-u_{n-2}=\left(-\dfrac12\right)^{n-3}\\...\\u_2-u_1=1\end{cases}$
Cộng vế với vế
$\to u_n-u_1=\left(-\dfrac12\right)^{n-2}+\left(-\dfrac12\right)^{n-3}+...+1$
$\to u_n-1=\left(-\dfrac12\right)^{n-2}+\left(-\dfrac12\right)^{n-3}+...+1$
$\to 2u_n-2=\left(-\dfrac12\right)^{n-1}+\left(-\dfrac12\right)^{n-2}+...+\left(-\dfrac12\right)$
$\to \left(2u_n-2\right)-\left(u_n-1\right)=\left(-\dfrac12\right)^{n-1}-1$
$\to u_n-1=\left(-\dfrac12\right)^{n-1}-1$
$\to u_n=\left(-\dfrac12\right)^{n-1}$
$\to \lim u_n=\lim \left(-\dfrac12\right)^{n-1}=0$
